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xSicfEWAfc7KEgJWOm7k9k2k5f4oskk | physics | dual-nature-of-radiation | matter-waves | An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio
of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c
= speed of light in vaccuum) | [{"identifier": "A", "content": "$${1 \\over c}{\\left( {{{2E} \\over m}} \\right)^{{1 \\over 2}}}$$"}, {"identifier": "B", "content": "$${1 \\over c}{\\left( {{E \\over {2m}}} \\right)^{{1 \\over 2}}}$$"}, {"identifier": "C", "content": "$${\\left( {{E \\over {2m}}} \\right)^{{1 \\over 2}}}$$"}, {"identifier": "D", "c... | ["B"] | null | $$\lambda $$<sub>e</sub> = $${h \over {{p_e}}}$$ = $${h \over {\sqrt {2mE} }}$$
<br><br>E = $${{hc} \over {{\lambda _{photon}}}}$$
<br><br>$$ \Rightarrow $$ $${{\lambda _{photon}}}$$ = $${{hc} \over E}$$
<br><br>$$ \therefore $$ $${{{\lambda _e}} \over {{\lambda _{photon}}}}$$ = $${1 \over c}{\left( {{E \over {2m}}} \r... | mcq | jee-main-2020-online-7th-january-evening-slot | 10,033 |
Ao715fGQj3WhooRFKt1klrnm25m | physics | dual-nature-of-radiation | matter-waves | An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be : | [{"identifier": "A", "content": "10<sup>$$-$$2</sup> nm"}, {"identifier": "B", "content": "10<sup>$$-$$1</sup> nm"}, {"identifier": "C", "content": "10<sup>$$-$$3</sup> nm"}, {"identifier": "D", "content": "10<sup>$$-$$4</sup> nm"}] | ["C"] | null | Given, V = 1.24 million volt = 1.24 $$\times$$ 10<sup>6</sup> volt<br/><br/>Since, energy (E) = eV<br/><br/>where, e is the charge of electron = 1.6 $$\times$$ 10<sup>$$-$$19</sup> C<br/><br/>$$\therefore$$ E = 1.6 $$\times$$ 10<sup>$$-$$19</sup> $$\times$$ 1.24 $$\times$$ 10<sup>6</sup> ..... (i)<br/><br/>As we know t... | mcq | jee-main-2021-online-24th-february-evening-slot | 10,034 |
JomtsgL1OYNuA3CLyC1klt2e3kk | physics | dual-nature-of-radiation | matter-waves | An electron of mass m<sub>e</sub> and a proton of mass m<sub>p</sub> = 1836 m<sub>e</sub> are moving with the same speed. The ratio of their de Broglie wavelength $${{{}^\lambda electron} \over {{}^\lambda proton}}$$ will be : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "1836"}, {"identifier": "C", "content": "$${1 \\over {1836}}$$"}, {"identifier": "D", "content": "918"}] | ["B"] | null | Given mass of electron = m<sub>e</sub><br><br>Mass of proton = m<sub>p</sub><br><br>$$ \therefore $$ given m<sub>p</sub> = 1836 m<sub>e</sub><br><br>From de-Broglie wavelength<br><br>$$\lambda = {h \over p} = {h \over {mv}}$$<br><br>$${{{\lambda _e}} \over {{\lambda _p}}} = {{{m_p}} \over {{m_e}}}$$<br><br>$$ = {{1836... | mcq | jee-main-2021-online-25th-february-evening-slot | 10,037 |
1ujXZEWsrvDZQdIIKR1kmip1y1j | physics | dual-nature-of-radiation | matter-waves | The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (m<sub>p</sub> = 1.00727u m<sub>e</sub> = 0.00055u) | [{"identifier": "A", "content": "41.4 : 1"}, {"identifier": "B", "content": "(1860)<sup>2</sup> : 1"}, {"identifier": "C", "content": "1860 : 1"}, {"identifier": "D", "content": "43 : 1"}] | ["D"] | null | $${\lambda _e} = {{12.27} \over {\sqrt V }}\mathop A\limits^o $$<br><br>$${\lambda _p} = {{0.286} \over {\sqrt V }}\mathop A\limits^o $$<br><br>$${{{\lambda _e}} \over {{\lambda _p}}} = {{12.27} \over {0.286}} = 43$$ | mcq | jee-main-2021-online-16th-march-evening-shift | 10,039 |
LOKAsIDhTSqq5LWhlE1kmj3vaqs | physics | dual-nature-of-radiation | matter-waves | An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light) | [{"identifier": "A", "content": "$${1 \\over c}{\\left( {{{2m} \\over E}} \\right)^{1/2}}$$"}, {"identifier": "B", "content": "$${1 \\over c}{\\left( {{E \\over {2m}}} \\right)^{1/2}}$$"}, {"identifier": "C", "content": "$${\\left( {{E \\over {2m}}} \\right)^{1/2}}$$"}, {"identifier": "D", "content": "$$c{(2mE)^{1/2}}$... | ["B"] | null | For photon, E = $${{hc} \over \lambda }$$<br><br>$${\lambda _p} = {{hc} \over E}$$ .... (i)<br><br>For electron, $${\lambda _e} = {{hc} \over {\sqrt {2mE} }}$$ .... (ii)<br><br>$${{{\lambda _e}} \over {{\lambda _p}}} = {{{{hc} \over {\sqrt {2mE} }}} \over {{{hc} \over E}}} = \sqrt {{E \over {2m{c^2}}}} = {1 \over c}{\... | mcq | jee-main-2021-online-17th-march-morning-shift | 10,040 |
rF1dBUOtxgJUOSj40M1kmkrbjo5 | physics | dual-nature-of-radiation | matter-waves | A particle is travelling 4 time as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2 : 1, the mass of the particle is : | [{"identifier": "A", "content": "$${1 \\over {16}}$$ times the mass of e<sup>$$-$$</sup>"}, {"identifier": "B", "content": "8 times the mass of e<sup>$$-$$</sup>"}, {"identifier": "C", "content": "16 times the mass of e<sup>$$-$$</sup>"}, {"identifier": "D", "content": "$${1 \\over {8}}$$ times the mass of e<sup>$$-$$<... | ["D"] | null | $$\lambda = {h \over p}$$<br><br>$${{{\lambda _p}} \over {{\lambda _e}}} = {{{p_e}} \over {{p_p}}} = {{{m_e}{v_e}} \over {{m_p}{v_p}}}$$<br><br>$$2 = {{{m_e}} \over {{m_p}}}\left( {{{{v_e}} \over {4{v_e}}}} \right)$$<br><br>$$ \therefore $$ $${m_p} = {{{m_e}} \over 8}$$ | mcq | jee-main-2021-online-18th-march-morning-shift | 10,041 |
1krw9v9n6 | physics | dual-nature-of-radiation | matter-waves | An electron moving with speed v and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is : | [{"identifier": "A", "content": "$${{3c} \\over v}$$"}, {"identifier": "B", "content": "$${v \\over {3c}}$$"}, {"identifier": "C", "content": "$${v \\over {2c}}$$"}, {"identifier": "D", "content": "$${{2c} \\over v}$$"}] | ["C"] | null | $${\lambda _e} = {\lambda _{Ph}}$$<br><br>$${h \over {{p_e}}} = {h \over {{p_{ph}}}}$$<br><br>$$\sqrt {2m{k_e}} = {{{E_{ph}}} \over c}$$<br><br>$$2m{k_e} = {{{{({E_{ph}})}^2}} \over {{c^2}}}$$<br><br>$${{{k_e}} \over {{E_{ph}}}} = {{{E_{ph}}} \over {{c^2}}}\left( {{1 \over {2m}}} \right)$$<br><br>$$ = {{{p_{ph}}} \ove... | mcq | jee-main-2021-online-25th-july-evening-shift | 10,046 |
1kryyie9u | physics | dual-nature-of-radiation | matter-waves | A particle of mass 9.1 $$\times$$ 10<sup>$$-$$31</sup> kg travels in a medium with a speed of 10<sup>6</sup> m/s and a photon of a radiation of linear momentum 10<sup>$$-$$27</sup> kg m/s travels in vacuum. The wavelength of photon is __________ times the wavelength of the particle. | [] | null | 910 | For photon $${\lambda _1} = {h \over P} = {{6.6 \times {{10}^{ - 34}}} \over {{{10}^{ - 27}}}}$$<br><br>For particle $${\lambda _2} = {h \over {mv}} = {{6.6 \times {{10}^{ - 34}}} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}$$<br><br>$$\therefore$$ $${{{\lambda _1}} \over {{\lambda _2}}} = 910$$ | integer | jee-main-2021-online-27th-july-morning-shift | 10,047 |
1ktbn77dl | physics | dual-nature-of-radiation | matter-waves | The de-Broglie wavelength of a particle having kinetic energy E is $$\lambda$$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value? | [{"identifier": "A", "content": "$${1 \\over 9}$$E"}, {"identifier": "B", "content": "$${7 \\over 9}$$E"}, {"identifier": "C", "content": "E"}, {"identifier": "D", "content": "$${16 \\over 9}$$E"}] | ["B"] | null | $$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$, $$mv = \sqrt {2mE} $$<br><br>$$\lambda \propto {1 \over {\sqrt E }}$$<br><br>$${{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}} = {3 \over 4}$$, $${\lambda _2} = 0.75{\lambda _1}$$<br><br>$${{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}}... | mcq | jee-main-2021-online-26th-august-evening-shift | 10,048 |
1ktmn3gnl | physics | dual-nature-of-radiation | matter-waves | The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is :<br/><br/>[m<sub>e</sub> = mass of electron = 9 $$\times$$ 10<sup>$$-$$31</sup> kg, h = Planck constant = 6.6 $$\times$$ 6.6 $$\times$$ 10<sup>$$-$$34</sup> Js, k<sub>B</sub> = B... | [{"identifier": "A", "content": "6.26 nm"}, {"identifier": "B", "content": "8.46 nm"}, {"identifier": "C", "content": "2.26 nm"}, {"identifier": "D", "content": "3.25 nm"}] | ["A"] | null | Given, Planck's constant, h = 6.6 $$\times$$ 10<sup>$$-$$34</sup> Js<br/><br/>Boltzmann constant, k<sub>B</sub> = 1.38 $$\times$$ 10<sup>$$-$$23</sup> J/K<br/><br/>Mass of an electron, m<sub>e</sub> = 9 $$\times$$ 10<sup>$$-$$31</sup> kg<br/><br/>Temperature of an ideal gas, T = 300 K<br/><br/>As we know that, de-Brogl... | mcq | jee-main-2021-online-1st-september-evening-shift | 10,051 |
1l58c7ivm | physics | dual-nature-of-radiation | matter-waves | <p>An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are E<sub>e</sub> and p<sub>e</sub> and that of photon are E<sub>ph</sub> and p<sub>ph</sub> respectively. Which of the following is correct?</p> | [{"identifier": "A", "content": "$${{{E_e}} \\over {{E_{ph}}}} = {{2c} \\over v}$$"}, {"identifier": "B", "content": "$${{{E_e}} \\over {{E_{ph}}}} = {v \\over {2c}}$$"}, {"identifier": "C", "content": "$${{{p_e}} \\over {{p_{ph}}}} = {{2c} \\over v}$$"}, {"identifier": "D", "content": "$${{{p_e}} \\over {{p_{ph}}}} = ... | ["B"] | null | <p>$$\lambda = {h \over p} \Rightarrow p = {h \over \lambda }$$</p>
<p>Now, A/Q, $${h \over {{P_e}}} = {h \over {{P_{photon}}}}$$</p>
<p>$$ \Rightarrow {P_e} = {P_{photon}}$$ ....... (i)</p>
<p>Now, $${K_e} = {1 \over 2}M{v^2} = {{Pv} \over 2}$$</p>
<p>$${K_{ph}} = m{c^2} = Pc$$ ..... (ii)</p>
<p>$${{{K_e}} \over {{K_... | mcq | jee-main-2022-online-26th-june-morning-shift | 10,054 |
1l6f4md7g | physics | dual-nature-of-radiation | matter-waves | <p>The ratio of wavelengths of proton and deuteron accelerated by potential V<sub>p</sub> and V<sub>d</sub> is 1 : $$\sqrt2$$. Then the ratio of V<sub>p</sub> to V<sub>d</sub> will be :</p> | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "$$\\sqrt2$$ : 1"}, {"identifier": "C", "content": "2 : 1"}, {"identifier": "D", "content": "4 : 1"}] | ["D"] | null | <p>$$\lambda = {h \over {mv}} = {h \over {\sqrt {2m\,eV} }}$$</p>
<p>so $${{{\lambda _p}} \over {{\lambda _d}}} = {{\sqrt {{m_d}{V_d}} } \over {\sqrt {{m_p}{V_p}} }} = {1 \over {\sqrt 2 }}$$</p>
<p>$${{2{V_d}} \over {{V_p}}} = {1 \over 2}$$</p>
<p>$${{{V_p}} \over {{V_d}}} = {4 \over 1}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift | 10,056 |
1l6jihwsz | physics | dual-nature-of-radiation | matter-waves | <p>An electron (mass $$\mathrm{m}$$) with an initial velocity $$\vec{v}=v_{0} \hat{i}\left(v_{0}>0\right)$$ is moving in an electric field $$\vec{E}=-E_{0} \hat{i}\left(E_{0}>0\right)$$ where $$E_{0}$$ is constant. If at $$\mathrm{t}=0$$ de Broglie wavelength is $$\lambda_{0}=\frac{h}{m v_{0}}$$, then its de Brog... | [{"identifier": "A", "content": "$$\\lambda_{0}$$"}, {"identifier": "B", "content": "$$\\lambda_{0}\\left(1+\\frac{e E_{0} t}{m v_{0}}\\right)$$"}, {"identifier": "C", "content": "$$\\lambda_{0} t$$"}, {"identifier": "D", "content": "$$\\frac{\\lambda_{0}}{\\left(1+\\frac{e E_{0} t}{m v_{0}}\\right)}$$"}] | ["D"] | null | $$
\text { At } t=0, \lambda_0=\frac{h}{m v_0}
$$
<br/><br/>Since $\vec{v}=v_0 \hat{i}$ and $\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{i}$
<br/><br/>its velocity $v$ at any time $t$ is given by
<br/><br/>$$
v=v_o+\frac{\varepsilon \mathrm{E}_{\mathrm{o}}}{m} t
$$
<br/><br/>De Broglie wavelength $\lambda$ at any tim... | mcq | jee-main-2022-online-27th-july-morning-shift | 10,057 |
1l6maxo0s | physics | dual-nature-of-radiation | matter-waves | <p>The equation $$\lambda=\frac{1.227}{x} \mathrm{~nm}$$ can be used to find the de-Brogli wavelength of an electron. In this equation $$x$$ stands for :</p>
<p>Where</p>
<p>$$\mathrm{m}=$$ mass of electron</p>
<p>$$\mathrm{P}=$$ momentum of electron</p>
<p>$$\mathrm{K}=$$ Kinetic energy of electron</p>
<p>$$\mathrm{V}... | [{"identifier": "A", "content": "$$\\sqrt{\\mathrm{mK}}$$"}, {"identifier": "B", "content": "$$\\sqrt{\\mathrm{P}}$$"}, {"identifier": "C", "content": "$$\\sqrt{\\mathrm{K}}$$"}, {"identifier": "D", "content": "$$\\sqrt{\\mathrm{V}}$$"}] | ["D"] | null | <p>The de Broglie wavelength of a particle can be expressed as:</p>
<p>$$ \lambda = \frac{h}{p} $$</p>
<p>where:</p>
<ul>
<li>$$ h $$ is Planck's constant, and</li>
<li>$$ p $$ is the momentum of the particle.</li>
</ul>
<p>For an electron accelerated through a potential difference of $$ V $$ volts, its kinetic ene... | mcq | jee-main-2022-online-28th-july-morning-shift | 10,058 |
1l6rh456f | physics | dual-nature-of-radiation | matter-waves | <p>An $$\alpha$$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:</p> | [{"identifier": "A", "content": "$$\\sqrt2$$ : 1"}, {"identifier": "B", "content": "2$$\\sqrt2$$ : 1"}, {"identifier": "C", "content": "4$$\\sqrt2$$ : 1"}, {"identifier": "D", "content": "8 : 1"}] | ["B"] | null | <p>We know,</p>
<p>Momentum $$(p) = \sqrt {2m{E_k}} $$</p>
<p>and $${E_k} = q{V_{acc}}$$</p>
<p>$$\therefore$$ $$p = \sqrt {2mq\,{V_{acc}}} $$</p>
<p>Both $$\alpha$$ particle and proton are passed through same potential difference.</p>
<p>$$\therefore$$ $${\left( {{V_{acc}}} \right)_\alpha } = {\left( {{V_{acc}}} \righ... | mcq | jee-main-2022-online-29th-july-evening-shift | 10,059 |
ldqux55z | physics | dual-nature-of-radiation | matter-waves | An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_{2}$, its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_{1}}{V_{2}}\right)$ is equal to | [{"identifier": "A", "content": "$\\frac{3}{2}$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$\\frac{9}{4}$"}] | ["D"] | null | <p>$$P = \sqrt {2\,eVm} $$</p>
<p>$$\lambda = \left( {{h \over {{P_1}}}} \right)$$ ..... (i)</p>
<p>$${{3\lambda } \over 2} = {h \over {{P_2}}}$$ ..... (ii)</p>
<p>Dividing (i) by (ii)</p>
<p>$$ \Rightarrow {2 \over 3} = \left( {{{{P_2}} \over {{P_1}}}} \right) = \sqrt {{{{v_2}} \over {{v_1}}}} $$</p>
<p>$$ \Rightarro... | mcq | jee-main-2023-online-30th-january-evening-shift | 10,061 |
1lds9f4ne | physics | dual-nature-of-radiation | matter-waves | <p>The ratio of de-Broglie wavelength of an $$\alpha$$ particle and a proton accelerated from rest by the same potential is $$\frac{1}{\sqrt m}$$, the value of m is -</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "4"}] | ["C"] | null | Here : $m_\alpha=4 m_P, q_\alpha=2 q_P$, potential $=V$
<br/><br/>$\lambda=\frac{h}{\sqrt{2 m q V}}$
<br/><br/>So, $\frac{\lambda_\alpha}{\lambda_P}=\sqrt{\frac{2 m_P q_P V}{2 m_\alpha q_\alpha V}}=\sqrt{\frac{m_P \cdot q_P}{4 m_P \times 2 q_P}}=\frac{1}{\sqrt{8}}$
<br/><br/>$$ \Rightarrow $$ $m=8$ | mcq | jee-main-2023-online-29th-january-evening-shift | 10,062 |
1lduhmrk0 | physics | dual-nature-of-radiation | matter-waves | <p>Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of $$\lambda_0$$. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :</p> | [{"identifier": "A", "content": "3 $$\\lambda_0$$"}, {"identifier": "B", "content": "9 $$\\lambda_0$$"}, {"identifier": "C", "content": "$$\\frac{\\lambda_0}{\\sqrt2}$$"}, {"identifier": "D", "content": "$$\\frac{\\lambda_0}{2}$$"}] | ["C"] | null | When electron is accelerated through potential difference $V$, then<br/><br/>
$$
\begin{aligned}
& \text { K.E. }=\mathrm{eV} \\\\
& \Rightarrow \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \\\\
& \therefore \lambda \alpha \frac{1}{\sqrt{\mathrm{V}}} \\\\
& \ther... | mcq | jee-main-2023-online-25th-january-morning-shift | 10,063 |
lgny1m5w | physics | dual-nature-of-radiation | matter-waves | The de Broglie wavelength of an electron having kinetic energy $\mathrm{E}$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its de-Broglie wavelength will be : | [{"identifier": "A", "content": "$\\sqrt{2} \\lambda$"}, {"identifier": "B", "content": "$2 \\lambda$"}, {"identifier": "C", "content": "$\\frac{\\lambda}{2}$"}, {"identifier": "D", "content": "$\\frac{\\lambda}{\\sqrt{2}}$"}] | ["B"] | null |
$$
\lambda = \frac{h}{\sqrt{2mE}}
$$
<br/><br/>
where $h$ is Planck's constant, $m$ is the mass of the particle, and $E$ is its kinetic energy.
<br/><br/>
We are given that the de Broglie wavelength of an electron with kinetic energy $E$ is $\lambda$, and we want to find the de Broglie wavelength of the same electron ... | mcq | jee-main-2023-online-15th-april-morning-shift | 10,065 |
1lgridy8u | physics | dual-nature-of-radiation | matter-waves | <p>A proton and an $$\alpha$$-particle are accelerated from rest by $$2 \mathrm{~V}$$ and $$4 \mathrm{~V}$$ potentials, respectively. The ratio of their de-Broglie wavelength is :</p> | [{"identifier": "A", "content": "4 : 1"}, {"identifier": "B", "content": "2 : 1"}, {"identifier": "C", "content": "8 : 1"}, {"identifier": "D", "content": "16 : 1"}] | ["A"] | null | The de-Broglie wavelength of a particle is given by:<br/><br/>
$$\lambda = \frac{h}{p}$$<br/><br/>
where $$h$$ is the Planck's constant and $$p$$ is the momentum of the particle.
<br/><br/>
The momentum of a particle of mass $$m$$ and charge $$q$$ accelerated through a potential difference $$V$$ is given by:<br/><br/>
... | mcq | jee-main-2023-online-12th-april-morning-shift | 10,066 |
1lgswtznp | physics | dual-nature-of-radiation | matter-waves | <p>The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy :</p>
<p>(Assume $$m_{p}=m_{e} \times 1849$$ )</p> | [{"identifier": "A", "content": "1:43"}, {"identifier": "B", "content": "1:62"}, {"identifier": "C", "content": "2:43"}, {"identifier": "D", "content": "1:30"}] | ["A"] | null | The de Broglie wavelength (λ) of a particle can be found using the formula:
<br/><br/>
$$
\lambda = \frac{h}{p}
$$
<br/><br/>
where h is the Planck constant and p is the momentum of the particle. The momentum of a particle can be expressed in terms of its kinetic energy (K) and mass (m) as follows:
<br/><br/>
$$
p = \s... | mcq | jee-main-2023-online-11th-april-evening-shift | 10,067 |
1lgxxq9qb | physics | dual-nature-of-radiation | matter-waves | <p>The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is $$\lambda_1$$. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes</p> | [{"identifier": "A", "content": "2 $$\\lambda_1$$"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$$$\\lambda_1$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt2}$$$$\\lambda_1$$"}, {"identifier": "D", "content": "$$\\sqrt2~\\lambda_1$$"}] | ["C"] | null | <p>The de Broglie wavelength of a particle is given by:</p>
<p>$$\lambda = \frac{h}{p}$$</p>
<p>where h is Planck's constant and p is the momentum of the particle. The momentum of a gas molecule can be related to its kinetic energy (which is related to the temperature of the gas) by:</p>
<p>$$p = \sqrt{2mK}$$</p>
<... | mcq | jee-main-2023-online-10th-april-morning-shift | 10,068 |
1lh02bc3h | physics | dual-nature-of-radiation | matter-waves | <p>Proton $$(\mathrm{P})$$ and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, $$\mathrm{m}_{\mathrm{p}}=1849 \mathrm{~m}_{\mathrm{e}}$$ ):</p> | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "1 : 43"}, {"identifier": "C", "content": "1 : 1849"}, {"identifier": "D", "content": "43 : 1"}] | ["A"] | null | <p>The de Broglie wavelength of a particle is given by the formula:</p>
<p>$$\lambda = \frac{h}{p}$$</p>
<p>where $h$ is Planck's constant and $p$ is the momentum of the particle.</p>
<p>If the de Broglie wavelengths of the proton and electron are the same, then:</p>
<p>$$\frac{h}{p_p} = \frac{h}{p_e}$$</p>
<p>wher... | mcq | jee-main-2023-online-8th-april-morning-shift | 10,069 |
lsbl1egu | physics | dual-nature-of-radiation | matter-waves | The de Broglie wavelengths of a proton and an $\alpha$ particle are $\lambda$ and $2 \lambda$ respectively. The ratio of the velocities of proton and $\alpha$ particle will be : | [{"identifier": "A", "content": "$8: 1$"}, {"identifier": "B", "content": "$1: 2$"}, {"identifier": "C", "content": "$1: 8$"}, {"identifier": "D", "content": "$4: 1$"}] | ["A"] | null | <p>To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength. The de Broglie's wavelength formula is given by:</p>
<p>$$ \lambda = \frac{h}{p} $$</p>
<p>where:<br>
<p>$\lambda$ is th... | mcq | jee-main-2024-online-1st-february-morning-shift | 10,071 |
jaoe38c1lsf11a0t | physics | dual-nature-of-radiation | matter-waves | <p>The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $$25 \%$$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:</p> | [{"identifier": "A", "content": "$$\\frac{1}{4}$$"}, {"identifier": "B", "content": "$$\\frac{8}{1}$$"}, {"identifier": "C", "content": "$$\\frac{1}{8}$$"}, {"identifier": "D", "content": "$$\\frac{1}{1}$$"}] | ["C"] | null | <p>We know that the de-Broglie wavelength $$\lambda$$ of a particle is given by:</p>
<p>$$\lambda = \frac{h}{p}$$</p>
<p>where:</p>
<ul>
<li>$$h$$ is Planck's constant,</li>
<li>$$p$$ is the momentum of the particle.</li>
</ul>
<p>For a photon (which has zero rest mass), its energy $$E$$ and momentum $$p$$ are r... | mcq | jee-main-2024-online-29th-january-morning-shift | 10,072 |
lv3vefzw | physics | dual-nature-of-radiation | matter-waves | <p>A proton and an electron have the same de Broglie wavelength. If $$\mathrm{K}_{\mathrm{p}}$$ and $$\mathrm{K}_{\mathrm{e}}$$ be the kinetic energies of proton and electron respectively, then choose the correct relation :</p> | [{"identifier": "A", "content": "$$\\mathrm{K_p>K_e}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{K_p=K_e}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{K}_{\\mathrm{p}}<\\mathrm{K}_{\\mathrm{e}}$$\n"}, {"identifier": "D", "content": "$$\\mathrm{K}_{\\mathrm{p}}=\\mathrm{K}_{\\mathrm{e}}{ }^2$$"}] | ["C"] | null | <p>To determine the correct relation between the kinetic energies of a proton ($$\mathrm{K}_{\mathrm{p}}$$) and an electron ($$\mathrm{K}_{\mathrm{e}}$$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:</p>
<p>
<p>$$\lambda = \frac{h}{p}$$</p>
</p>
<p>where $$\lambda$... | mcq | jee-main-2024-online-8th-april-evening-shift | 10,074 |
lv5gs90p | physics | dual-nature-of-radiation | matter-waves | <p>A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:</p>
<p>(Assume h = 6.63 $$\times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}$$ and $$\mathrm{m}_{\mathrm{p}}=1836$$ times $$\mathrm{m}_{\mathrm{e}}$$ )</p> | [{"identifier": "A", "content": "$$1: \\frac{1}{1836}$$\n"}, {"identifier": "B", "content": "$$1: \\sqrt{1836}$$\n"}, {"identifier": "C", "content": "$$1: 1836$$\n"}, {"identifier": "D", "content": "$$1: \\frac{1}{\\sqrt{1836}}$$"}] | ["C"] | null | <p>To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.</p>
<p>The de-Broglie wavelength $$\lambda$$ is given by:</p>
<p>$$\lambda = \frac{h}{p}$$</p>
<p>wh... | mcq | jee-main-2024-online-8th-april-morning-shift | 10,075 |
Ve4Dqcl01VnIEBou | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | Photon of frequency $$v$$ has a momentum associated with it. If $$c$$ is the velocity of light, the momentum is | [{"identifier": "A", "content": "$$hv/c$$ "}, {"identifier": "B", "content": "$$v/c$$ "}, {"identifier": "C", "content": "$$h$$ $$v$$ $$c$$ "}, {"identifier": "D", "content": "$$hv/{c^2}$$ "}] | ["A"] | null | Energy of a photon of frequency $$v$$ is given by $$E = hv.$$
<br><br>Also, $$E = m{c^2},\,\,m{c^2} = hv$$
<br><br>$$ \Rightarrow mc = {{hv} \over C} \Rightarrow p = {{hv} \over c}$$ | mcq | aieee-2007 | 10,076 |
2cqPDULK6vwwWYVp | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | If a source of power $$4kW$$ produces $${10^{20}}$$ photons/second, the radiation belongs to a part of the spectrum called | [{"identifier": "A", "content": "$$X$$ -rays "}, {"identifier": "B", "content": "ultraviolet rays "}, {"identifier": "C", "content": "microwaves "}, {"identifier": "D", "content": "$$\\gamma $$ - rays "}] | ["A"] | null | Power, $$P = {{nhv} \over t}$$
<br><br>$$ \Rightarrow v = {{P \times t} \over {nh}}$$
<br><br>$$ = {{4 \times {{10}^3} \times 1} \over {{{10}^{20}} \times 6.63 \times {{10}^{ - 34}}}} = 6 \times {10^{16}}Hz$$ | mcq | aieee-2010 | 10,077 |
cAS8qWFNS1IIclz7QlAnD | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are :
<br/><br/>[Take Planck's constant h $$=$$ 6.62 $$ \times $$ 10<sup>$$-$$34</sup> Js] | [{"identifier": "A", "content": "10<sup>20</sup> "}, {"identifier": "B", "content": "10<sup>18</sup>"}, {"identifier": "C", "content": "10<sup>22</sup>"}, {"identifier": "D", "content": "10<sup>19</sup>"}] | ["A"] | null | <p>The power of the given laser light is expressed as</p>
<p>$$P = {{nhc} \over {\lambda t}}$$</p>
<p>from which the number of photons per second is given by</p>
<p>$$n = P\left( {{{\lambda t} \over {hc}}} \right) = (5 \times {10^2}) \times \left[ {{{(660 \times {{10}^{ - 9}})(60 \times {{10}^{ - 3}})} \over {(6.6 \tim... | mcq | jee-main-2017-online-9th-april-morning-slot | 10,078 |
xhjABBvSWgvzpuiAtKafb | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to : | [{"identifier": "A", "content": "2020 nm"}, {"identifier": "B", "content": "250 nm"}, {"identifier": "C", "content": "1700 nm"}, {"identifier": "D", "content": "220 nm"}] | ["B"] | null | The minimum wavelength of emitted photons is
<br><br>$$\lambda $$ = $${{1240} \over {5.6 - 0.7}}nm$$ = 250 nm | mcq | jee-main-2019-online-12th-january-evening-slot | 10,079 |
EIMP9g235QDa5kzPAj3rsa0w2w9jwzmh7cr | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is :
[Given Planck's constant h = 6.6 × 10<sup>–34</sup> Js, speed of light c = 3.0 × 10<sup>8</sup>
m/s] | [{"identifier": "A", "content": "5 \u00d7 10<sup>15</sup>"}, {"identifier": "B", "content": "1.5 \u00d7 10<sup>16</sup>"}, {"identifier": "C", "content": "1 \u00d7 10<sup>16</sup>"}, {"identifier": "D", "content": "2 \u00d7 10<sup>16</sup>"}] | ["A"] | null | $$2 \times {10^{ - 3}} = {{hc} \over \lambda }{{dn} \over {dt}}$$<br><br>
$${{dn} \over {dt}} = {{2 \times {{10}^{ - 3}}\lambda } \over {hc}}$$<br><br>
$$ = {{2 \times {{10}^{ - 3}} \times 500 \times {{10}^{ - 9}}} \over {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$$<br><br>
$$ = {{1000} \over {6.6 \times 3}} ... | mcq | jee-main-2019-online-10th-april-evening-slot | 10,080 |
4qtPfpdLJwpftYngWp7k9k2k5dvnumz | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | A beam of electromagnetic radiation of intensity 6.4 × 10<sup>–5</sup> W/cm<sup>2</sup> is comprised of wavelength,
$$\lambda $$ = 310 nm. It falls normally on a metal (work function $$\phi $$ = 2eV) of surface area of 1 cm<sup>2</sup>. If one
in 10<sup>3</sup> photons ejects an elctron, total number of electrons eject... | [] | null | 11 | Energy of photon = $${{1240} \over {310}}$$ = 4 eV
<br><br>Energy is greater than work function so photoelectric effect will take place.
<br><br>Number of photons = <span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;">Intensity</div>
<div... | integer | jee-main-2020-online-7th-january-morning-slot | 10,081 |
8tswxI2jopQMLVFddKjgy2xukf3r3te7 | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of
photons of X-rays to the number density of photons of the visible light of the given wavelengths is : | [{"identifier": "A", "content": "$${1 \\over {500}}$$"}, {"identifier": "B", "content": "500"}, {"identifier": "C", "content": "250"}, {"identifier": "D", "content": "$${1 \\over {250}}$$"}] | ["A"] | null | Given, wavelength of x ray ($$\lambda $$<sub>1</sub>) = 1 nm
<br><br>And wavelength of visible light ($$\lambda $$<sub>2</sub>) = 500 nm
<br><br>we know, Power$$(P) = {{nhc} \over \lambda }$$
<br><br>As P = constant and h, c also constant
<br><br>So, $${n \over \lambda } = constant$$
<br><br>$$ \Rightarrow $$ $${{{n_1... | mcq | jee-main-2020-online-3rd-september-evening-slot | 10,082 |
jv7KIdATlIqsD0vR9V1klrjq8yv | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | Given below are two statements : <br/><br/>Statement I : Two photons having equal linear momenta have equal wavelengths.<br/><br/>Statement II : If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease.<br/><br/>In the light of the above statements, choose the correct answer... | [{"identifier": "A", "content": "Statement I is false but Statement II is true"}, {"identifier": "B", "content": "Both Statement I and Statement II are false"}, {"identifier": "C", "content": "Both Statement I and Statement II are true"}, {"identifier": "D", "content": "Statement I is true but Statement II is false"}] | ["D"] | null | As we know, $$\lambda = {h \over p} = {h \over {\sqrt {2mK} }}$$<br/><br/>If linear momenta of two photons are equal, then their wavelengths is also equal.<br/><br/>Also, if the wavelength is decreased, then the momentum and energy of photon will increase.<br/><br/>Hence, option (d) is correct. | mcq | jee-main-2021-online-24th-february-morning-slot | 10,083 |
hetGVoncSv4VpsJYOq1klule2e4 | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be : | [{"identifier": "A", "content": "4.34 m/s"}, {"identifier": "B", "content": "2.19 m/s"}, {"identifier": "C", "content": "3.25 m/s"}, {"identifier": "D", "content": "4.17 m/s"}] | ["D"] | null | ($$\Delta$$E) Releases when photon going from n = 5 to n = 1<br><br>$$\Delta$$E = (13.6 $$-$$ 0.54) eV = 13.06 eV.<br><br>P<sub>i</sub> = P<sub>f</sub> (By linear momentum conservation)<br><br>$$0 = {h \over \lambda } - Mv = {V_{{\mathop{\rm Re}\nolimits} coil}} = {h \over {\lambda M}}$$ ..... (i)<br><br>& $$\Delta... | mcq | jee-main-2021-online-26th-february-evening-slot | 10,084 |
1ktjni7lb | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | A free electron of 2.6 eV energy collides with a H<sup>+</sup> ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 $$\times$$ 10<sup>$$-$$34</sup> Js) | [{"identifier": "A", "content": "1.45 $$\\times$$ 10<sup>16</sup> MHz"}, {"identifier": "B", "content": "0.19 $$\\times$$ 10<sup>15</sup> MHz"}, {"identifier": "C", "content": "1.45 $$\\times$$ 10<sup>9</sup> MHz"}, {"identifier": "D", "content": "9.0 $$\\times$$ 10<sup>27</sup> MHz"}] | ["C"] | null | For every large distance P.E. = 0<br><br>& total energy = 2.6 + 0 = 2.6 eV<br><br>Finally in first excited state of H atom total energy = $$-$$3.4 eV<br><br>Loss in total energy = 2.6 $$-$$ ($$-$$3.4) = 6 eV<br><br>It is emitted as photon<br><br>$$\lambda = {{1240} \over 6} = 206$$ nm<br><br>$$f = {{3 \times {{10}... | mcq | jee-main-2021-online-31st-august-evening-shift | 10,085 |
1l5w35hlf | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>A source of monochromatic light liberates 9 $$\times$$ 10<sup>20</sup> photon per second with wavelength 600 nm when operated at 400 W. The number of photons emitted per second with wavelength of 800 nm by the source of monochromatic light operating at same power will be :</p> | [{"identifier": "A", "content": "12 $$\\times$$ 10<sup>20</sup>"}, {"identifier": "B", "content": "6 $$\\times$$ 10<sup>20</sup>"}, {"identifier": "C", "content": "9 $$\\times$$ 10<sup>20</sup>"}, {"identifier": "D", "content": "24 $$\\times$$ 10<sup>20</sup>"}] | ["A"] | null | As we know
<br/><br/>
$$
\begin{aligned}
&I=\frac{E}{A t}=\frac{n h v}{A t} \Rightarrow \frac{n}{t}=\frac{I A \lambda}{h C} \Rightarrow \frac{n}{t}=\frac{\rho \lambda}{h c} \Rightarrow \frac{n}{t}=\rho \lambda \\\\
&\Rightarrow\left(\frac{n}{t}\right)_{2}=\left(\frac{n}{t}\right)_{1} \times \frac{p_{2} \lambda_{2}}{p_{... | mcq | jee-main-2022-online-30th-june-morning-shift | 10,086 |
1l6gk1twe | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>A parallel beam of light of wavelength $$900 \mathrm{~nm}$$ and intensity $$100 \,\mathrm{Wm}^{-2}$$ is incident on a surface perpendicular to the beam. The number of photons crossing $$1 \mathrm{~cm}^{2}$$ area perpendicular to the beam in one second is :</p> | [{"identifier": "A", "content": "$$3 \\times 10^{16}$$"}, {"identifier": "B", "content": "$$4.5 \\times 10^{16}$$"}, {"identifier": "C", "content": "$$4.5 \\times 10^{17}$$"}, {"identifier": "D", "content": "$$4.5 \\times 10^{20}$$"}] | ["B"] | null | <p>$$\lambda$$ = 900 nm</p>
<p>I = 100 W/m<sup>2</sup></p>
<p>A = 10<sup>$$-$$4</sup></p>
<p>$$\Rightarrow$$ P = 10<sup>$$-$$2</sup> W</p>
<p>$$\Rightarrow$$ Number of photons incident per second</p>
<p>$$ = {{{{10}^{ - 2}}\lambda } \over {hc}}$$</p>
<p>$$ = {{9 \times {{10}^{ - 11}} \times {{10}^2}} \over {6.63 \times... | mcq | jee-main-2022-online-26th-july-morning-shift | 10,087 |
1ldpmm2fe | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>If a source of electromagnetic radiation having power $$15 \mathrm{~kW}$$ produces $$10^{16}$$ photons per second, the radiation belongs to a part of spectrum is.</p>
<p>(Take Planck constant $$h=6 \times 10^{-34} \mathrm{Js}$$ )</p> | [{"identifier": "A", "content": "Gamma rays"}, {"identifier": "B", "content": "Radio waves"}, {"identifier": "C", "content": "Micro waves"}, {"identifier": "D", "content": "Ultraviolet rays"}] | ["A"] | null | $$
\begin{aligned}
& \text { Energy of one photon }=\frac{\text { Power }}{\text { Photon frequency }} \\\\
& \mathrm{E}=\mathrm{h} v=\frac{15 \times 10^3}{10^{16}} \\\\
& \Rightarrow v=\frac{15 \times 10^{-13}}{6 \times 10^{-34}}=2.5 \times 10^{21}
\end{aligned}
$$
<br/><br/>So gamma Rays. | mcq | jee-main-2023-online-31st-january-morning-shift | 10,088 |
1ldr0wjkf | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>A small object at rest, absorbs a light pulse of power $$20 \mathrm{~mW}$$ and duration $$300 \mathrm{~ns}$$. Assuming speed of light as $$3 \times 10^{8} \mathrm{~m} / \mathrm{s}$$, the momentum of the object becomes equal to :</p> | [{"identifier": "A", "content": "$$1 \\times 10^{-17} \\mathrm{~kg} \\mathrm{~m} / \\mathrm{s}$$"}, {"identifier": "B", "content": "$$0.5 \\times 10^{-17} \\mathrm{~kg} \\mathrm{~m} / \\mathrm{s}$$"}, {"identifier": "C", "content": "$$3 \\times 10^{-17} \\mathrm{~kg} \\mathrm{~m} / \\mathrm{s}$$"}, {"identifier": "D", ... | ["D"] | null | <p>Assuming the small object as photon.</p>
<p>Momentum $$(p)=\frac{E}{C}$$</p>
<p>$$=\frac{20\times10^{-3}\times300\times10^{-9}}{3\times10^8}$$</p>
<p>$$=2\times10^{-17}$$ kg m/s</p> | mcq | jee-main-2023-online-30th-january-morning-shift | 10,089 |
1lguywia1 | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is $$x \times 10^{15} \mathrm{~Hz}$$. The value of $$x$$ is ____________.</p>
<p>(Given h $$=4.25 \times 10^{... | [] | null | 3 | <p>When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy leve... | integer | jee-main-2023-online-11th-april-morning-shift | 10,090 |
lsamftqb | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | Monochromatic light of frequency $6 \times 10^{14} \mathrm{~Hz}$ is produced by a laser. The power emitted is $2 \times 10^{-3} \mathrm{~W}$. <br/><br/>How many photons per second on an average, are emitted by the source ?<br/><br/>
(Given $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$ ) | [{"identifier": "A", "content": "$5 \\times 10^{15}$"}, {"identifier": "B", "content": "$7 \\times 10^{16}$"}, {"identifier": "C", "content": "$6 \\times 10^{15}$"}, {"identifier": "D", "content": "$9 \\times 10^{18}$"}] | ["A"] | null | <p>To find out the number of photons emitted per second by the laser, we can use the relationship between the energy of a single photon, the total energy emitted per second (power), and the number of photons emitted per second. The energy $E$ of a single photon is given by Planck's equation:</p>
<p>$$ E = hf $$</p>
<... | mcq | jee-main-2024-online-1st-february-evening-shift | 10,091 |
lsan4w1v | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | Conductivity of a photodiode starts changing only if the wavelength of incident light is less than $660 \mathrm{~nm}$. The band gap of photodiode is found to be $\left(\frac{\mathrm{X}}{8}\right) \mathrm{eV}$. The value of $\mathrm{X}$ is :<br/><br/>
(Given, $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{e}=1.6 \... | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "21"}] | ["C"] | null | <p>To find the value of $$ X $$ in the band gap $$ \left(\frac{\mathrm{X}}{8}\right) \mathrm{eV} $$, we need to understand the relationship between the wavelength of light that can result in changes in the conductivity of a photodiode and the photodiode's band gap energy. </p>
<p>The band gap energy ($$ E_{g} $$) o... | mcq | jee-main-2024-online-1st-february-evening-shift | 10,092 |
jaoe38c1lsfm1x4p | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>Two sources of light emit with a power of $$200 \mathrm{~W}$$. The ratio of number of photons of visible light emitted by each source having wavelengths $$300 \mathrm{~nm}$$ and $$500 \mathrm{~nm}$$ respectively, will be :</p> | [{"identifier": "A", "content": "$$5: 3$$\n"}, {"identifier": "B", "content": "$$3: 5$$\n"}, {"identifier": "C", "content": "$$1: 5$$\n"}, {"identifier": "D", "content": "$$1: 3$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \mathrm{n}_1 \times \frac{\mathrm{hc}}{\lambda_1}=200 \\
& \mathrm{n}_2 \times \frac{\mathrm{hc}}{\lambda_2}=200 \\
& \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_1}{\lambda_2}=\frac{300}{500} \\
& \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{5}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift | 10,093 |
1lsg6tja7 | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>If the total energy transferred to a surface in time $$\mathrm{t}$$ is $$6.48 \times 10^5 \mathrm{~J}$$, then the magnitude of the total momentum delivered to this surface for complete absorption will be:</p> | [{"identifier": "A", "content": "$$2.16 \\times 10^{-3} \\mathrm{~kg} \\mathrm{~m} / \\mathrm{s}$$\n"}, {"identifier": "B", "content": "$$2.46 \\times 10^{-3} \\mathrm{~kg} \\mathrm{~m} / \\mathrm{s}$$\n"}, {"identifier": "C", "content": "$$1.58 \\times 10^{-3} \\mathrm{~kg} \\mathrm{~m} / \\mathrm{s}$$\n"}, {"identifi... | ["A"] | null | <p>$$\mathrm{p=\frac{E}{C}=\frac{6.48 \times 10^5}{3 \times 10^8}=2.16 \times 10^{-3}}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift | 10,094 |
lvb29gzg | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at $$10.2 \mathrm{~V}$$. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ________ nm. (Given hc $$=1245 \mathrm{~eV} \mathrm{~nm}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$$... | [] | null | 122 | <p>The Franck-Hertz experiment provides evidence for quantized energy levels within atoms. When atoms are excited by electrons with a specific kinetic energy, they can jump to higher energy levels. Upon returning to lower levels, they emit photons whose energies correspond to the difference between these levels. The fi... | integer | jee-main-2024-online-6th-april-evening-shift | 10,095 |
lvc58e61 | physics | dual-nature-of-radiation | particle-nature-of-light-the-photon | <p>Which of the following phenomena does not explain by wave nature of light.</p>
<p>A. reflection</p>
<p>B. diffraction</p>
<p>C. photoelectric effect</p>
<p>D. interference</p>
<p>E. polarization</p>
<p>Choose the most appropriate answer from the options given below:</p> | [{"identifier": "A", "content": "C only\n"}, {"identifier": "B", "content": "B, D only\n"}, {"identifier": "C", "content": "A, C only\n"}, {"identifier": "D", "content": "E only"}] | ["A"] | null | <p>The correct answer is Option A: C only.</p>
<p>Reflection, diffraction, interference, and polarization are all phenomena that can be explained by the wave nature of light. These phenomena are evidence that light behaves as a wave, evident through various experimental observations:</p>
<ul>
<li><strong>Reflection<... | mcq | jee-main-2024-online-6th-april-morning-shift | 10,096 |
WitG0xTNyFoJBMVH | physics | dual-nature-of-radiation | photoelectric-effect | Two identical photo-cathodes receive light of frequencies $${f_1}$$ and $${f_2}$$. If the velocities of the photo electrons (of mass $$m$$ ) coming out are respectively $${v_1}$$ and $${v_2},$$ then | [{"identifier": "A", "content": "$$v_1^2 - v_2^2 = {{2h} \\over m}\\left( {{f_1} - {f_2}} \\right)$$ "}, {"identifier": "B", "content": "$${v_1} + {v_2} = {\\left[ {{{2h} \\over m}\\left( {{f_1} + {f_2}} \\right)} \\right]^{1/2}}$$"}, {"identifier": "C", "content": "$$v_1^2 + v_2^2 = {{2h} \\over m}\\left( {{f_1} + {f_... | ["A"] | null | For one photo cathode
<br><br>$$h{f_1} - W = {1 \over 2}mv_1^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>For another photo cathode
<br><br>$$h{f_2} - W = {1 \over 2}mv_2^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Subtracting $$(ii)$$ from $$(i)$$ we get
<br><br>$$\left( {h{f_1} - W} \right... | mcq | aieee-2003 | 10,098 |
rPQL3Teyas56lGkW | physics | dual-nature-of-radiation | photoelectric-effect | The work function of a substance is $$4.0$$ $$eV.$$ The longest wavelength of light that can cause photo-electron emission from this substance is approximately. | [{"identifier": "A", "content": "$$310$$ $$nm$$ "}, {"identifier": "B", "content": "$$400$$ $$nm$$"}, {"identifier": "C", "content": "$$540$$ $$nm$$ "}, {"identifier": "D", "content": "$$220$$ $$nm$$ "}] | ["A"] | null | For the longest wavelength to emit photo electron
<br><br>$${{hc} \over \lambda } = \phi \Rightarrow \lambda = {{hc} \over \phi }$$
<br><br>$$ \Rightarrow \lambda = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {40 \times 1.6 \times {{10}^{ - 16}}}} = 310nm$$ | mcq | aieee-2004 | 10,099 |
nrbseZ6OuEZem1EX | physics | dual-nature-of-radiation | photoelectric-effect | According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal $$Vs$$ the frequency, of the incident radiation gives as straight the whose slope | [{"identifier": "A", "content": "depends both on the intensity of the radiation and the metal used"}, {"identifier": "B", "content": "depends on the intensity of the radiation "}, {"identifier": "C", "content": "depends on the nature of the metal used "}, {"identifier": "D", "content": "is the same for the all metals a... | ["D"] | null | <p>Einstein's photoelectric equation is given by</p>
<p>$$K_{\max} = h\nu - \phi,$$</p>
<p>where $K_{\max}$ is the maximum kinetic energy of the emitted photoelectrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident radiation, and $\phi$ is the work function of the metal (the minimum energy ne... | mcq | aieee-2004 | 10,100 |
FFDQfsoCPCowNhlI | physics | dual-nature-of-radiation | photoelectric-effect | A photocell is illuminated by a small bright source placed $$1$$ $$m$$ away. When the same source of light is placed $${1 \over 2}$$ $$m$$ away, the number of electrons emitted by photo-cathode would | [{"identifier": "A", "content": "increases by a factor of $$4$$ "}, {"identifier": "B", "content": "decreases by a factor of $$4$$ "}, {"identifier": "C", "content": "increases by a factor of $$2$$ "}, {"identifier": "D", "content": "decreases by a factor of $$2$$ "}] | ["A"] | null | $$I \propto {1 \over {{r^2}}};{{{I_1}} \over {{I_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^2} = {1 \over 4}$$
<br><br>$${I_2} \to 4\,\,$$ times $${I_1}$$
<br><br>When intensity becomes 4 times, no. of photoelectrons emitted would increase by $$4$$ times, since number of electrons emitted per second is directly p... | mcq | aieee-2005 | 10,101 |
vC6HfFy8Ml3p14QH | physics | dual-nature-of-radiation | photoelectric-effect | The anode voltage of a photocell is kept fixed. The wavelength $$\lambda $$ of the light falling on the cathode is gradually changed. The plate current $${\rm I}$$ of the photocell varies as follows | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265340/exam_images/fmw8wvbqgpbpnsksiwn1.webp\" loading=\"lazy\" alt=\"AIEEE 2006 Physics - Dual Nature of Radiation Question 156 English Option 1\"> "}, {"identifier": "B", "content": "<img class=... | ["B"] | null | As $$\lambda $$ decreases, $$v$$ increases and hence the speed of photo electron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases. | mcq | aieee-2006 | 10,102 |
EPhe4z1UCmTDtBHI | physics | dual-nature-of-radiation | photoelectric-effect | The time taken by a photoelectron to come out after the photon strikes is approximately | [{"identifier": "A", "content": "$${10^{ - 4}}\\,s$$ "}, {"identifier": "B", "content": "$${10^{ - 10}}\\,s$$ "}, {"identifier": "C", "content": "$${10^{ - 16}}\\,s$$ "}, {"identifier": "D", "content": "$${10^{ - 1}}\\,s$$ "}] | ["B"] | null | Emission of photo-electron starts from the surface
after incidence of photons in about $${10^{ - 10}}s.$$ | mcq | aieee-2006 | 10,103 |
GXDVcfVoBoEdGtOH | physics | dual-nature-of-radiation | photoelectric-effect | The threshold frequency for a metallic surface corresponds to an energy of $$6.2$$ $$eV$$ and the stopping potential for a radiation incident on this surface is $$5V.$$ The incident radiation lies in | [{"identifier": "A", "content": "ultra-violet region "}, {"identifier": "B", "content": "infra-red region "}, {"identifier": "C", "content": "visible region "}, {"identifier": "D", "content": "$$x$$-ray region"}] | ["A"] | null | <p>The energy of the incident radiation that causes photoelectrons to be emitted can be calculated using the stopping potential, V, via the equation </p>
<p>$$E = eV,$$ </p>
<p>where e is the elementary charge. </p>
<p>The elementary charge, e, is approximately equal to $$1.602 \times 10^{-19}$$ C. So the energy of the... | mcq | aieee-2006 | 10,104 |
ZCDp3u5CQLvArb4N | physics | dual-nature-of-radiation | photoelectric-effect | <b>Statement - $$1$$ :</b> When ultraviolet light is incident on a photocell, its stopping potential is $${V_0}$$ and the maximum kinetic energy of the photoelectrons is $${K_{\max }}$$. When the ultraviolet light is replaced by $$X$$-rays, both $${V_0}$$ and $${K_{\max }}$$ increase.
<p><b>Statement - $$2$$ :</b> Ph... | [{"identifier": "A", "content": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is the correct explanation of Statement - $$1$$ "}, {"identifier": "B", "content": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is not the correct explanation of Statement - $$1$$ "}, {... | ["B"] | null | <b>Statement 1</b> is true. The energy of an incident photon (from the ultraviolet light or X-rays) on a photocell is given by Planck's equation, $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of the light. X-rays have a higher frequency than ultraviolet light, so they deliver more energy to the ... | mcq | aieee-2010 | 10,106 |
AS1zkE4MlKRcGlPg | physics | dual-nature-of-radiation | photoelectric-effect | This question has Statement - $$1$$ and Statement - $$2$$. Of the four choices given after the statements, choose the one that best describes the two statements.
<p><b>Statement - $$1$$ : </b> A metallic surface is irradiated by a monochromatic light of frequency $$v > {v_0}$$ (the threshold frequency). The maximum ... | [{"identifier": "A", "content": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is the correct explanation of Statement - $$1$$."}, {"identifier": "B", "content": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is not the correct explanation of Statement - $$1$$."}, {"i... | ["C"] | null | By Einstein photoelectric equation,
<br><br>$${K_{\max }} = e{V_0} = hv - h{v_0}$$
<br><br>When $$v$$ is doubled, $${K_{\max }}$$ and $${V_0}$$ become more than double. | mcq | aieee-2011 | 10,107 |
EfuANHkP3EUeGXxQ | physics | dual-nature-of-radiation | photoelectric-effect | The anode voltage of a photocell is kept fixed. The wavelength $$\lambda $$ of the light falling on the cathode is gradually changed. The plate current $$I$$ of the photocell varies as follows : | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264737/exam_images/fliokk5lo5hjgjoospr5.webp\" loading=\"lazy\" alt=\"JEE Main 2013 (Offline) Physics - Dual Nature of Radiation Question 153 English Option 1\"> "}, {"identifier": "B", "content":... | ["D"] | null | As $$\lambda $$ is increased, there will be a value of $$\lambda $$ above which photo electrons will be cases to come out so photo current will become zero. Hence $$(d)$$ is correct answer. | mcq | jee-main-2013-offline | 10,108 |
Xik8Lah1nucIfYoB | physics | dual-nature-of-radiation | photoelectric-effect | Radiation of wavelength $$\lambda ,$$ is incident on a photocell. The fastest emitted electron has speed $$v.$$ If the wavelength is changed to $${{3\lambda } \over 4},$$ the speed of the fastest emitted electron will be: | [{"identifier": "A", "content": "$$ = v{\\left( {{4 \\over 3}} \\right)^{{1 \\over 2}}}$$ "}, {"identifier": "B", "content": "$$ = v{\\left( {{3 \\over 4}} \\right)^{{1 \\over 2}}}$$ "}, {"identifier": "C", "content": "$$ > v{\\left( {{4 \\over 3}} \\right)^{{1 \\over 2}}}$$ "}, {"identifier": "D", "content": "$$ &l... | ["C"] | null | $$h{v_0}^2 - h{v_0} = {1 \over 2}m{v^2}$$
<br><br>$$\therefore$$ $${4 \over 3}h{v_0} - h{v_0} = {1 \over 2}mv{'^2}$$
<br><br>$$\therefore$$ $${{v{'^2}} \over {{v^2}}} = {{{4 \over 3}v - {v_0}} \over {v - {v_0}}}$$
<br><br>$$\therefore$$ $$v' = v\sqrt {{{{4 \over 3}v - {v_0}} \over {v - {v_0}}}} $$
<br><br>$$\therefore$... | mcq | jee-main-2016-offline | 10,109 |
nKS3T7KHG0Trb0oXXc7Mw | physics | dual-nature-of-radiation | photoelectric-effect | When photons of wavelength $${\lambda _1}$$ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength $${\lambda _2}$$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength $${\lambda _3}$$ is used then find ... | [{"identifier": "A", "content": "$${{hc} \\over e}\\left[ {{1 \\over {{\\lambda _3}}} - {1 \\over {{\\lambda _2}}} - {1 \\over {{\\lambda _1}}}} \\right]$$"}, {"identifier": "B", "content": "$${{hc} \\over e}\\left[ {{1 \\over {{\\lambda _3}}} + {1 \\over {{\\lambda _2}}} - {1 \\over {{\\lambda _1}}}} \\right]$$"}, {"i... | ["C"] | null | We know,
<br><br>Einstein's photoelectric equation,
<br><br>$$eV = {{hc} \over \lambda } - {\phi _0}$$
<br><br>and $${\phi _0}$$, $${{hc} \over {{\lambda _0}}}$$, where $${{\lambda _0}}$$ is the threashhold wavelength.
<br><br>$$ \therefore $$ In first case,
<br><br>eV $$ = {{hc} \over {{\lambda _1}}... | mcq | jee-main-2016-online-9th-april-morning-slot | 10,110 |
b6r7lKqAd52ksTxDlvZYH | physics | dual-nature-of-radiation | photoelectric-effect | A photoelectric surface is illuminated successively by monochromatic light of wavelengths $$\lambda $$ and $${\lambda \over 2}.$$ If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is : | [{"identifier": "A", "content": "$${{hc} \\over {3\\lambda }}$$ "}, {"identifier": "B", "content": "$${{hc} \\over {2\\lambda }}$$"}, {"identifier": "C", "content": "$${{hc} \\over {\\lambda }}$$"}, {"identifier": "D", "content": "$${3\\,{hc} \\over {\\lambda }}$$"}] | ["B"] | null | We know,
<br><br>Einstein's photo electric equation,
<br><br>(KE)<sub>max</sub> = $${{hc} \over \lambda }$$ $$-$$ $$\phi $$<sub>0</sub>
<br><br>In first case,
<br><br>K = $${{hc} \over \lambda }$$ $$-$$ $$\phi $$<sub>0</sub> . . .(1)
<br><br>In second case,
... | mcq | jee-main-2016-online-10th-april-morning-slot | 10,111 |
sdlfBQAkq7QkYgbIpK1YA | physics | dual-nature-of-radiation | photoelectric-effect | The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be : | [{"identifier": "A", "content": "less than $$\\sqrt 3 $$ v"}, {"identifier": "B", "content": "v"}, {"identifier": "C", "content": "more than $$\\sqrt 3 \\,v$$ "}, {"identifier": "D", "content": "equal to $$\\sqrt 3 \\,v$$"}] | ["C"] | null | <p>The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by</p>
<p>$$KE = {1 \over 2}m{v^2} = hn - \phi $$</p>
<p>where h is Planck's constant and $$\phi$$ is the work function. Therefore, the kinetic energy of the incident light is</p>
<p>$${E_1} = hn - \phi $$ ..... (1)</p>
<p>W... | mcq | jee-main-2017-online-8th-april-morning-slot | 10,112 |
yfE4GlkMrqBbtt2xt24Au | physics | dual-nature-of-radiation | photoelectric-effect | The electric field of light wave is given as
$$$\overrightarrow E = {10^{ - 3}}\cos \left( {{{2\pi x} \over {5 \times {{10}^{ - 7}}}} - 2\pi \times 6 \times {{10}^{14}}t} \right)\mathop x\limits^ \wedge {{\rm N} \over C}$$$
This
light falls on a metal plate of work function
2eV. The stopping potential of the photoe... | [{"identifier": "A", "content": "2.48 V"}, {"identifier": "B", "content": "0.48 V"}, {"identifier": "C", "content": "0.72 V"}, {"identifier": "D", "content": "2.0 V"}] | ["B"] | null | $$\omega = 6 \times {10^{14}} \times 2\pi $$<br>
f = 6 × 10<sup>14</sup><br>
C = f $$\lambda $$<br><br>
$$\lambda = {C \over f} = {{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}} = 5000$$ Å<br>
Energy of photon $$ \Rightarrow {{12375} \over {5000}} = 2.475\,eV$$<br><br>
From Einstein’s equation<br>
KEmax = E – $$\p... | mcq | jee-main-2019-online-9th-april-morning-slot | 10,113 |
MDdSKwLbuYhxn5umLa3rsa0w2w9jx3ezsx1 | physics | dual-nature-of-radiation | photoelectric-effect | The stopping potential V<sub>0</sub> (in volt) as a function of frequency ($$\upsilon $$) for a sodium emitter, is shown in the figure.
The work function of sodium, from the data plotted in the figure, will be: <br/>(Given: Planck’s constant (h) = 6.63 × 10<sup>–34</sup> Js, electron charges e = 1.6 × 10<sup>–19</sup> ... | [{"identifier": "A", "content": "1.95 eV"}, {"identifier": "B", "content": "2.12 eV"}, {"identifier": "C", "content": "1.82 eV"}, {"identifier": "D", "content": "1.66 eV"}] | ["D"] | null | $$hv = \varphi + e{v_0}$$<br><br>
$${v_0} = {{hv} \over e} - {\varphi \over e}$$<br><br>
v<sub>0</sub> is zero for v = 4 × 10<sup>14</sup> Hz<br><br>
$$0 = {{hv} \over e} - {\varphi \over e}$$<br><br>
$$ \Rightarrow \phi = hv$$<br><br>
$$ = {{6.63 \times {{10}^{ - 34}} \times 4 \times {{10}^{14}}} \over {1.6 \times... | mcq | jee-main-2019-online-12th-april-morning-slot | 10,114 |
XwoeKLRmjgbzuA46oV18hoxe66ijvzt2m15 | physics | dual-nature-of-radiation | photoelectric-effect | In a photoelectric effect experiment the
threshold wavelength of the light is 380 nm. If
the wavelentgh of incident light is 260 nm, the
maximum kinetic energy of emitted electrons
will be:<br/>
Given E (in eV) = 1237/$$\lambda $$ (in nm) | [{"identifier": "A", "content": "4.5 eV"}, {"identifier": "B", "content": "15.1 eV"}, {"identifier": "C", "content": "3.0 eV"}, {"identifier": "D", "content": "1.5 eV"}] | ["D"] | null | $${K_{\max }} = {{hc} \over \lambda } - {{hc} \over {{\lambda _o}}}$$<br><br>
$$ \Rightarrow {K_{\max }} = hc\left( {{{{\lambda _o} - \lambda } \over {\lambda {\lambda _o}}}} \right)$$<br><br>
$$ \Rightarrow {K_{\max }} = \left( {1237} \right)\left( {{{380 - 260} \over {380 \times 260}}} \right)$$
= 1.5 eV | mcq | jee-main-2019-online-10th-april-morning-slot | 10,115 |
rlPSZjNPZ5tiIiyRtIMl8 | physics | dual-nature-of-radiation | photoelectric-effect | When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping
potential for the current is –V<sub>0</sub>/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is – V<sub>0</sub>. The threshold frequency for photoelectric emission... | [{"identifier": "A", "content": "2$$v$$"}, {"identifier": "B", "content": "$${4 \\over 3}v$$"}, {"identifier": "C", "content": "$${{3v} \\over 2}$$"}, {"identifier": "D", "content": "$${{5v} \\over 3}$$"}] | ["C"] | null | Einstein’s photoelectric equation in the two cases
is given by
<br><br>$${{e{V_0}} \over {\Delta E}} = h\upsilon - h{\upsilon _0}$$ ......(i)
<br><br>and $$e{V_0} = {{h\upsilon } \over 2} - h{\upsilon _0}$$ .....(ii)
<br><br>From eqn. (i) and (ii),
<br><br>$${1 \over 2} = {{h\upsilon - h{\upsilon _0}} \over {{{h\upsi... | mcq | jee-main-2019-online-12th-january-evening-slot | 10,116 |
jGjrJ5raJzrdb982WUc10 | physics | dual-nature-of-radiation | photoelectric-effect | The magnetic field associated with a light wave is given, at the origin, by B = B<sub>0</sub> [sin(3.14 $$ \times $$ 10<sup>7</sup>)ct + sin(6.28 $$ \times $$ 10<sup>7</sup>)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
<br... | [{"identifier": "A", "content": "6.82 eV"}, {"identifier": "B", "content": "12.5 eV"}, {"identifier": "C", "content": "8.52 eV"}, {"identifier": "D", "content": "7.72 eV"}] | ["D"] | null | Given that,
<br><br>B = B<sub>0</sub>[sin (3.14 $$ \times $$ 10<sup>7</sup>) ct + sin(6.28 $$ \times $$ 10<sup>7</sup>) ct]
<br><br>This light wave is non-monochromotic wave as here is two different frequency in the magnetic field.
<br><br>Given work function ($$\phi $$) = 4.7 eV
<br><br>Question says to find the maxim... | mcq | jee-main-2019-online-9th-january-evening-slot | 10,118 |
7aPbDyHmLKoT8vTkhgQU8 | physics | dual-nature-of-radiation | photoelectric-effect | Surface of certain metal is first illuminated with light of wavelength $$\lambda $$<sub>1</sub> = 350 nm and then, by light of wavelength $$\lambda $$<sub>2</sub> = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close... | [{"identifier": "A", "content": "1.8"}, {"identifier": "B", "content": "2.5"}, {"identifier": "C", "content": "5.6"}, {"identifier": "D", "content": "1.4"}] | ["A"] | null | Let speed of photon electron in first case is 2v
<br><br>then in the second case speed is v.
<br><br>For first case
<br><br>$${{hc} \over {{\lambda _1}}} = \phi + {1 \over 2}$$m(2v)<sup>2</sup>
<br><br>For second case,
<br><br>$${{hc} \over {{\lambda _2}}} = \phi + {1 \over 2}$$mv<sup>2</sup>
<br><br>$$ \therefore $... | mcq | jee-main-2019-online-9th-january-morning-slot | 10,119 |
HYoxdPryi2r6KQbvsZIj5 | physics | dual-nature-of-radiation | photoelectric-effect | In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: ($${{{hc} \over e}}$$ = 1240 nm eV) | [{"identifier": "A", "content": "0.5 V"}, {"identifier": "B", "content": "1.0 V"}, {"identifier": "C", "content": "2.0 V"}, {"identifier": "D", "content": "1.5 V"}] | ["B"] | null | $${{hc} \over {{\lambda _1}}} = \phi + e$$V<sub>1</sub> . . . (i)
<br><br>$${{hc} \over {{\lambda _2}}} = \phi + e$$V<sub>2</sub> . . . (ii)
<br><br>(i) $$-$$ (ii)
<br><br>hc$$\left( {{1 \over {{\lambda _1}}} - {1 \over {{\l... | mcq | jee-main-2019-online-11th-january-evening-slot | 10,120 |
IhQrnGathkLSr9hztojgy2xukfoteqep | physics | dual-nature-of-radiation | photoelectric-effect | The surface of a metal is illuminated alternately
with photons of energies E<sub>1</sub> = 4 eV and E<sub>2</sub> = 2.5 eV
respectively. The ratio of maximum speeds of the
photoelectrons emitted in the two cases is 2. The
work function of the metal in (eV) is _____. | [] | null | 2 | Given,
<br><br>E<sub>1</sub> = 4 eV
<br><br>E<sub>2</sub> = 2.5 eV
<br><br>and $${{{{\left( {{V_1}} \right)}_{\max }}} \over {{{\left( {{V_2}} \right)}_{\max }}}} = 2$$
<br><br>$${{{1 \over 2}m{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{1 \over 2}m{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \ri... | integer | jee-main-2020-online-5th-september-evening-slot | 10,121 |
iIc2gsD3llQtIpvI9Njgy2xukfaxpie7 | physics | dual-nature-of-radiation | photoelectric-effect | In a photoelectric effect experiment, the graph of stopping potential V versus reciprocal of wavelength
obtained is shown in the figure. As the intensity of incident radiation is increased :
<img src="data:image/png;base64,UklGRtAEAABXRUJQVlA4IMQEAABwJACdASrbAIoAPm02l0ikIyIhI5W5uIANiWlu4WxhG/Nj8Z/2btq/tfRVeT/aH01KVXtpP... | [{"identifier": "A", "content": "Slope of the straight line get more steep"}, {"identifier": "B", "content": "Graph does not change"}, {"identifier": "C", "content": "Straight line shifts to left"}, {"identifier": "D", "content": "Straight line shifts to right"}] | ["B"] | null | eV = $${{hc} \over \lambda } - \phi $$
<br><br>$$ \Rightarrow $$ V = $$\left( {{{hc} \over e}} \right){1 \over \lambda } - {\phi \over e}$$
<br><br>Slope of the line in above equation and all other terms are independent of intensity. The graph does not
change. | mcq | jee-main-2020-online-4th-september-evening-slot | 10,122 |
q9BbKmltZkNaO6DOjnjgy2xukf6iub58 | physics | dual-nature-of-radiation | photoelectric-effect | Given figure shows few data points in a phot electric effect experiment for a certain metal. The
minimum energy for ejection of electron from its surface is: <br/>(Plancks constant h = 6.62 × 10<sup>–34</sup> J.s)
<img src="data:image/png;base64,UklGRiYLAABXRUJQVlA4IBoLAABwcACdASr0AZcBP4G41mY2LSwnIbF5QsAwCWlu/EsXY5fnZ1... | [{"identifier": "A", "content": "2.10 eV"}, {"identifier": "B", "content": "2.27 eV"}, {"identifier": "C", "content": "2.59 eV"}, {"identifier": "D", "content": "1.93 eV"}] | ["B"] | null | K<sub>max</sub> = hf - $$\phi $$
<br><br>$$ \Rightarrow $$ eV<sub>0</sub> = hf - $$\phi $$
<br><br>at B V<sub>0</sub>
= 0, f = 5.5
<br><br>$$ \therefore $$ $$\phi $$ = hf
<br><br>= $${6.62 \times {{10}^{ - 34}} \times 5.5 \times {{10}^{14}}}$$ J
<br><br>= $${{6.62 \times {{10}^{ - 34}} \times 5.5 \times {{10}^{14}}} ... | mcq | jee-main-2020-online-4th-september-morning-slot | 10,123 |
8mkVsgrBxoOjNALFxOjgy2xukf14jl8g | physics | dual-nature-of-radiation | photoelectric-effect | When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the
maximum kinetic energy of the photoelectrons becomes three times larger. The work function of
the metal is close to : | [{"identifier": "A", "content": "1.02 eV"}, {"identifier": "B", "content": "0.81 eV"}, {"identifier": "C", "content": "0.61 eV"}, {"identifier": "D", "content": "0.52 eV"}] | ["C"] | null | K<sub>1</sub> = $${{hc} \over {500}} - {\phi _0}$$
<br><br>K<sub>2</sub> = $${{hc} \over {200}} - {\phi _0}$$
<br><br>$$ \because $$ K<sub>2</sub> = 3K<sub>1</sub>
<br><br>$$ \Rightarrow $$ $$3\left[ {{{hc} \over {500}} - {\phi _0}} \right] = \left[ {{{hc} \over {200}} - {\phi _0}} \right]$$
<br><br>$$ \Rightarrow $$ $... | mcq | jee-main-2020-online-3rd-september-morning-slot | 10,124 |
2lAOp8CWialPwm69UD7k9k2k5ielbrc | physics | dual-nature-of-radiation | photoelectric-effect | Radiation, with wavelength 6561 $$\mathop A\limits^o $$ falls on a
metal surface to produce photoelectrons. The
electrons are made to enter a uniform magnetic
field of 3 × 10<sup>–4</sup> T. If the radius of the largest
circular path followed by the electrons is
10 mm, the work function of the metal is
close to : | [{"identifier": "A", "content": "1.8eV"}, {"identifier": "B", "content": "0.8eV"}, {"identifier": "C", "content": "1.1eV"}, {"identifier": "D", "content": "1.6eV"}] | ["C"] | null | Let the work function be $$\phi $$.
<br><br>$$ \therefore $$ KE<sub>max</sub> = $${{hc} \over \lambda } - \phi $$
<br><br>We know r = $${{mv} \over {qB}}$$
<br><br>and p = mv = rqB
<br><br>$$ \therefore $$ KE<sub>max</sub> = $${{{p^2}} \over {2m}}$$ = $${{{q^2}{r^2}{B^2}} \over {2m}}$$
<br><br> = $${{{{\left( {1.6 \tim... | mcq | jee-main-2020-online-9th-january-morning-slot | 10,126 |
yDaeUpXTllB32t3iK67k9k2k5gv3pf7 | physics | dual-nature-of-radiation | photoelectric-effect | When photon of energy 4.0 eV strikes the
surface of a metal A, the ejected photoelectrons
have maximum kinetic energy T<sub>A</sub> eV end
de-Broglie wavelength $$\lambda _A$$. The maximum
kinetic energy of photoelectrons liberated from
another metal B by photon of energy 4.50 eV
is T<sub>B</sub> = (T<sub>A</sub> – 1.5... | [{"identifier": "A", "content": "1.5eV"}, {"identifier": "B", "content": "4eV"}, {"identifier": "C", "content": "2eV"}, {"identifier": "D", "content": "3eV"}] | ["B"] | null | We know, de-Broglie wavelength
<br><br>$$\lambda $$ = $${h \over p} = {h \over {\sqrt {2m{K_e}} }}$$
<br><br>$$ \therefore $$ $$\lambda \propto {1 \over {\sqrt {{K_e}} }}$$
<br><br>So $${{{\lambda _A}} \over {{\lambda _B}}} = \sqrt {{{{T_B}} \over {{T_A}}}} $$
<br><br>$$ \Rightarrow $$ $${\left( {{1 \over 2}} \right)^... | mcq | jee-main-2020-online-8th-january-morning-slot | 10,127 |
DlwRrPTsfuVvPXBTfA1klt3cieu | physics | dual-nature-of-radiation | photoelectric-effect | The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is : | [{"identifier": "A", "content": "400 nm"}, {"identifier": "B", "content": "329 nm"}, {"identifier": "C", "content": "309 nm"}, {"identifier": "D", "content": "382 nm"}] | ["D"] | null | From the photoelectric effect equation<br><br>$${{hc} \over \lambda } = \phi + e{v_s}$$<br><br>so, $$e{v_{{s_1}}} = {{hc} \over {{\lambda _1}}} - \phi $$ .....(i)<br><br>$$e{v_{{s_2}}} = {{hc} \over {{\lambda _2}}} - \phi $$ ......(ii)<br><br>Subtract equation (i) from equation (ii)<br><br>$$e{v_{{s_1}}} - e{v_{{s_2}}... | mcq | jee-main-2021-online-25th-february-evening-slot | 10,128 |
uUegg21QpmCaMVSTWy1klunx457 | physics | dual-nature-of-radiation | photoelectric-effect | Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is ___________. | [] | null | 1 | $$K{E_{\max }} = hv - \phi $$<br><br>$${1 \over 2}m{v^2} = hv - \phi $$<br><br>$$v = \sqrt {{{2(hv - \phi )} \over m}} $$<br><br>Given $$h{v_1} = 2\phi $$<br><br>$$h{v_2} = 10\phi $$<br><br>$$ \therefore $$ $${{{v_1}} \over {{v_2}}} = \sqrt {{{h{v_1} - \phi } \over {h{v_2} - \phi }}} $$<br><br>$${{{v_1}} \over {{v_2}}}... | integer | jee-main-2021-online-26th-february-evening-slot | 10,129 |
63J8F1m6HiqR6xSyPO1kmhownx8 | physics | dual-nature-of-radiation | photoelectric-effect | The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation : | [{"identifier": "A", "content": "Phase"}, {"identifier": "B", "content": "Frequency"}, {"identifier": "C", "content": "Amnplitude"}, {"identifier": "D", "content": "Intensity"}] | ["B"] | null | Stopping potential depends on frequency, according to Einstein's photoelectric equation.<br><br>$$hv - h{v_0} = eV$$<br><br>$$ \Rightarrow V = {h \over e}v - {h \over e}{v_0}$$ | mcq | jee-main-2021-online-16th-march-morning-shift | 10,130 |
RvTWNmgapIuHf2mMkm1kmkag4h0 | physics | dual-nature-of-radiation | photoelectric-effect | Two identical photocathodes receive the light of frequencies f<sub>1</sub> and f<sub>2</sub> respectively. If the velocities of the photo-electrons coming out are v<sub>1</sub> and v<sub>2</sub> respectively, then | [{"identifier": "A", "content": "$${v_1} - {v_2} = {\\left[ {{{2h} \\over m}({f_1} - {f_2})} \\right]^{{1 \\over 2}}}$$"}, {"identifier": "B", "content": "$$v_1^2 + v_2^2 = {{2h} \\over m}[{f_1} + {f_2}]$$"}, {"identifier": "C", "content": "$${v_1} + {v_2} = {\\left[ {{{2h} \\over m}({f_1} + {f_2})} \\right]^{{1 \\over... | ["D"] | null | $${1 \over 2}mv_1^2 = h{f_1} - \phi $$ ___________(1)<br><br>$${1 \over 2}mv_2^2 = h{f_2} - \phi $$ ___________(2)<br><br>Subtracting equation (1) by equation (2)<br><br>$${1 \over 2}mv_1^2 - {1 \over 2}mv_2^2 = h{f_1} - h{f_2}$$<br><br>$$v_1^2 - v_2^2 = {{2h} \over m}({f_1} - {f_2})$$ | mcq | jee-main-2021-online-17th-march-evening-shift | 10,131 |
1krwc7got | physics | dual-nature-of-radiation | photoelectric-effect | A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular are of radius 30 cm. The value of B is ___________ $$\times$$ 10<sup>$$-$$7... | [] | null | 125 | By photoelectric equation<br><br>$${{hc} \over \lambda } - \phi = {k_{\max }}$$<br><br>$${k_{\max }} = {{1240} \over {500}} - 1.25 \approx 1.25$$<br><br>$$r = {{\sqrt {2mk} } \over {eB}}$$<br><br>$$B = {{\sqrt {2mk} } \over {er}}$$<br><br>$$ = 125 \times {10^{ - 7}}T$$ | integer | jee-main-2021-online-25th-july-evening-shift | 10,135 |
1ks0jshx1 | physics | dual-nature-of-radiation | photoelectric-effect | An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000$$\mathop A\limits^o $$. What ... | [{"identifier": "A", "content": "7.61 eV"}, {"identifier": "B", "content": "1.41 eV"}, {"identifier": "C", "content": "3.3 eV"}, {"identifier": "D", "content": "No photoelectron would be emitted"}] | ["B"] | null | initially, energy of electron = + 3eV<br><br>finally, in 2<sup>nd</sup> excited state,<br><br>energy of electron = $$ - {{(13.6eV)} \over {{3^2}}}$$<br><br>$$ = - 1.51eV$$<br><br>Loss in energy is emitted as photon, <br><br>So, photon energy $${{hc} \over \lambda } = 4.51eV$$<br><br>Now, photoelectric effect equation ... | mcq | jee-main-2021-online-27th-july-evening-shift | 10,136 |
1ktac8sij | physics | dual-nature-of-radiation | photoelectric-effect | In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function $$\phi$$ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 $$\times$$ 10<sup>$$-$$34</sup> Js, c = 3 $$\times$$ 10<sup>8</sup... | [{"identifier": "A", "content": "1.3 V"}, {"identifier": "B", "content": "1.1 V"}, {"identifier": "C", "content": "1.9 V"}, {"identifier": "D", "content": "0.6 V"}] | ["A"] | null | $$K{E_{\max }} = e{V_s} = {{hc} \over \lambda } - \phi $$<br><br>$$ \Rightarrow e{V_s} = {{1240} \over {280}} - 2.5 = $$ 1.93 eV<br><br>$$ \Rightarrow {V_{{s_1}}} = $$ 1.93 V .... (i)<br><br>$$ \Rightarrow e{V_{{s_2}}} = {{1240} \over {400}} - 2.5 = $$ 0.6 eV<br><br>$$ \Rightarrow {V_{{s_2}}} = $$ 0.6 V .... (ii)<br><b... | mcq | jee-main-2021-online-26th-august-morning-shift | 10,137 |
1kte0pcwe | physics | dual-nature-of-radiation | photoelectric-effect | In a photoelectric experiment, increasing the intensity of incident light : | [{"identifier": "A", "content": "increases the number of photons incident and also increases the K.E. of the ejected electrons"}, {"identifier": "B", "content": "increases the frequency of photons incident and increases the K.E. of the ejected electrons"}, {"identifier": "C", "content": "increases the frequency of phot... | ["D"] | null | $$\to$$ Increasing intensity means number of incident photons are increased.<br><br>$$\to$$ Kinetic energy of ejected electrons depend on the frequency of incident photons, not the intensity. | mcq | jee-main-2021-online-27th-august-morning-shift | 10,138 |
1l5474aqk | physics | dual-nature-of-radiation | photoelectric-effect | <p>Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :</p>
<p>Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal.</p>
<p>Reason R : Kinetic energy of the photoelectrons is zero... | [{"identifier": "A", "content": "Both A and R are correct and R is the correct explanation of A."}, {"identifier": "B", "content": "Both A and R are correct but R is not the correct explanation of A."}, {"identifier": "C", "content": "A is correct but R is not correct."}, {"identifier": "D", "content": "A is not correc... | ["B"] | null | <p>When energy of incident radiation is equal to the work function of the metal, then the KE of photoelectrons would be zero. But this statement does not comment on the situation when energy is less than the work function.</p> | mcq | jee-main-2022-online-29th-june-morning-shift | 10,140 |
1l54uuc8l | physics | dual-nature-of-radiation | photoelectric-effect | <p>The electric field at a point associated with a light wave is given by</p>
<p>E = 200 [sin (6 $$\times$$ 10<sup>15</sup>)t + sin (9 $$\times$$ 10<sup>15</sup>)t] Vm<sup>$$-$$1</sup></p>
<p>Given : h = 4.14 $$\times$$ 10<sup>$$-$$15</sup> eVs</p>
<p>If this light falls on a metal surface having a work function of 2.5... | [{"identifier": "A", "content": "1.90 eV"}, {"identifier": "B", "content": "3.27 eV"}, {"identifier": "C", "content": "3.60 eV"}, {"identifier": "D", "content": "3.42 eV"}] | ["D"] | null | <p>Frequency of EM waves = $${6 \over {2\pi }} \times {10^{15}}$$ and $${9 \over {2\pi }} \times {10^{15}}$$</p>
<p>Energy of one photon of these waves</p>
<p>$$ = \left( {4.14 \times {{10}^{ - 15}} \times {6 \over {2\pi }} \times {{10}^{15}}} \right)$$ eV</p>
<p>and $$\left( {4.14 \times {{10}^{ - 15}} \times {9 \over... | mcq | jee-main-2022-online-29th-june-evening-shift | 10,141 |
1l55km7fv | physics | dual-nature-of-radiation | photoelectric-effect | <p>Let K<sub>1</sub> and K<sub>2</sub> be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $$\lambda$$<sub>1</sub> and $$\lambda$$<sub>2</sub>, respectively are incident on a metallic surface. If $$\lambda$$<sub>1</sub> = 3$$\lambda$$<sub>2</sub> then :</p> | [{"identifier": "A", "content": "$${K_1} > {{{K_2}} \\over 3}$$"}, {"identifier": "B", "content": "$${K_1} < {{{K_2}} \\over 3}$$"}, {"identifier": "C", "content": "$${K_1} = {{{K_2}} \\over 3}$$"}, {"identifier": "D", "content": "$${K_2} = {{{K_1}} \\over 3}$$"}] | ["B"] | null | <p>$${K_1} = {{hc} \over {{\lambda _1}}} - \phi = {{hc} \over {3{\lambda _2}}} - \phi $$ ..... (i)</p>
<p>and $${K_2} = {{hc} \over {{\lambda _2}}} - \phi $$ ..... (ii)</p>
<p>from (i) and (ii) we can say</p>
<p>$$3{K_1} = {K_2} - 2\phi $$</p>
<p>$${K_1} < {{{K_2}} \over 3}$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift | 10,142 |
1l58i6qtp | physics | dual-nature-of-radiation | photoelectric-effect | <p>A metal surface is illuminated by a radiation of wavelength 4500 $$\mathop A\limits^o $$. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90$$^\circ$$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximat... | [{"identifier": "A", "content": "1.36 eV"}, {"identifier": "B", "content": "1.69 eV"}, {"identifier": "C", "content": "2.78 eV"}, {"identifier": "D", "content": "2.23 eV"}] | ["A"] | null | <p>$${{hc} \over \lambda } - \phi = KE$$ ...... (i)</p>
<p>$$R = {{mv} \over {Bq}} = {{\sqrt {2m(KE)} } \over {Bq}}$$ ...... (ii)</p>
<p>Putting the values,</p>
<p>$$\phi \simeq 1.36$$ eV</p> | mcq | jee-main-2022-online-26th-june-evening-shift | 10,143 |
1l58ir3vh | physics | dual-nature-of-radiation | photoelectric-effect | <p>The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $$\mathop A\limits^o $$ is 0.42 V. If the threshold frequency is x $$\times$$ 10<sup>13</sup> /s, where x is _________ (nearest integer).</p>
<p>(Given, speed light = 3 $$\times$$ 10<sup>8</sup> m/s, Planck's con... | [] | null | 35 | <p>$${{hc} \over \lambda } - \phi = KE = e{V_0}$$</p>
<p>$$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {6630 \times {{10}^{ - 10}}}} - 6.63 \times {10^{ - 34}}{f_{th}} = 1.6 \times {10^{ - 19}} \times 0.4$$</p>
<p>$$ \Rightarrow {f_{th}} \simeq 35.11 \times {10^{13}}\,H$$</p> | integer | jee-main-2022-online-26th-june-evening-shift | 10,144 |
1l5bc7yy3 | physics | dual-nature-of-radiation | photoelectric-effect | <p>The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :</p> | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "2 : 1"}, {"identifier": "C", "content": "4 : 1"}, {"identifier": "D", "content": "1 : 4"}] | ["B"] | null | <p>$$3.8 = 0.6 + {1 \over 2}mv_1^2$$</p>
<p>$$1.4 = 0.6 + {1 \over 2}mv_2^2$$</p>
<p>$$ \Rightarrow {{v_1^2} \over {v_2^2}} = {{3.2} \over {0.8}} = {4 \over 1}$$</p>
<p>$$ \Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 1}$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift | 10,145 |
1l5c4lla0 | physics | dual-nature-of-radiation | photoelectric-effect | <p>When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v<sub>1</sub>. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v<sub>2</sub>. If v<sub>2</sub> = ... | [] | null | 2 | <p>Let us say that work function is $$\phi$$</p>
<p>$$ \Rightarrow 2\phi = \phi + {1 \over 2}mv_1^2$$ ...... (1)</p>
<p>and $$5\phi = \phi + {1 \over 2}mv_2^2$$ ..... (2)</p>
<p>From (1) and (2)</p>
<p>$${{v_2^2} \over {v_1^2}} = {4 \over 1}$$ or $${{{v_2}} \over {{v_1}}} = 2$$</p> | integer | jee-main-2022-online-24th-june-morning-shift | 10,146 |
1l6dz7o9g | physics | dual-nature-of-radiation | photoelectric-effect | <p>A metal exposed to light of wavelength $$800 \mathrm{~nm}$$ and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength $$500 \mathrm{~nm}$$ is used. The workfunction of the metal is : (Take hc $$=1230 \,\mathrm{eV}-\mathrm{nm}$$ ).</p> | [{"identifier": "A", "content": "1.537 eV"}, {"identifier": "B", "content": "2.46 eV"}, {"identifier": "C", "content": "0.615 eV"}, {"identifier": "D", "content": "1.23 eV"}] | ["C"] | null | <p>$$\because$$ $${K_m} = {{hc} \over \lambda } - \phi $$</p>
<p>$$ \Rightarrow K = {{1230} \over {800}} - \phi $$</p>
<p>and, $$2K = {{1230} \over {500}} - \phi $$</p>
<p>$$ \Rightarrow 2 \times {{1230} \over {800}} - 2\phi = {{1230} \over {500}} - \phi $$</p>
<p>$$ \Rightarrow \phi = 0.615\,eV$$</p> | mcq | jee-main-2022-online-25th-july-morning-shift | 10,147 |
1l6knp3m1 | physics | dual-nature-of-radiation | photoelectric-effect | <p>With reference to the observations in photo-electric effect, identify the correct statements from below :</p>
<p>(A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light.</p>
<p>(B) The value of saturation current increases on moving the source of light away from the meta... | [{"identifier": "A", "content": "(A) and (B) only"}, {"identifier": "B", "content": "(A) and (E) only"}, {"identifier": "C", "content": "(C) and (E) only"}, {"identifier": "D", "content": "(D) and (E) only"}] | ["B"] | null | <p>$$\because$$ $${1 \over 2}mv_m^2 = h\nu - \phi $$</p>
<p>$$ \Rightarrow v_m^2$$ varies linearly with frequency.</p>
<p>And, threshold wavelength can be explained by particle nature of light.</p> | mcq | jee-main-2022-online-27th-july-evening-shift | 10,148 |
1l6nsyeeo | physics | dual-nature-of-radiation | photoelectric-effect | <p>Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be</p> | [{"identifier": "A", "content": "1 : 2"}, {"identifier": "B", "content": "1 : 3"}, {"identifier": "C", "content": "2 : 3"}, {"identifier": "D", "content": "3 : 2"}] | ["C"] | null | <p>$${1 \over 2}mv_1^2 = 5\phi - \phi $$</p>
<p>And, $${1 \over 2}mv_2^2 = 10\phi - \phi $$</p>
<p>$$ \Rightarrow {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {4 \over 9}$$</p>
<p>$$ \Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 3}$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift | 10,149 |
1l6p5cftg | physics | dual-nature-of-radiation | photoelectric-effect | <p>The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $$\lambda$$. To double the kinetic energy, the incident light must have wavelength:</p> | [{"identifier": "A", "content": "$$\\frac{\\mathrm{hc}}{\\mathrm{E} \\lambda-\\mathrm{hc}}$$"}, {"identifier": "B", "content": "$$\\frac{\\mathrm{hc} \\lambda}{\\mathrm{E} \\lambda+\\mathrm{hc}}$$"}, {"identifier": "C", "content": "$$\\frac{\\mathrm{h} \\lambda}{\\mathrm{E} \\lambda+\\mathrm{hc}}$$"}, {"identifier": "D... | ["B"] | null | <p>$$k = {{hc} \over \lambda } - \phi = E$$</p>
<p>and, $$2k = {{hc} \over {{\lambda _2}}} - \phi = 2E$$</p>
<p>$$ \Rightarrow {{hc} \over \lambda } - E = {{hc} \over {{\lambda _2}}} - 2E$$</p>
<p>$$ \Rightarrow {{hc} \over {{\lambda _2}}} = {{hc} \over \lambda } + E$$</p>
<p>$$ \Rightarrow {\lambda _2} = {{hc\lambda... | mcq | jee-main-2022-online-29th-july-morning-shift | 10,150 |
ldo68faz | physics | dual-nature-of-radiation | photoelectric-effect | If the two metals $\mathrm{A}$ and $\mathrm{B}$ are exposed to radiation of wavelength $350 \mathrm{~nm}$. The work functions of metals $\mathrm{A}$ and $\mathrm{B}$ are $4.8 \mathrm{eV}$ and $2.2 \mathrm{eV}$. Then choose the correct option. | [{"identifier": "A", "content": "Metal B will not emit photo-electrons"}, {"identifier": "B", "content": "Both metals $\\mathrm{A}$ and $\\mathrm{B}$ will not emit photo-electrons"}, {"identifier": "C", "content": "Metal A will not emit photo-electrons\n"}, {"identifier": "D", "content": "Both metals A and B will emit ... | ["C"] | null | $$
\phi=\frac{h c}{\lambda}=\frac{1240}{350} \mathrm{eV}=3.54 \mathrm{eV}
$$
<br/><br/>$\therefore$ Only metal B will emit photoelectron. | mcq | jee-main-2023-online-31st-january-evening-shift | 10,152 |
1ldspqit2 | physics | dual-nature-of-radiation | photoelectric-effect | <p>The threshold wavelength for photoelectric emission from a material is 5500 $$\mathop A\limits^o $$. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a</p>
<p>A. 75 W infra-red lamp</p>
<p>B. 10 W infra-red lamp</p>
<p>C. 75 W ultra-violet lamp</p>
<p>D. 10 W ultra-... | [{"identifier": "A", "content": "C only"}, {"identifier": "B", "content": "A and D only"}, {"identifier": "C", "content": "C and D only"}, {"identifier": "D", "content": "B and C only"}] | ["C"] | null | Wavelength of infra-red $=700 \mathrm{~nm}$ (minimum)
<br/><br/>
Wavelength of UV $=100-400 \mathrm{~nm}$
<br/><br/>
Since we need $\lambda<5000$ Å
<br/><br/>
$\Rightarrow$ Only UV would be able to emit photoelectrons. | mcq | jee-main-2023-online-29th-january-morning-shift | 10,153 |
1ldtzi30b | physics | dual-nature-of-radiation | photoelectric-effect | <p>Given below are two statements :</p>
<p>Statement I : Stopping potential in photoelectric effect does not depend on the power of the light source.</p>
<p>Statement II : For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.</p>
<p>In the light of above sta... | [{"identifier": "A", "content": "Both Statement I and Statement II are incorrect"}, {"identifier": "B", "content": "Statement I is correct but Statement II is incorrect"}, {"identifier": "C", "content": "Both Statement I and Statement II are correct"}, {"identifier": "D", "content": "Statement I is incorrect but Statem... | ["C"] | null | <b>Statement I</b> is correct as stopping potential is
independent of power of light used.<br/><br/>
<b>Statement II</b> is correct as maximum kinetic energy
of photoelectron depends on wavelength of light. | mcq | jee-main-2023-online-25th-january-evening-shift | 10,154 |
1lgp0f0jj | physics | dual-nature-of-radiation | photoelectric-effect | <p>Given below are two statements:</p>
<p>Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.</p>
<p>Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely pro... | [{"identifier": "A", "content": "Both Statement I and Statement II are true"}, {"identifier": "B", "content": "Statement I is true but statement II is false"}, {"identifier": "C", "content": "Statement I is false but statement II is true"}, {"identifier": "D", "content": "Both Statement I and Statement II are false"}] | ["B"] | null | Now let's analyze both statements:
<br/><br/>
Statement I is correct. According to the photoelectric effect, the ability to emit electrons from a metallic surface depends on the energy of the incident light. The energy of a photon is given by $$E = h\nu$$, where $$h$$ is Planck's constant and $$\nu$$ is the frequency o... | mcq | jee-main-2023-online-13th-april-evening-shift | 10,156 |
1lgp0tkqs | physics | dual-nature-of-radiation | photoelectric-effect | <p>An atom absorbs a photon of wavelength $$500 \mathrm{~nm}$$ and emits another photon of wavelength $$600 \mathrm{~nm}$$. The net energy absorbed by the atom in this process is $$n \times 10^{-4} ~\mathrm{eV}$$. The value of n is __________. [Assume the atom to be stationary during the absorption and emission process... | [] | null | 4125 | The energy $$E$$ of a photon is related to its wavelength $$\lambda$$ by the formula:
<br/><br/>
$$E=\frac{hc}{\lambda}$$
<br/><br/>
where $$h$$ is Planck's constant and $$c$$ is the speed of light.
In this problem, we are given that an atom absorbs a photon of wavelength $$\lambda_1=500~\mathrm{nm}$$ and emits anoth... | integer | jee-main-2023-online-13th-april-evening-shift | 10,157 |
1lguy18zk | physics | dual-nature-of-radiation | photoelectric-effect | <p>A metallic surface is illuminated with radiation of wavelength $$\lambda$$, the stopping potential is $$V_{0}$$. If the same surface is illuminated with radiation of wavelength $$2 \lambda$$. the stopping potential becomes $$\frac{V_{o}}{4}$$. The threshold wavelength for this metallic surface will be</p> | [{"identifier": "A", "content": "$$3 \\lambda$$"}, {"identifier": "B", "content": "$$4 \\lambda$$"}, {"identifier": "C", "content": "$$\\frac{3}{2} \\lambda$$"}, {"identifier": "D", "content": "$$\\frac{\\lambda}{4}$$"}] | ["A"] | null | <p>The photoelectric effect occurs when light (or more generally, electromagnetic radiation) incident on a metallic surface causes the ejection of electrons from the surface. The energy of the incident photons must be greater than the work function of the metal (denoted by $$\phi_0$$) for the electrons to be ejected. <... | mcq | jee-main-2023-online-11th-april-morning-shift | 10,159 |
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