question_id stringlengths 8 35 | subject stringclasses 3
values | chapter stringclasses 90
values | topic stringclasses 459
values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
values | answer stringclasses 460
values | explanation stringlengths 1 10.6k | question_type stringclasses 3
values | paper_id stringclasses 154
values | __index_level_0__ int64 2 13.4k |
|---|---|---|---|---|---|---|---|---|---|---|---|
CSmROBamUkhdHJ1Oe1ewW | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | A particle of mass m moves in a circular orbit in a central potential field U(r) = $${1 \over 2}$$ kr<sup>2</sup>. If Bohr 's
quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as :
| [{"identifier": "A", "content": "r<sub>n</sub> $$ \\propto $$ $$\\sqrt n $$, E<sub>n</sub> $$ \\propto $$ n"}, {"identifier": "B", "content": "r<sub>n</sub> $$ \\propto $$ $$\\sqrt n $$, E<sub>n</sub> $$ \\propto $$ $${1 \\over n}$$"}, {"identifier": "C", "content": "r<sub>n</sub> $$ \\propto $$ n, E<sub>n</sub> $$ \\p... | ["A"] | null | Force due to this field, F = $$ - {{\partial U} \over {\partial r}}$$
<br><br>F = $$ - {\partial \over {\partial r}}\left( {{1 \over 2}k{r^2}} \right)$$ = -kr
<br><br>For circular orbit, $${{m{v^2}} \over r}$$ = -kr
<br><br>$$ \Rightarrow $$ v $$ \propto $$ r ..... (1)
<br><br>Fron Bohr’s quantization condition
<br><b... | mcq | jee-main-2019-online-12th-january-morning-slot | 9,039 |
VDKk7Z1f4AiNiWyhELJ4I | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $$\lambda $$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be: | [{"identifier": "A", "content": "$${{25} \\over {16}}$$ $$\\lambda $$"}, {"identifier": "B", "content": "$${{27} \\over {20}}$$ $$\\lambda $$"}, {"identifier": "C", "content": "$${{16} \\over {25}}$$ $$\\lambda $$"}, {"identifier": "D", "content": "$${{20} \\over {27}}$$ $$\\lambda $$"}] | ["D"] | null | For M $$ \to $$ L steel
<br><br>$${1 \over \lambda }$$ = K $$\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}$$
<br><br>for N $$ \to $$ L
<br><br>$${1 \over {\lambda '}}$$ = K$$\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}$$
<br><br>$$\lambda ' =... | mcq | jee-main-2019-online-11th-january-evening-slot | 9,040 |
huPUPLVIjXYGZr30XjUD0 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | A He<sup>+</sup> ion is in its first excited state. Its
ionization energy is :- | [{"identifier": "A", "content": "13.60 eV"}, {"identifier": "B", "content": "6.04 eV"}, {"identifier": "C", "content": "48.36 eV"}, {"identifier": "D", "content": "54.40 eV"}] | ["A"] | null | $$T.E. = - \left( {13.6} \right)\left( {{{{Z^2}} \over {{n^2}}}} \right)eV$$<br><br>
z = n = 2<br><br>
$$ \Rightarrow $$ Ionisation energy = –T.E.
= 13.6 eV | mcq | jee-main-2019-online-9th-april-evening-slot | 9,042 |
qHvrWxRT4A6DGYIGXN7k9k2k5dl1t8p | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 $$ \times $$ 10<sup>-16</sup> s. The frequency of revolution of the electron in its first excited state (in s<sup>-1</sup>) is :
| [{"identifier": "A", "content": "5.6 $$ \\times $$ 10<sup>12</sup>"}, {"identifier": "B", "content": "1.6 $$ \\times $$ 10<sup>14</sup>"}, {"identifier": "C", "content": "7.8 $$ \\times $$ 10<sup>14</sup>"}, {"identifier": "D", "content": "6.2 $$ \\times $$ 10<sup>15</sup>"}] | ["C"] | null | Time period of revolution of electron in n<sup>th</sup> orbit
<br><br>V = $${{2\pi r} \over V}$$
<br><br>= $${{2\pi {a_0}\left( {{{{n^2}} \over Z}} \right)} \over {{V_0}\left( {{Z \over n}} \right)}}$$
<br><br>$$ \Rightarrow $$ T $$ \propto $$ $${{{{n^3}} \over {{Z^2}}}}$$
<br><br>$$ \therefore $$ $${{{T_1}} \over {{T_... | mcq | jee-main-2020-online-7th-january-morning-slot | 9,043 |
blZKlWL9p6XXiMFZJH7k9k2k5hi38jp | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The first member of the Balmer series of
hydrogen atom has a wavelength of 6561 Å.
The wavelength of the second member of the
Balmer series (in nm) is: | [] | null | 486 | $${1 \over {{\lambda _1}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)$$ = $${5 \over {36}}$$RZ<sup>2</sup>
<br><br>$${1 \over {{\lambda _2}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$$ = $${{12} \over {64}}$$RZ<sup>2</sup>
<br><br>$$ \therefore $$ $${{{\lambda _2}} \over {{\lam... | integer | jee-main-2020-online-8th-january-evening-slot | 9,044 |
ytzgc8giNCiRqXJuIPjgy2xukexs9ya6 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | In a hydrogen atom the electron makes a
transition from (n + 1)<sup>th</sup> level to the n<sup>th</sup> level.
If n >> 1, the frequency of radiation emitted is
proportional to : | [{"identifier": "A", "content": "$${1 \\over n}$$"}, {"identifier": "B", "content": "$${1 \\over {{n^2}}}$$"}, {"identifier": "C", "content": "$${1 \\over {{n^3}}}$$"}, {"identifier": "D", "content": "$${1 \\over {{n^4}}}$$"}] | ["C"] | null | In hydrogen atom,
<br><br>E<sub>n</sub> = $${{ - {E_0}} \over {{n^2}}}$$
<br><br>Where E<sub>0</sub> is Ionisation Energy of H.
<br><br>For transition from (n + 1) to n, the energy of emitted radiation is equal to the difference in energies of
levels.
<br><br>$$\Delta $$E = E<sub>n+1</sub> - E<sub>n</sub>
<br><br>= $${... | mcq | jee-main-2020-online-2nd-september-evening-slot | 9,046 |
Dah43PgzkZABWneNm1jgy2xukf6jbc66 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | In the line spectra of hydrogen atoms, difference between the largest and the shortest wavelengths of the Lyman series is 304 $$\mathop A\limits^0 $$. The corresponding difference for the Paschan series in $$\mathop A\limits^0 $$ is :
___________. | [] | null | 10553 | For Lyman series :
<br><br>Shortest wavelength,
<br><br>$$\lambda $$<sub>s</sub> = $$R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$$
<br><br>longest wavelength,
<br><br>$$\lambda $$<sub>l</sub> = $$R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$$
<br><br>Given, $$\lambda $$<sub>l<... | integer | jee-main-2020-online-4th-september-morning-slot | 9,047 |
lr57s97192hOCZF0fLjgy2xukfhqxenz | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | A particle of mass 200 MeV/c<sup>2</sup> collides with a
hydrogen atom at rest. Soon after the collision
the particle comes to rest, and the atom
recoils and goes to its first excited state. The
initial kinetic energy of the particle (in eV) is
<br/>$${N \over 4}$$. The value of N is :
<br/>(Given the mass of the hydro... | [] | null | 51 | Given, m<sub>particle</sub> = 200 MeV/c<sup>2</sup> = m(Assume)
<br><br>and m<sub>H</sub> = 1
GeV/c<sup>2</sup> = 1000 MeV/c<sup>2</sup> = 5 $$ \times $$ 200 = 5m
<br><br>Applying momentum conservation,
<br><br>p<sub>i</sub> = p<sub>f</sub>
<br><br>$$ \Rightarrow $$ mv<sub>0</sub>
+ 0 = 0 + 5 mv'
<br><br>$$ \Rightarr... | integer | jee-main-2020-online-5th-september-morning-slot | 9,048 |
m8auc4dTdUq8DERoB71klri2550 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E.<br/><br/>The transitions A, B and C respectively represent :<br/><br/>
<img src="data:image/png;base64,UklGRhASAABXRUJQVlA4IAQSAABw+QCdASoAA+4BP4G+2GM2MTkooxKqkyAwCWlu1SbZibn+V4rXDFdA/8y1hY/Yz/h/... | [{"identifier": "A", "content": "The series limit of Lyman series, second member of Balmer series and second member of Paschen series."}, {"identifier": "B", "content": "The ionization potential of hydrogen, second member of Balmer series and third member of Paschen series."}, {"identifier": "C", "content": "The series... | ["C"] | null | In transition A, hydrogen atom comes from higher energy state n = $$\infty$$ to lower energy state n = 1. Hence, transition A represents series limit of Lyman series. In transition B, hydrogen atom comes from higher energy state n = 5 to lower energy state n = 2. Hence, transition B represents 3rd line of Balmer series... | mcq | jee-main-2021-online-24th-february-morning-slot | 9,049 |
d2ySfBgbnFTePmpauK1klroo58e | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | According to Bohr atomic model, in which of the following transitions will the frequency be maximum? | [{"identifier": "A", "content": "n = 2 to n = 1"}, {"identifier": "B", "content": "n = 3 to n = 2"}, {"identifier": "C", "content": "n = 4 to n = 3"}, {"identifier": "D", "content": "n = 5 to n = 4"}] | ["A"] | null | Let, n<sub>f</sub>, n<sub>i</sub> be the final and initial orbit.<br/><br/>As we know that,<br/><br/>$${1 \over \lambda } = 1.09 \times {10^7}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$$<br/><br/>Now, checking for each option, we get<br/><br/>(a) $${1 \over \lambda } \propto \left[ {{1 \over {{3^2}}} - {1 \... | mcq | jee-main-2021-online-24th-february-evening-slot | 9,050 |
ldWYWX064gd7GPEDzu1klt2u91x | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is : | [{"identifier": "A", "content": "194.8 nm"}, {"identifier": "B", "content": "490.7 nm"}, {"identifier": "C", "content": "913.3 nm"}, {"identifier": "D", "content": "121.8 nm"}] | ["D"] | null | $$\Delta$$E = 10.2 eV<br><br>$${{hc} \over \lambda }$$ = 10.2 eV<br><br>$$\lambda$$ = $${{hc} \over {(10.2)e}}$$<br><br>= $${{12400} \over {10.2}}\mathop A\limits^o $$<br><br>= 121.56 nm<br><br>$$ \simeq $$ 121.8 nm | mcq | jee-main-2021-online-25th-february-evening-slot | 9,051 |
Zyqnpg3XXS9z2CjBVr1kltjcayj | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | If $$\lambda$$<sub>1</sub> and $$\lambda$$<sub>2</sub> are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of $$\lambda$$<sub>1</sub> : $$\lambda$$<sub>2</sub> is : | [{"identifier": "A", "content": "7 : 135"}, {"identifier": "B", "content": "7 : 108"}, {"identifier": "C", "content": "1 : 9"}, {"identifier": "D", "content": "1 : 3"}] | ["A"] | null | For Lyman series
<br><br>n<sub>1</sub> = 1, n<sub>2</sub> = 4
<br><br>$${1 \over {{\lambda _1}}} = R\left[ {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right]$$
<br><br>For paschen series
<br><br>n<sub>1</sub> = 3, n<sub>2</sub> = 4
<br><br>$${1 \over {{\lambda _2}}} = R\left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right... | mcq | jee-main-2021-online-26th-february-morning-slot | 9,052 |
Ku3f7NOWoEVufpImbq1kmj1e3vm | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | If an electron is moving in the n<sup>th</sup> orbit of the hydrogen atom, then its velocity (v<sub>n</sub>) for the n<sup>th</sup> orbit is given as : | [{"identifier": "A", "content": "$${v_n} \\propto {1 \\over n}$$"}, {"identifier": "B", "content": "v<sub>n</sub> $$ \\propto $$ n<sup>2</sup>"}, {"identifier": "C", "content": "v<sub>n</sub> $$ \\propto $$ n"}, {"identifier": "D", "content": "$${v_n} \\propto {1 \\over {{n^2}}}$$"}] | ["A"] | null | We know velocity of electron in n<sup>th</sup> shell of hydrogen atom is given by<br><br>$$v = {{2\pi kZ{e^2}} \over {nh}}$$<br><br>$$ \therefore $$ $$v \propto {1 \over n}$$ | mcq | jee-main-2021-online-17th-march-morning-shift | 9,054 |
txvpfizTT1K3LQq3xX1kmj3t6ql | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom? | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["B"] | null | $${E_n} = - 13.6{{{Z^2}} \over {{n^2}}}$$<br><br>E<sub>n<sup>th</sup></sub> of Carbon = E<sub>1<sup>st</sup></sub> of Hydrogen<br><br>$$ \Rightarrow $$ $$ - 13.6 \times {{{6^2}} \over {{n^2}}} = - 13.6 \times {{{1^2}} \over {{1^2}}}$$<br><br>$$ \Rightarrow $$ n = 6 | mcq | jee-main-2021-online-17th-march-morning-shift | 9,055 |
o9VorCr6eZN0xXMKZA1kmka86pl | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region? | [{"identifier": "A", "content": "Brackett series"}, {"identifier": "B", "content": "Balmer series"}, {"identifier": "C", "content": "Paschen series"}, {"identifier": "D", "content": "Lyman series"}] | ["B"] | null | Balmer series of hydrogen atomic spectrum is lying in the visible region, when electron jumps from a higher energy level to n = 2 orbit. | mcq | jee-main-2021-online-17th-march-evening-shift | 9,056 |
0ZVeUApTSswzwSzpZo1kmkcev2q | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | A particle of mass m moves in a circular orbit in a <br/>central potential field U(r) = U<sub>0</sub>r<sup>4</sup>. If Bohr's quantization conditions <br/>are applied, radii of possible <br/>orbitals r<sub>n</sub> vary with $${n^{{1 \over \alpha }}}$$, where $$\alpha$$ is ____________. | [] | null | 3 | $$\overrightarrow F = - {{d\overrightarrow u } \over {dr}}$$<br><br>$$ = - {d \over {dr}}({U_0}{r^4})$$<br><br>$$\overrightarrow F = - 4{U_0}{r^3}$$<br><br>$$ \because $$ $${{m{v^2}} \over r} = 4{U_0}{r^3}$$<br><br>$$m{v^2} = 4{U_0}{r^4}$$<br><br>Then $$v \propto {r^2}$$<br><br>$$ \because $$ $$mvr = {{nh} \over {... | integer | jee-main-2021-online-17th-march-evening-shift | 9,057 |
8KP7rLahTFLiWpJbQ71kmkqwv4e | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Imagine that the electron in a hydrogen atom is replaced by a muon ($$\mu$$). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be : | [{"identifier": "A", "content": "13.6 eV"}, {"identifier": "B", "content": "2815.2 eV"}, {"identifier": "C", "content": "331.2 eV"}, {"identifier": "D", "content": "27.2 eV"}] | ["B"] | null | m<sub>m</sub> = 207 m<sub>e</sub>
<br><br>In hydrogen atom one electron present. Now that the electron in hydrogen atom is replaced by a muon ($$\mu$$).
<br><br>We know, Energy(E) = $$ - {{{e^4}m} \over {8\varepsilon _0^2{n^2}{h^2}}}$$
<br><br>For electron,
<br><br>E<sub>e</sub> = $$ - {{{e^4}{m_e}} \over {8\varepsilon... | mcq | jee-main-2021-online-18th-march-morning-shift | 9,058 |
1ks1awgcg | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The K<sub>$$\alpha$$</sub> X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be __________ keV. (Round off to the nearest integer)<br/><br/>[h = 4.14 $$\times$$ 10<sup>$$-$$15</sup> e... | [] | null | 10 | $${E_{{k_\alpha }}} = {E_k} - {E_L}$$<br><br>$${{hc} \over {{\lambda _{{k_\alpha }}}}} = {E_k} - {E_L}$$<br><br>$${E_L} = {E_k} - {{hc} \over {{\lambda _{{k_\alpha }}}}}$$<br><br>= 27.5 KeV $$ - {{12.42 \times {{10}^{ - 7}}eVm} \over {0.071 \times {{10}^{ - 9}}m}}$$<br><br>E<sub>L</sub> = (27.5 $$-$$ 17.5) keV<br><br>=... | integer | jee-main-2021-online-27th-july-evening-shift | 9,060 |
1ktac194x | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | A particular hydrogen like ion emits radiation of frequency 2.92 $$\times$$ 10<sup>15</sup> Hz when it makes transition from n = 3 to n = 1. The frequency in Hz of radiation emitted in transition from n = 2 to n = 1 will be : | [{"identifier": "A", "content": "0.44 $$\\times$$ 10<sup>15</sup>"}, {"identifier": "B", "content": "6.57 $$\\times$$ 10<sup>15</sup>"}, {"identifier": "C", "content": "4.38 $$\\times$$ 10<sup>15</sup>"}, {"identifier": "D", "content": "2.46 $$\\times$$ 10<sup>15</sup>"}] | ["D"] | null | $$n{f_1} = k\left( {{1 \over 1} - {1 \over {{3^2}}}} \right)$$<br><br>$$n{f_2} = k\left( {1 - {1 \over {{2^2}}}} \right)$$<br><br>$${{{f_1}} \over {{f_2}}} = {{8/9} \over {3/4}} \Rightarrow {f_2} = 2.46 \times {10^{15}}$$ | mcq | jee-main-2021-online-26th-august-morning-shift | 9,061 |
1ktfnxoj2 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | X different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are exited to states with principal quantum number n = 6 ? The value of X is ______________. | [] | null | 15 | No. of different wavelengths = $${{n(n - 1)} \over 2}$$<br><br>$$ = {{6 \times (6 - 1)} \over 2} = {{6 \times 5} \over 2} = 15$$ | integer | jee-main-2021-online-27th-august-evening-shift | 9,062 |
1l56vfq7c | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>Given below are two statements :</p>
<p>Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E<sub>1</sub>) to higher energy orbit (E<sub>2</sub>), is given as hf = E<sub>1</sub> $$-$$ E<sub>2</sub></p>
<p>Statement II : The jumping of electron from highe... | [{"identifier": "A", "content": "Both Statement I and Statement II are true."}, {"identifier": "B", "content": "Both Statement I and Statement II are false."}, {"identifier": "C", "content": "Statement I is correct but Statement II is false."}, {"identifier": "D", "content": "Statement I is incorrect but Statement II i... | ["D"] | null | <p>Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.</p>
<p>Also, E<sub>2</sub> $$-$$ E<sub>1</sub> is the energy of photon</p>
<p>$$\Rightarrow$$ E<sub>2</sub> $$-$$ E<sub>1</sub> = hf</p>
<p>$$ \Rightarrow f = {{{E_2} - {E_1}} \over h}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 9,063 |
1l57qgiqx | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of :</p>
<p>(Given, Planck's constant = 6.6 $$\times$$ 10<sup>$$-$$34</sup> Js).</p> | [{"identifier": "A", "content": "2.10 $$\\times$$ 10<sup>$$-$$34</sup> Js"}, {"identifier": "B", "content": "1.05 $$\\times$$ 10<sup>$$-$$34</sup> Js"}, {"identifier": "C", "content": "3.15 $$\\times$$ 10<sup>$$-$$34</sup> Js"}, {"identifier": "D", "content": "4.2 $$\\times$$ 10<sup>$$-$$34</sup> Js"}] | ["B"] | null | <p>$$ - 13.6 + 10.2 = {{ - 13.6} \over {{n^2}}}$$</p>
<p>$$ \Rightarrow {{13.6} \over {{n^2}}} = 3.4$$</p>
<p>$$ \Rightarrow n = 2$$</p>
<p>$$ \Rightarrow \Delta L = 2 \times {h \over {2\lambda }} - 1 \times {h \over {2\lambda }}$$</p>
<p>$$ = {h \over {2\lambda }}$$</p>
<p>$$ \Rightarrow \Delta L \simeq 1.05 \times {1... | mcq | jee-main-2022-online-27th-june-morning-shift | 9,064 |
1l5akyfa2 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The ratio for the speed of the electron in the 3<sup>rd</sup> orbit of He<sup>+</sup> to the speed of the electron in the 3<sup>rd</sup> orbit of hydrogen atom will be :</p> | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "1 : 2"}, {"identifier": "C", "content": "4 : 1"}, {"identifier": "D", "content": "2 : 1"}] | ["D"] | null | <p>We know that $$v \propto {Z \over n}$$</p>
<p>$$\Rightarrow$$ Required ratio $$ = {{{2 \over 3}} \over {{1 \over 3}}}$$</p>
<p>$$ = 2:1$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift | 9,066 |
1l5bcb0bc | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>In Bohr's atomic model of hydrogen, let K, P and E are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level :</p> | [{"identifier": "A", "content": "All K, P and E increase."}, {"identifier": "B", "content": "K decreases, P and E increase."}, {"identifier": "C", "content": "P decreases, K and E increase."}, {"identifier": "D", "content": "K increases, P and E decrease."}] | ["B"] | null | <p>$$T.E. = {{ - {Z^2}m{e^4}} \over {8{{\left( {nh{\varepsilon _0}} \right)}^2}}}$$</p>
<p>$$P.E. = {{ - {Z^2}m{e^4}} \over {4{{\left( {nh{\varepsilon _0}} \right)}^2}}}$$</p>
<p>$$K.E. = {{{Z^2}m{e^4}} \over {8{{\left( {nh{\varepsilon _0}} \right)}^2}}}$$</p>
<p>As electron makes transition to higher level, total ener... | mcq | jee-main-2022-online-24th-june-evening-shift | 9,067 |
1l5w38ipy | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :</p> | [{"identifier": "A", "content": "1.05 $$\\times$$ 10<sup>$$-$$34</sup> Js"}, {"identifier": "B", "content": "2.11 $$\\times$$ 10<sup>$$-$$34</sup> Js"}, {"identifier": "C", "content": "3.16 $$\\times$$ 10<sup>$$-$$34</sup> Js"}, {"identifier": "D", "content": "4.22 $$\\times$$ 10<sup>$$-$$34</sup> Js"}] | ["B"] | null | Change in energy
<br/><br/>
$$
\begin{aligned}
&\Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{2} \Rightarrow 12.09=\mathrm{E}_{\mathrm{f}}-(-13.6) \Rightarrow \mathrm{E}_{\mathrm{f}}=-1.51 \mathrm{eV} \\\\
&\Rightarrow \frac{-13.6}{n^{2}}=-1.51 \Rightarrow \mathrm{n}^{2}=9 \Rightarrow \mathrm{n}=3 \\\\
&\text {... | mcq | jee-main-2022-online-30th-june-morning-shift | 9,068 |
1l5w3rgcn | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A hydrogen atom in its first excited state absorbs a photon of energy x $$\times$$ 10<sup>$$-$$2</sup> eV and excited to a higher energy state where the potential energy of electron is $$-$$1.08 eV. The value of x is ______________.</p> | [] | null | 286 | As, $E_{n}=\frac{P \cdot E_{n}}{2}=-\frac{1.08}{2}=-0.544$
<br/><br/>
$$
\begin{aligned}
& \text { So, } \Delta \mathrm{E}, \mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}=-0.544-\left(-\frac{13.6}{2^{2}}\right)=3.4-0.544 \\\\
& \approx 2.86 \mathrm{eV}=286 \times 10^{-2} \mathrm{eV}
\end{aligned}
$$ | integer | jee-main-2022-online-30th-june-morning-shift | 9,069 |
1l6f5mlpu | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>$${x \over {x + 4}}$$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its</p>
<p>(i) third permitted energy level to the second level and</p>
<p>(ii) the highest permitted energy level to the second permitted level.</p>
<p>The value of x will be ____________.</p> | [] | null | 5 | <p>$${E_n} = - {{13.6} \over {{n^2}}}\,eV$$</p>
<p>$${{{1 \over {{2^2}}} - {1 \over {{3^2}}}} \over {{1 \over {{2^2}}}}} = {x \over {x + 4}}$$</p>
<p>$$ \Rightarrow {{9 - 4} \over {9 \times 4 \times {1 \over 4}}} = {x \over {x + 4}} = {5 \over 9}$$</p>
<p>$$x = 5$$</p> | integer | jee-main-2022-online-25th-july-evening-shift | 9,073 |
1l6got1jf | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>In the hydrogen spectrum, $$\lambda$$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "a$$\lambda$$'' between the wavelength of $$3^{\text {rd }}$$ transition line of Paschen series and that of $$2^{\text {nd }}$$ transition line of Balmer series where $$\mathrm{a}=$$ __... | [] | null | 5 | <p>$${1 \over \lambda } = {R_H}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$$</p>
<p>$${1 \over {{\lambda _3}}} = {R_H}\left( {{1 \over {{3^2}}} - {1 \over {{6^2}}}} \right)$$</p>
<p>$${1 \over {{\lambda _2}}} = {R_H}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$$</p>
<p>$$\therefore$$ $${\lambda _3}... | integer | jee-main-2022-online-26th-july-morning-shift | 9,074 |
1l6p5g0cg | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.</p> | [{"identifier": "A", "content": "3 : 4"}, {"identifier": "B", "content": "4 : 3"}, {"identifier": "C", "content": "1 : 4"}, {"identifier": "D", "content": "4 : 1"}] | ["A"] | null | <p>$${E_1} = {E_0}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right) = {E_0} \times {3 \over 4}$$</p>
<p>$${E_2} = {E_0}$$</p>
<p>$$\therefore$$ $${{{E_1}} \over {{E_2}}} = {3 \over 4}$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift | 9,075 |
1ldny12o3 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>An electron of a hydrogen like atom, having $$Z=4$$, jumps from $$4^{\text {th }}$$ energy state to $$2^{\text {nd }}$$ energy state. The energy released in this process, will be :</p>
<p>(Given Rch = $$13.6~\mathrm{eV}$$)</p>
<p>Where R = Rydberg constant</p>
<p>c = Speed of light in vacuum</p>
<p>h = Planck's cons... | [{"identifier": "A", "content": "$$10.5 ~\\mathrm{eV}$$"}, {"identifier": "B", "content": "$$40.8 ~\\mathrm{eV}$$"}, {"identifier": "C", "content": "$$13.6 ~\\mathrm{eV}$$"}, {"identifier": "D", "content": "$$3.4 ~\\mathrm{eV}$$"}] | ["B"] | null | The energy difference between the 4th and 2nd energy states of a hydrogen-like atom can be calculated using the formula for the energy levels of a hydrogen-like atom:
<br/><br/>$\begin{aligned} \Delta \mathrm{E} & =13.6 \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\\\ \mathrm{Z} & =4 \te... | mcq | jee-main-2023-online-1st-february-evening-shift | 9,076 |
ldo7nir0 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | If the binding energy of ground state electron in a hydrogen atom is $13.6\, \mathrm{eV}$, then, the energy required to remove the electron from the second excited state of $\mathrm{Li}^{2+}$ will be : $x \times 10^{-1} \mathrm{eV}$. The value of $x$ is ________.<br/><br/> | [] | null | 136 | $\mathrm{E}_{H}=13.6$
<br/><br/>$$
\begin{aligned}
& \mathrm{E}_{\mathrm{Li}^{2+}}=13.6 \frac{Z^{2}}{n^{2}}=13.6 \times \frac{9}{9}=13.6 \mathrm{eV} \\\\
& =136 \times 10^{-1} \mathrm{eV}
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-evening-shift | 9,078 |
1ldoh7zt6 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A light of energy $$12.75 ~\mathrm{eV}$$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $$\frac{x}{\pi} \times 10^{-17} ~\mathrm{eVs}$$. The value of $$x$$ is ___________ (use $$h=4.... | [] | null | 828 | Let the electron jumps to $n^{\text {th }}$ orbit so
<br/><br/>$$
\begin{aligned}
& 12.75=13.6\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}\right] \\\\
& \Rightarrow n=4 \\\\
& \text { So, Angular momentum } L=\frac{n h}{2 \pi}=\frac{2 h}{\pi}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\text { Angular momentum }=\frac{2}{... | integer | jee-main-2023-online-1st-february-morning-shift | 9,079 |
1ldpmqijh | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>For hydrogen atom, $$\lambda_{1}$$ and $$\lambda_{2}$$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $$\lambda_{1}$$ and $$\lambda_{2}$$ is $$\frac{x}{32}$$. The value of $$x$$ is __________.</p>
<p><img src="data:image/png;base64,UklGRogIAABXRUJQVlA4IHwIA... | [] | null | 27 | $\frac{1}{\lambda}=\mathrm{RZ}^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$
<br/><br/>$$
\begin{aligned}
& \frac{1}{\lambda_{1}}=\mathrm{RZ}^{2}\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]=\frac{8}{9} \mathrm{RZ}^{2}
........(1)\\\\
& \frac{1}{\lambda_{2}}=\mathrm{RZ}^{2}\left[\frac{1}... | integer | jee-main-2023-online-31st-january-morning-shift | 9,080 |
1ldr1y5cy | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>Speed of an electron in Bohr's $$7^{\text {th }}$$ orbit for Hydrogen atom is $$3.6 \times 10^{6} \mathrm{~m} / \mathrm{s}$$. The corresponding speed of the electron in $$3^{\text {rd }}$$ orbit, in $$\mathrm{m} / \mathrm{s}$$ is :</p> | [{"identifier": "A", "content": "$$\\left(1.8 \\times 10^{6}\\right)$$"}, {"identifier": "B", "content": "$$\\left(7.5 \\times 10^{6}\\right)$$"}, {"identifier": "C", "content": "$$\\left(8.4 \\times 10^{6}\\right)$$"}, {"identifier": "D", "content": "$$\\left(3.6 \\times 10^{6}\\right)$$"}] | ["C"] | null | <p>$$v\,\alpha {z \over n}$$</p>
<p>$${{{v_1}} \over {{v_2}}} = \left( {{{{n_2}} \over {{n_1}}}} \right)$$</p>
<p>$$ \Rightarrow {{3.6 \times {{10}^6}} \over {{v_2}}} = {3 \over 7}$$</p>
<p>$$ \Rightarrow {v_2} = {7 \over 3} \times 3.6 \times {10^6}$$ m/s</p>
<p>$$ = 8.4 \times {10^6}$$ m/s</p> | mcq | jee-main-2023-online-30th-january-morning-shift | 9,081 |
1ldujlgca | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The wavelength of the radiation emitted is $$\lambda_0$$ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will $$\frac{20}{x}\lambda_0$... | [] | null | 27 | Second excited state $$ \to $$ first excited state , $n=3$ to $n=2$
<br/><br/>
$\frac{1}{\lambda_{0}}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\left(\frac{5 R}{36}\right)$
<br/><br/>
Third excited state $$ \to $$ second orbit , $n=4$ to $n=2$
<br/><br/>
$\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{16}\right)=\left(\fra... | integer | jee-main-2023-online-25th-january-morning-shift | 9,083 |
lgnyzx7k | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | As per given figure $A, B$ and $C$ are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengths $\left(\right.$ i.e. $\left.\frac{\lambda_{1}}{\lambda_{2}}\right)$ is $\frac{7}{4 n}$, then the value of $n$ will be __________.<br/><br/>
<img src="data:image/png... | [] | null | 5 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrdqvw1/469e8610-a609-4793-9320-89d617a6f046/ffe7c610-e0b7-11ed-bb10-15704e76115a/file-1lgrdqvw2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrdqvw1/469e8610-a609-4793-9320-89d617a6f046/ffe7c610-e0b7-11ed-bb10-15704e76115a/fi... | integer | jee-main-2023-online-15th-april-morning-shift | 9,085 |
1lgrjh1y3 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A $$12.5 \mathrm{~eV}$$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["C"] | null | The energy of an electron in an excited state of hydrogen is given by the equation:
<br/><br/>
Code snippet<br/><br/>
$$E = -13.6 \frac{1}{n^2} \mathrm{~eV}$$<br/><br/>
where n is the principal quantum number of the state.
<br/><br/>
The energy of the electron beam is 12.5 eV, which is enough to excite the electron to ... | mcq | jee-main-2023-online-12th-april-morning-shift | 9,087 |
1lgyf0gm1 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be</p> | [{"identifier": "A", "content": "L"}, {"identifier": "B", "content": "$$\\frac{L}{2}$$"}, {"identifier": "C", "content": "zero"}, {"identifier": "D", "content": "2 L"}] | ["A"] | null | <p>According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck's constant). This can be expressed as:</p>
<p>$$ L = n \frac{h}{2\pi} $$</p>
<p>where $n$ is the principal q... | mcq | jee-main-2023-online-10th-april-morning-shift | 9,088 |
1lh25ib5c | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is :</p>
<p><img src="data:image/png;base64,UklGRlgLAABXRUJQVlA4IEwLAACQwACdASoAAzUCP4G+2GW2L6ynILEJ4sAwCWlu+EQaAdZXd+/d/AvdNHf2exgnDmz67dZj9vnzpjJT//9VZcmdFw6tYR9/zoOrWEff86Dq1hH3/Og6effpdsqKcZOXph... | [{"identifier": "A", "content": "A"}, {"identifier": "B", "content": "C"}, {"identifier": "C", "content": "B"}, {"identifier": "D", "content": "D"}] | ["D"] | null | We know that, $E=\frac{h c}{\lambda}$
<br/><br/>$$
\therefore \lambda \propto \frac{1}{E}
$$
<br/><br/>For minimum wavelength, energy must be maximum which is possible from transition $n=3$ to $n=1$. | mcq | jee-main-2023-online-6th-april-morning-shift | 9,089 |
1lh25mxeg | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The radius of fifth orbit of the $$\mathrm{Li}^{++}$$ is __________ $$\times 10^{-12} \mathrm{~m}$$.</p>
<p>Take: radius of hydrogen atom $$ = 0.51\,\mathop A\limits^o $$</p> | [] | null | 425 | <p>The formula to calculate the radius of an orbit for a hydrogen-like atom/ion is:</p>
<p>$$
r_n = r_0 \frac{n^2}{Z}
$$</p>
<p>where:</p>
<ul>
<li>$r_n$ is the radius of the nth orbit,</li>
<li>$n$ is the principal quantum number (the orbit number),</li>
<li>$r_0$ is the Bohr radius (radius of the first Bohr orbit in ... | integer | jee-main-2023-online-6th-april-morning-shift | 9,090 |
1lh30r2ot | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A small particle of mass $$m$$ moves in such a way that its potential energy $$U=\frac{1}{2} m ~\omega^{2} r^{2}$$ where $$\omega$$ is constant and $$r$$ is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of $$n^{\text {th }}$$ orbit will be proportio... | [{"identifier": "A", "content": "$$\\sqrt{n}$$"}, {"identifier": "B", "content": "$$n^{2}$$"}, {"identifier": "C", "content": "$$\\frac{1}{n}$$"}, {"identifier": "D", "content": "$$n$$"}] | ["A"] | null | <p>According to Bohr's quantization of angular momentum, the angular momentum $$L$$ of a particle in a circular orbit is given by:</p>
<p>$$L = n\hbar$$</p>
<p>Where $$n$$ is an integer and $$\hbar$$ is the reduced Planck's constant. The angular momentum $$L$$ can also be expressed as:</p>
<p>$$L = mvr$$</p>
<p... | mcq | jee-main-2023-online-6th-april-evening-shift | 9,091 |
1lh30xaes | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>Experimentally it is found that $$12.8 ~\mathrm{eV}$$ energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is $$\frac{9}{x} \times 10^{-10} \mathrm{~m}$$. The value of the $$x$$ is __________.</p>
<p>$$\left(1 \mathrm{eV}=1.6 \times 10... | [] | null | 16 | <p>The binding energy of an electron in a hydrogen atom is given by the formula:</p>
<p>$$ E = \frac{k e^2}{2 r} $$</p>
<p>where:</p>
<ul>
<li>$E$ is the energy of the electron,</li>
<li>$k$ is Coulomb's constant ($9 \times 10^9 \, \text{Nm}^2/\text{C}^2$),</li>
<li>$e$ is the charge of the electron ($1.6 \times 10... | integer | jee-main-2023-online-6th-april-evening-shift | 9,092 |
lsamb8bl | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | From the statements given below :<br/><br/>
(A) The angular momentum of an electron in $n^{\text {th }}$ orbit is an integral multiple of $\hbar$.<br/><br/>
(B) Nuclear forces do not obey inverse square law.<br/><br/>
(C) Nuclear forces are spin dependent.<br/><br/>
(D) Nuclear forces are central and charge independent... | [{"identifier": "A", "content": "(B), (C), (D), (E) only"}, {"identifier": "B", "content": "(A), (C), (D), (E) only"}, {"identifier": "C", "content": "(A), (B), (C), (E) only"}, {"identifier": "D", "content": "(A), (B), (C), (D) only"}] | ["C"] | null | <p>Let's analyze each of the given statements to determine which ones are correct:</p>
<p><b>(A) The angular momentum of an electron in $n^{\text{th}}$ orbit is an integral multiple of $\hbar$.</b></p>
<p>This statement reflects Bohr's quantization rule for angular momentum in the Bohr model of the hydrogen atom. Acco... | mcq | jee-main-2024-online-1st-february-evening-shift | 9,093 |
lsbkxrye | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly : | [{"identifier": "A", "content": "$13.6 \\mathrm{eV}$"}, {"identifier": "B", "content": "$1.5 \\mathrm{eV}$"}, {"identifier": "C", "content": "$12.1 \\mathrm{eV}$"}, {"identifier": "D", "content": "$1.9 \\mathrm{eV}$"}] | ["C"] | null | <p>The energy $E_n$ of an electron in the $n$th energy level of a hydrogen atom is given by the formula:</p>
<p>$ E_n = -\frac{13.6 \, \text{eV}}{n^2} $</p>
<p>where:</p>
<ul>
<li>$E_n$ is the energy in electron volts (eV),</li><br>
<li>$13.6 \, \text{eV}$ is the ionization energy of hydrogen (the energy required to... | mcq | jee-main-2024-online-1st-february-morning-shift | 9,095 |
jaoe38c1lsc3gmgr | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The radius of third stationary orbit of electron for Bohr's atom is R. The radius of fourth stationary orbit will be:</p> | [{"identifier": "A", "content": "$$\\frac{4}{3} \\mathrm{R}$$\n"}, {"identifier": "B", "content": "$$\\frac{16}{9} R$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{4} R$$\n"}, {"identifier": "D", "content": "$$\\frac{9}{16} \\mathrm{R}$$"}] | ["B"] | null | <p>The radius of the nth stationary orbit in the Bohr model of an atom is directly proportional to the square of its principal quantum number n and inversely proportional to the atomic number Z. This relationship is represented by the formula $$r_n = \frac{n^2h^2}{4\pi^2kme^2Z},$$ where :</p>
<ul>
<li>
<p>$n$ is the... | mcq | jee-main-2024-online-27th-january-morning-shift | 9,096 |
jaoe38c1lscq53oy | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>If Rydberg's constant is $$R$$, the longest wavelength of radiation in Paschen series will be $$\frac{\alpha}{7 R}$$, where $$\alpha=$$ ________.</p> | [] | null | 144 | <p>Longest wavelength corresponds to transition between $$\mathrm{n}=3$$ and $$\mathrm{n}=4$$</p>
<p>$$\begin{aligned}
& \frac{1}{\lambda}=\mathrm{RZ}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\mathrm{RZ}^2\left(\frac{1}{9}-\frac{1}{16}\right) \\
& =\frac{7 \mathrm{RZ}^2}{9 \times 16} \\
& \Rightarrow \lambda=\frac{144... | integer | jee-main-2024-online-27th-january-evening-shift | 9,097 |
jaoe38c1lse6f47l | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>If the wavelength of the first member of Lyman series of hydrogen is $$\lambda$$. The wavelength of the second member will be</p> | [{"identifier": "A", "content": "$$\\frac{27}{5} \\lambda$$\n"}, {"identifier": "B", "content": "$$\\frac{5}{27} \\lambda$$\n"}, {"identifier": "C", "content": "$$\\frac{27}{32} \\lambda$$\n"}, {"identifier": "D", "content": "$$\\frac{32}{27} \\lambda$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] ...... \text{(i)}\\
& \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] ...... \text{(ii)}
\end{aligned}$$</p>
<p>On dividing (i) & (ii)</p>
<p>$$... | mcq | jee-main-2024-online-31st-january-morning-shift | 9,098 |
1lsg5gkdv | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>An electron revolving in $$n^{\text {th }}$$ Bohr orbit has magnetic moment $$\mu_n$$. If $$\mu_n \propto n^x$$, the value of $$x$$ is</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <p>To determine the relationship between the magnetic moment and the principal quantum number $ n $, we need to understand the formula for the magnetic moment of an electron in a Bohr orbit.</p>
<p>The magnetic moment ($ \mu_n $) of an electron in the nth Bohr orbit is given by:</p>
<p>$$ \mu_n = \frac{n \cdot e}{2m}... | mcq | jee-main-2024-online-30th-january-evening-shift | 9,101 |
1lsgcx4ob | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5<sup>th</sup> excited state of a hydrogen atom is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$\\frac{1}{4}$$"}] | ["C"] | null | <p>$$\frac{1}{2}|P E|=K E$$ for each value of $$\mathrm{n}$$ (orbit)</p>
<p>$$\therefore \frac{K E}{|P E|}=\frac{1}{2}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 9,102 |
1lsgx9b3x | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A electron of hydrogen atom on an excited state is having energy $$\mathrm{E}_{\mathrm{n}}=-0.85 \mathrm{~eV}$$. The maximum number of allowed transitions to lower energy level is _________.</p> | [] | null | 6 | <p>$$\begin{aligned}
& E_n=-\frac{13.6}{n^2}=-0.85 \\
& \Rightarrow n=4
\end{aligned}$$</p>
<p>No of transition</p>
<p>$$=\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6$$</p> | integer | jee-main-2024-online-30th-january-morning-shift | 9,103 |
luxwco24 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A hydrogen atom in ground state is given an energy of $$10.2 \mathrm{~eV}$$. How many spectral lines will be emitted due to transition of electrons?</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <p>To determine how many spectral lines will be emitted due to transitions of electrons in a hydrogen atom when it is given an energy of $10.2 \, \text{eV}$, we first need to ascertain which energy level the electron will reach with this energy and then count the possible transitions (spectral lines) as it returns to t... | mcq | jee-main-2024-online-9th-april-evening-shift | 9,104 |
lv0vy072 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>A hydrogen atom changes its state from $$n=3$$ to $$n=2$$. Due to recoil, the percentage change in the wave length of emitted light is approximately $$1 \times 10^{-n}$$. The value of $$n$$ is _______.</p> <p>[Given Rhc $$=13.6 \mathrm{~eV}, \mathrm{hc}=1242 \mathrm{~eV} \mathrm{~nm}, \mathrm{h}=6.6 \times 10^{-34} ... | [] | null | 7 | <p>$$\begin{aligned}
\Delta E & =13.6 \mathrm{eV}\left(\frac{1}{4}-\frac{1}{9}\right) \\
& =\frac{68}{36} \mathrm{eV}=1.89 \mathrm{eV}
\end{aligned}$$</p>
<p>Due to recoil of hydrogen atom, the energy of emitted photon will decrease by very small amount.</p>
<p>So for approximate calculations,</p>
<p>$$\begin{aligned}
... | integer | jee-main-2024-online-4th-april-morning-shift | 9,105 |
lv2es2v5 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>According to Bohr's theory, the moment of momentum of an electron revolving in $$4^{\text {th }}$$ orbit of hydrogen atom is:</p> | [{"identifier": "A", "content": "$$2 \\frac{h}{\\pi}$$\n"}, {"identifier": "B", "content": "$$\\frac{h}{2 \\pi}$$\n"}, {"identifier": "C", "content": "$$\\frac{h}{\\pi}$$\n"}, {"identifier": "D", "content": "$$8 \\frac{h}{\\pi}$$"}] | ["A"] | null | <p>According to Bohr's theory, one of the postulates specifies that the angular momentum of an electron in orbit around a nucleus is quantized. This quantization can be expressed by the formula:</p>
<p>$$ L = n\frac{h}{2\pi} $$</p>
<p>Where:</p>
<ul>
<li>$L$ is the angular momentum of the electron,</li>
<li>$n$ is... | mcq | jee-main-2024-online-4th-april-evening-shift | 9,106 |
lv9s23rt | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The angular momentum of an electron in a hydrogen atom is proportional to : (Where $$\mathrm{r}$$ is the radius of orbit of electron)</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\mathrm{r}}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{\\sqrt{\\mathrm{r}}}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{\\mathrm{r}}$$"}, {"identifier": "D", "content": "r"}] | ["C"] | null | <p>$$\begin{aligned}
& L=m v r \propto n \\
& \therefore \quad m v r \propto \sqrt{r}
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift | 9,108 |
lv9s25mk | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $$915\mathop A\limits^o$$. The longest wavelength of spectral lines in the Balmer series will be _______ $$\mathop A\limits^o$$.</p> | [] | null | 6588 | <p>$$\frac{1}{915}=R_H\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)\quad$$ (For Lyman)</p>
<p>$$\Rightarrow \frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\quad$$ (For Balmer)</p>
<p>$$\Rightarrow \lambda=6588$$ $$\mathop A\limits^o $$</p> | integer | jee-main-2024-online-5th-april-evening-shift | 9,109 |
lvb29eed | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The longest wavelength associated with Paschen series is : (Given $$\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{SI}$$ unit)</p> | [{"identifier": "A", "content": "$$2.973 \\times 10^{-6} \\mathrm{~m}$$\n"}, {"identifier": "B", "content": "$$1.876 \\times 10^{-6} \\mathrm{~m}$$\n"}, {"identifier": "C", "content": "$$1.094 \\times 10^{-6} \\mathrm{~m}$$\n"}, {"identifier": "D", "content": "$$3.646 \\times 10^{-6} \\mathrm{~m}$$"}] | ["B"] | null | <p>To determine the longest wavelength associated with the Paschen series, we need to understand what the Paschen series is and how its wavelength can be calculated. The Paschen series pertains to the spectral line emissions of the hydrogen atom as an electron transitions from higher energy levels (n > 3) down to n = 3... | mcq | jee-main-2024-online-6th-april-evening-shift | 9,110 |
lvc587n8 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | <p>The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :</p> | [{"identifier": "A", "content": "$$1: 2$$\n"}, {"identifier": "B", "content": "$$1: 4$$\n"}, {"identifier": "C", "content": "$$2: 1$$\n"}, {"identifier": "D", "content": "$$4: 1$$"}] | ["D"] | null | <p>The wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom is given by the Rydberg formula:</p>
<p>$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$</p>
<p>where:</p>
<ul>
<li>$$\lambda$$ is the wavelength of the emitted light,</li>
<li>$$R$$ i... | mcq | jee-main-2024-online-6th-april-morning-shift | 9,111 |
YnV25A2VbItpU38U | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | In the nuclear fusion reaction
$$${}_1^2H + {}_1^3H \to {}_2^4He + n$$$
<br/>given that the repulsive potential energy between the two nuclei is $$ \sim 7.7 \times {10^{ - 14}}J$$, the temperature at which the gases must be heated to initiate the reaction is nearly
<br/>[ Boltzmann's Constant $$k = 1.38 \times {10^{ ... | [{"identifier": "A", "content": "$${10^7}\\,\\,K$$ "}, {"identifier": "B", "content": "$${10^5}\\,\\,K$$ "}, {"identifier": "C", "content": "$${10^3}\\,\\,K$$ "}, {"identifier": "D", "content": "$${10^9}\\,\\,K$$ "}] | ["D"] | null | The average kinetic energy per molecule $$ = {3 \over 2}kT$$
<br><br>This kinetic energy should be able to provide the repulsive potential energy
<br><br>$$\therefore$$ $${3 \over 2}kT = 7.7 \times {10^{ - 14}}$$
<br><br>$$ \Rightarrow T = {{2 \times 7.7 \times {{10}^{ - 14}}} \over {3 \times 1.38 \times {{10}^{ - 23}... | mcq | aieee-2003 | 9,113 |
1BXDWJYutoMAJLBe | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | The binding energy per nucleon of deuteron $$\left( {{}_1^2\,H} \right)$$ and helium nucleus $$\left( {{}_2^4\,He} \right)$$ is $$1.1$$ $$MeV$$ and $$7$$ $$MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is | [{"identifier": "A", "content": "$$23.6\\,\\,MeV$$ "}, {"identifier": "B", "content": "$$26.9\\,\\,MeV$$"}, {"identifier": "C", "content": "$$13.9\\,\\,MeV$$"}, {"identifier": "D", "content": "$$19.2\\,\\,MeV$$"}] | ["A"] | null | The nuclear reaction of process is $$2_1^2H \to {4 \over 2}$$ He
<br><br>Energy released $$ = 4 \times \left( 7 \right) - 4\left( {1.1} \right) = 23.6\,MeV$$ | mcq | aieee-2004 | 9,114 |
94qQQCVosZrezE7R | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | A nuclear transformation is denoted by $$X\left( {n,\alpha } \right)\matrix{
7 \cr
3 \cr
} Li.$$ Which of the following is the nucleus of element $$X$$ ? | [{"identifier": "A", "content": "$${}_5^{10}B$$ "}, {"identifier": "B", "content": "$${}^{12}{C_6}$$ "}, {"identifier": "C", "content": "$${}_4^{11}Be$$ "}, {"identifier": "D", "content": "$${}_5^9B$$ "}] | ["A"] | null | $${}_z{X^A} + {}_0{n^1} \to {}_3L{i^7} + {}_2H{e^4}$$
<br><br>On comparison,
<br><br>$$A = 7 + 4 - 1 = 10,\,\,\,z = 3 + 2 - 0 = 5$$
<br><br>It is boron $${}_5{B^{10}}$$ | mcq | aieee-2005 | 9,115 |
jWgpEmdreM2ZVJUe | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | When $${}_3L{i^7}$$ nuclei are bombarded by protons, and the resultant nuclei are $${}_4B{e^8}$$, the emitted particles will be | [{"identifier": "A", "content": "alpha particles "}, {"identifier": "B", "content": "beta particles "}, {"identifier": "C", "content": "gamma photons "}, {"identifier": "D", "content": "neutrons "}] | ["C"] | null | $${}_3^7Li + {}_1^1p \to {}_4^8Be + {}_0^0\gamma $$ | mcq | aieee-2006 | 9,116 |
lSthkiS5WymUs0th | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | If the binding energy per nucleon in $${}_3^7Li$$ and $${}_2^4He$$ nuclei are $$5.60$$ $$MeV$$ and $$7.06$$ $$MeV$$ respectively, then in the reaction
$$$p + {}_3^7Li \to 2\,{}_2^4He$$$
<br/>energy of proton must be | [{"identifier": "A", "content": "$$28.24$$ $$MeV$$ "}, {"identifier": "B", "content": "$$17.28$$ $$MeV$$ "}, {"identifier": "C", "content": "$$1.46$$ $$MeV$$ "}, {"identifier": "D", "content": "$$39.2$$ $$MeV$$ "}] | ["B"] | null | Let $$E$$ be the energy of proton, then
<br><br>$$E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right]$$
<br><br>$$ \Rightarrow E = 56.48 - 39.2 = 17.28MeV$$ | mcq | aieee-2006 | 9,117 |
N2Mpt56VPTA7TUtF | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | If $${M_O}$$ is the mass of an oxygen isotope $${}_8{O^{17}}$$ , $${M_p}$$ and $${M_N}$$ are the masses of a proton and neutron respectively, the nuclear binding energy of the isotope is | [{"identifier": "A", "content": "$$\\left( {{M_O} - 17{M_N}} \\right){C^2}$$ "}, {"identifier": "B", "content": "$$\\left( {{M_O} - 8{M_P}} \\right){C^2}$$ "}, {"identifier": "C", "content": "$$\\left( {{M_O} - 8{M_P} - 9{M_N}} \\right){C^2}$$ "}, {"identifier": "D", "content": "$${{M_O}{c^2}}$$ "}] | ["C"] | null | Binding energy
<br><br>$$ = \left[ {Z{M_p} + \left( {A - Z} \right){M_N} - M} \right]{c^2}$$
<br><br>$$ = \left[ {8{M_p} + \left( {17 - 8} \right){M_N} - M} \right]{c^2}$$
<br><br>$$ = \left[ {8{M_p} + 9{M_N} - M} \right]{c^2}$$
<br><br>$$ = \left[ {8{M_p} + 9{M_N} - {M_O}} \right]{c^2}$$ | mcq | aieee-2007 | 9,118 |
Hs6rDLDFbOs0tuUh | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | A nucleus of mass $$M+$$$$\Delta m$$ is at rest and decays into two daughter nuclei of equal mass $${M \over 2}$$ each. Speed of light is $$c.$$
<br/><br/><p>The binding energy per nucleon for the parent nucleus is $${E_1}$$ and that for the daughter nuclei is $${E_2}.$$ Then </p> | [{"identifier": "A", "content": "$${E_2} = 2{E_1}$$ "}, {"identifier": "B", "content": "$${E_1} > {E_2}$$ "}, {"identifier": "C", "content": "$${E_2} > {E_1}$$ "}, {"identifier": "D", "content": "$${E_1} = 2{E_2}$$ "}] | ["C"] | null | In nuclear fission, the binding energy per nucleon of daughter nuclei is greater than the parent nucleon of daughter nuclei is greater than the parent nucleus. | mcq | aieee-2010 | 9,121 |
bLLjw7KFn61oLENm | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | A nucleus of mass $$M+$$$$\Delta m$$ is at rest and decays into two daughter nuclei of equal mass $${M \over 2}$$ each. Speed of light is $$c.$$
<br/><br/><p>The speed of daughter nuclei is </p> | [{"identifier": "A", "content": "$$c{{\\Delta m} \\over {M + \\Delta m}}$$ "}, {"identifier": "B", "content": "$$c\\sqrt {{{2\\Delta m} \\over M}} $$ "}, {"identifier": "C", "content": "$$c\\sqrt {{{\\Delta m} \\over M}} $$ "}, {"identifier": "D", "content": "$$c\\sqrt {{{\\Delta m} \\over {M + \\Delta m}}} $$ "}] | ["B"] | null | By conservation of energy,
<br><br>$$\left( {M + \Delta m} \right){c^2} = {{2M} \over 2}{c^2} + {1 \over 2}.{{2M} \over 2}{v^2},$$
<br><br>where $$v$$ is the speed of the daughter nuclei
<br><br>$$ \Rightarrow \Delta m{c^2} = {M \over 2}{v^2}$$
<br><br>$$\therefore$$ $$v = c\sqrt {{{2\Delta m} \over M}} $$ | mcq | aieee-2010 | 9,122 |
e4iI1qyGqr7MvABX | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron $$ = 1.6725 \times {10^{ - 27}}kg,$$ mass of proton $$ = 1.6725 \times {10^{ - 27}}\,kg,$$ mass of electron $$ = 9 \times {10^{ - 31}}\,kg$$ ). | [{"identifier": "A", "content": "$$0.51$$ $$MeV$$ "}, {"identifier": "B", "content": "$$7.10\\,MeV$$ "}, {"identifier": "C", "content": "$$6.30\\,MeV$$"}, {"identifier": "D", "content": "$$5.4\\,MeV$$"}] | ["A"] | null | $${}_0^1n \to {}_1^1H + {}_{ - 1}{e^0} + \overrightarrow v + Q$$
<br><br>The mass defect during the process
<br><br>$$\Delta m = {m_n} - {m_H} - {m_e}$$
<br><br>$$ = 1.6725 \times {10^{ - 27}} - \left( {1.6725 \times {{10}^{ - 27}} + 9 \times {{10}^{ - 31}}kg} \right)$$
<br><br>$$ = - 9 \times {10^{ - 31}}kg$$
<br><b... | mcq | aieee-2012 | 9,123 |
ibbQ63XVYABiB0w04ZycX | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | Imagine that a reactor converts all given mass into energy and that it operates at a power level of 10<sup>9</sup> watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is
3×10<sup>8</sup> m/s)
| [{"identifier": "A", "content": "0.96 gm"}, {"identifier": "B", "content": "0.8 gm "}, {"identifier": "C", "content": "4 $$ \\times $$ 10<sup>$$-$$2</sup> gm"}, {"identifier": "D", "content": "6.6 $$ \\times $$ 10<sup>$$-$$5</sup> gm"}] | ["C"] | null | <p>The power can be calculated by the relation</p>
<p>$$P = {E \over {\Delta t}} = {{\Delta m{c^2}} \over {\Delta t}}$$ ...... (1)</p>
<p>Therefore, from Eq. (1), the mass of the fuel consumed per hour in the reactor is</p>
<p>$${{\Delta m} \over {\Delta t}} = {P \over {{c^2}}} = {{{{10}^9}} \over {{{(3 \times {{10}^8}... | mcq | jee-main-2017-online-9th-april-morning-slot | 9,124 |
E5VoaWfI1qQHQgOntLuRH | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | Consider the nuclear fission
<br/><br/>Ne<sup>20</sup> $$ \to $$ 2He<sup>4</sup> + C<sup>12</sup>
<br/><br/>Given that the binding energy/ nucleon of Ne<sup>20</sup>, He<sup>4</sup> and C<sup>12</sup> are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -
| [{"identifier": "A", "content": "8.3 MeV energy will be released"}, {"identifier": "B", "content": "energy of 11.9 MeV has to be supplied"}, {"identifier": "C", "content": "energy of 12.4 MeV will be supplied"}, {"identifier": "D", "content": "energy of 3.6 MeV will be released"}] | ["B"] | null | Ne<sup>20</sup>$$~ \to $$ 2He<sup>4</sup> + C<sup>12</sup>
<br><br>Q – value, E<sub>B</sub> = (BE)<sub>react</sub> $$-$$ (BE)<sub>product</sub>
<br><br>= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)
<br><br>= 9.72 MeV | mcq | jee-main-2019-online-10th-january-evening-slot | 9,126 |
38RYFv8N6PByPCAYlfjgy2xukfakyrna | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | Find the Binding energy per neucleon for $${}_{50}^{120}Sn$$. Mass of proton m<sub>p</sub>
= 1.00783 U, mass of neutron
m<sub>n</sub>
= 1.00867 U and mass of tin nucleus m<sub>Sn</sub> = 119.902199 U. (take 1U = 931 MeV) | [{"identifier": "A", "content": "9.0 MeV"}, {"identifier": "B", "content": "8.5 MeV"}, {"identifier": "C", "content": "8.0 MeV"}, {"identifier": "D", "content": "7.5 MeV\n"}] | ["B"] | null | $$B.E. = \Delta m{c^2}$$<br><br>$$ = \Delta m \times 931$$<br><br>$$\Delta m = \left( {50 \times 1.00783} \right) + \left( {70 \times 1.00867} \right) - \left\{ {119.902199} \right\}$$<br><br>$$ = \left\{ {120.9984 - 119.902199} \right\}\,U$$<br><br>$$ = 1.1238\,U$$<br><br>$$BE = 1.1238\, \times 931 = 1046.2578\,MeV$$<... | mcq | jee-main-2020-online-4th-september-evening-slot | 9,128 |
NnxEMtzu85vK9Mat50jgy2xukfrnqxbb | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | You are given that Mass of $${}_3^7Li$$ = 7.0160u,
<br/>Mass of $${}_2^4He$$ = 4.0026u
<br/>and Mass of $${}_1^1H$$ = 1.0079u.
<br/>When 20 g of $${}_3^7Li$$ is converted into $${}_2^4He$$ by proton capture, the energy liberated, (in kWh), is :
<br/>[Mass of nucleon = 1 GeV/c<sup>2</sup>]
| [{"identifier": "A", "content": "6.82 $$ \\times $$ 10<sup>5</sup>"}, {"identifier": "B", "content": "4.5 $$ \\times $$ 10<sup>5</sup>"}, {"identifier": "C", "content": "8 $$ \\times $$ 10<sup>6</sup>"}, {"identifier": "D", "content": "1.33 $$ \\times $$ 10<sup>6</sup>"}] | ["D"] | null | $${}_3^7Li$$ + $${}_1^1H$$ $$ \to $$ 2($${}_2^4He$$)
<br><br>$$\Delta $$m = $$\left[ {{M_{Li}} + {M_H}} \right] - 2\left[ {{M_{He}}} \right]$$
<br><br>= (7.0160 + 1.0079) - 2 $$ \times $$ 4.0003
<br><br>= 0.0187
<br><br>Energy released in 1 reaction = $$\Delta $$mc<sup>2</sup>
<br><br>In use of 7.016 u Li energy is $$\... | mcq | jee-main-2020-online-6th-september-morning-slot | 9,129 |
LcnA0r5yQ4XacLMF8o1krpna7dr | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | A nucleus of mass M emits $$\gamma$$ -ray photon of frequency 'v'. The loss of internal energy by the nucleus is :<br/><br/>[Take 'c' as the speed of electromagnetic wave] | [{"identifier": "A", "content": "hv"}, {"identifier": "B", "content": "$$hv\\left[ {1 + {{hv} \\over {2M{c^2}}}} \\right]$$"}, {"identifier": "C", "content": "$$hv\\left[ {1 - {{hv} \\over {2M{c^2}}}} \\right]$$"}, {"identifier": "D", "content": "0"}] | ["B"] | null | Energy of $$\gamma$$-ray, E<sub>$$\gamma$$</sub> = hv<br/><br/> and momentum of $$\gamma$$-ray, $${p_\gamma } = {h \over \lambda }$$ .... (i)<br/><br/>As, $${p_\gamma } = {{{E_\gamma }} \over c} = {{hv} \over c}$$ ..... (ii)<br/><br/>From Eqs. (i) and (ii), we get<br/><br/>$${p_\gamma } = {{hv} \over c} = {h \over \lam... | mcq | jee-main-2021-online-20th-july-morning-shift | 9,130 |
1l569n6dx | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>The Q-value of a nuclear reaction and kinetic energy of the projectile particle, K<sub>p</sub> are related as :</p> | [{"identifier": "A", "content": "Q = K<sub>p</sub>"}, {"identifier": "B", "content": "(K<sub>p</sub> + Q) < 0"}, {"identifier": "C", "content": "Q < K<sub>p</sub>"}, {"identifier": "D", "content": "(K<sub>p</sub> + Q) > 0"}] | ["D"] | null | <p>K<sub>p</sub> > 0</p>
<p>If Q is released $$\Rightarrow$$ Q > 0</p>
<p>$$\Rightarrow$$ K<sub>p</sub> + Q > 0</p>
<p>Even the particle has to be given kinetic energy greater than magnitude of Q to maintain momentum conservation.</p>
<p>$$\Rightarrow$$ K + Q > 0</p> | mcq | jee-main-2022-online-28th-june-morning-shift | 9,133 |
1l5c43tt2 | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments 'B' and 'C' of mass numbers 105 and 115. The binding energy of nucleons in 'B' and 'C' is 6.4 MeV per nucleon. The energy Q released per fission will be :</p> | [{"identifier": "A", "content": "0.8 MeV"}, {"identifier": "B", "content": "275 MeV"}, {"identifier": "C", "content": "220 MeV"}, {"identifier": "D", "content": "176 MeV"}] | ["D"] | null | <p><sup>220</sup>A $$\to$$ <sup>105</sup>B + <sup>115</sup>C</p>
<p>$$\Rightarrow$$ Q = [105 $$\times$$ 6.4 + 115 $$\times$$ 6.4] $$-$$ [220 $$\times$$ 5.6] MeV</p>
<p>$$\Rightarrow$$ Q = 176 MeV</p> | mcq | jee-main-2022-online-24th-june-morning-shift | 9,134 |
1l6i0yn6o | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>A nucleus of mass $$M$$ at rest splits into two parts having masses $$\frac{M^{\prime}}{3}$$ and $${{2M'} \over 3}(M' < M)$$. The ratio of de Broglie wavelength of two parts will be :</p> | [{"identifier": "A", "content": "1 : 2"}, {"identifier": "B", "content": "2 : 1"}, {"identifier": "C", "content": "1 : 1"}, {"identifier": "D", "content": "2 : 3"}] | ["C"] | null | <p>Linear momentum is conserved</p>
<p>so, $${p_{M'/3}} = {p_{2M'/3}}$$</p>
<p>so, $${{{\lambda _{M'/3}}} \over {{\lambda _{2M'/3}}}} = {1 \over 1}$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift | 9,135 |
1l6i3jcsl | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below :</p>
<p>$${ }_{1}^{2} X+{ }_{1}^{2} X={ }_{2}^{4} Y$$</p>
<p>The binding energies per nucleon for $$\frac{2}{1} X$$ and $${ }_{2}^{4} Y$$ are $$1.1 \,\mathrm{MeV}$$ and $$7.6 \,\mathrm{MeV}$$ respectively. The energy rele... | [] | null | 26 | <p>Energy released = Change in B.E.</p>
<p>(7.6 $$\times$$ 4) $$-$$ [4 $$\times$$ 1.1] = 26 MeV</p> | integer | jee-main-2022-online-26th-july-evening-shift | 9,136 |
1ldnz2rx6 | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>Nucleus A having $$Z=17$$ and equal number of protons and neutrons has $$1.2 ~\mathrm{MeV}$$ binding energy per nucleon.</p>
<p>Another nucleus $$\mathrm{B}$$ of $$Z=12$$ has total 26 nucleons and $$1.8 ~\mathrm{MeV}$$ binding energy per nucleons.</p>
<p>The difference of binding energy of $$\mathrm{B}$$ and $$\math... | [] | null | 6 | <b>For Nucleus A :</b>
<br/><br/>$$
\begin{aligned}
& \mathrm{Z}=17=\text { Number of protons } \\\\
& Given, Z = N \\\\
& \therefore N = 17 \\\\
& A=34=Z+N \\\\
& E_{b n}=1.2 \mathrm{MeV} \\\\
& \frac{\left(E_B\right)_1}{A}=1.2 \mathrm{MeV} \\\\
& \left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times \math... | integer | jee-main-2023-online-1st-february-evening-shift | 9,137 |
1ldogd6pr | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>The mass of proton, neutron and helium nucleus are respectively $$1.0073~u,1.0087~u$$ and $$4.0015~u$$. The binding energy of helium nucleus is :</p> | [{"identifier": "A", "content": "$$28.4~\\mathrm{MeV}$$"}, {"identifier": "B", "content": "$$56.8~\\mathrm{MeV}$$"}, {"identifier": "C", "content": "$$7.1~\\mathrm{MeV}$$"}, {"identifier": "D", "content": "$$14.2~\\mathrm{MeV}$$"}] | ["A"] | null | Mass defect $=2($ Mass of $p+$ mass of $n)-$ mass of $\mathrm{He}$ nucleus
<br/><br/>$$
\begin{aligned}
& \Delta m=0.0305 u \\\\
& \text { B.E }=931.5 \times \Delta m=931.5 \times 0.0305 \\\\
& =28.4 \mathrm{MeV}
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift | 9,138 |
1ldws038l | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>The energy released per fission of nucleus of $$^{240}$$X is 200 MeV. The energy released if all the atoms in 120g of pure $$^{240}$$X undergo fission is ____________ $$\times$$ 10$$^{25}$$ MeV.</p>
<p>(Given $$\mathrm{N_A=6\times10^{23}}$$)</p> | [] | null | 6 | $120 \mathrm{~g}$ of $^{240}X$ will have $\frac{1}{2}$ mole of $X$
<br/><br/>
Number of atom of $X=\frac{1}{2} \times N_{A}=3 \times 10^{23}$ atom
<br/><br/>
Energy released $=3 \times 10^{23} \times 200 ~ \mathrm{MeV}$
<br/><br/>
$$
=6 \times 10^{25} ~\mathrm{MeV}
$$ | integer | jee-main-2023-online-24th-january-evening-shift | 9,139 |
1lgrjybw0 | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>A common example of alpha decay is $${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2} \mathrm{He}^{4}+\mathrm{Q}$$</p>
<p>Given :</p>
<p>$${ }_{92}^{238} \mathrm{U}=238.05060 ~\mathrm{u}$$,</p>
<p>$${ }_{90}^{234} \mathrm{Th}=234.04360 ~\mathrm{u}$$,</p>
<p>$${ }_{2}^{4} \mathrm{He}=4.0026... | [] | null | 4 | To find the energy released during the alpha decay, we first need to calculate the mass difference between the reactants and the products.
<br/><br/>
Mass difference = Mass of Uranium-238 - (Mass of Thorium-234 + Mass of Helium-4)
<br/><br/>
Mass difference = $$238.05060 \mathrm{u} - (234.04360 \mathrm{u} + 4.00260 \ma... | integer | jee-main-2023-online-12th-april-morning-shift | 9,141 |
1lgsxnhso | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $$1: 2^{1 / 3}$$. Their respective speed have a ratio of $$n: 1$$. The value of $n$ is __________.</p> | [] | null | 2 | Let the masses of the two nuclear parts be $$m_1$$ and $$m_2$$, and their respective speeds be $$v_1$$ and $$v_2$$. According to the problem, the ratio of their nuclear sizes is $$1 : 2^{1/3}$$. Since the nuclear size is proportional to the cube root of the mass, we can write:
<br/><br/>
$$
\frac{m_1}{m_2} = \left(\fra... | integer | jee-main-2023-online-11th-april-evening-shift | 9,142 |
1lh022hep | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>For a nucleus $${ }_{\mathrm{A}}^{\mathrm{A}} \mathrm{X}$$ having mass number $$\mathrm{A}$$ and atomic number $$\mathrm{Z}$$</p>
<p>A. The surface energy per nucleon $$\left(b_{\mathrm{s}}\right)=-a_{1} A^{2 / 3}$$.</p>
<p>B. The Coulomb contribution to the binding energy $$\mathrm{b}_{\mathrm{c}}=-a_{2} \frac{Z(Z-... | [{"identifier": "A", "content": "C, D only"}, {"identifier": "B", "content": "B, C, E only"}, {"identifier": "C", "content": "B, C only"}, {"identifier": "D", "content": "A, B, C, D only"}] | ["A"] | null | <p>In the semi-empirical mass formula, the terms have the following forms:</p>
<ul>
<li>The volume term (binding energy) is proportional to the volume of the nucleus, and therefore to the number of nucleons $A$.</li><br/>
<li>The surface term takes into account that the nucleons at the surface of the nucleus have fewer... | mcq | jee-main-2023-online-8th-april-morning-shift | 9,143 |
1lh02hzv1 | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>A nucleus with mass number 242 and binding energy per nucleon as $$7.6~ \mathrm{MeV}$$ breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as $$8.1 ~\mathrm{MeV}$$, the total gain in binding energy is _________ $$\mathrm{MeV}$$.</p> | [] | null | 121 | <p>The total binding energy of a nucleus is the binding energy per nucleon multiplied by the number of nucleons (protons and neutrons), which is the mass number.</p>
<p>The initial total binding energy of the nucleus is $242 \times 7.6 \, \text{MeV}$.</p>
<p>After the break, each fragment has a total binding energy of ... | integer | jee-main-2023-online-8th-april-morning-shift | 9,144 |
jaoe38c1lsc4cekv | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>In a nuclear fission process, a high mass nuclide $$(A \approx 236)$$ with binding energy $$7.6 \mathrm{~MeV} /$$ Nucleon dissociated into middle mass nuclides $$(\mathrm{A} \approx 118)$$, having binding energy of $$8.6 \mathrm{~MeV} / \mathrm{Nucleon}$$. The energy released in the process would be ______ $$\mathrm... | [] | null | 236 | <p>To determine the energy released in a nuclear fission process, we use the difference in binding energy (BE) before and after the fission. The formula for energy released ($$Q$$ value) in the process is given by:</p><p>$$Q = (\text{Total BE of products}) - (\text{Total BE of reactants})$$</p><p>In this case, the reac... | integer | jee-main-2024-online-27th-january-morning-shift | 9,145 |
jaoe38c1lscp9q8q | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>The atomic mass of $${ }_6 \mathrm{C}^{12}$$ is $$12.000000 \mathrm{~u}$$ and that of $${ }_6 \mathrm{C}^{13}$$ is $$13.003354 \mathrm{~u}$$. The required energy to remove a neutron from $${ }_6 \mathrm{C}^{13}$$, if mass of neutron is $$1.008665 \mathrm{~u}$$, will be :</p> | [{"identifier": "A", "content": "62.5 MeV"}, {"identifier": "B", "content": "6.25 MeV"}, {"identifier": "C", "content": "4.95 MeV"}, {"identifier": "D", "content": "49.5 MeV"}] | ["C"] | null | <p>$$\begin{aligned}
& { }_6 \mathrm{C}^{13}+\text { Energy } \rightarrow{ }_6 \mathrm{C}^{12}+{ }_0 \mathrm{n}^1 \\
& \Delta \mathrm{m}=(12.000000+1.008665)-13.003354 \\
& =-0.00531 \mathrm{u} \\
& \therefore \text { Energy required }=0.00531 \times 931.5 \mathrm{~MeV} \\
& =4.95 \mathrm{~MeV}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift | 9,146 |
jaoe38c1lse6vfvg | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>The mass defect in a particular reaction is $$0.4 \mathrm{~g}$$. The amount of energy liberated is $$n \times 10^7 \mathrm{~kWh}$$, where $$n=$$ __________. (speed of light $$\left.=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$$</p> | [] | null | 1 | <p>$$\begin{aligned}
& \mathrm{E}=\Delta \mathrm{mc}^2 \\
& =0.4 \times 10^{-3} \times\left(3 \times 10^8\right)^2 \\
& =3600 \times 10^7 \mathrm{kWs} \\
& =\frac{3600 \times 10^7}{3600} \mathrm{kWh}=1 \times 10^7 \mathrm{kWh}
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-morning-shift | 9,147 |
luxwdqsr | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of $$2: 1$$. After disintegration they will move :</p> | [{"identifier": "A", "content": "in opposite directions with speed in the ratio of $$1: 2$$ respectively.\n"}, {"identifier": "B", "content": "in the same direction with same speed.\n"}, {"identifier": "C", "content": "in opposite directions with speed in the ratio of $$2: 1$$ respectively.\n"}, {"identifier": "D", "co... | ["A"] | null | <p>In nuclear disintegration, the conservation of momentum plays a crucial role in determining the motion of the resulting fragments. Since the original nucleus is at rest, its total initial momentum is zero. After the disintegration, the total momentum of the system must still be zero to conserve momentum.</p>
<p>Giv... | mcq | jee-main-2024-online-9th-april-evening-shift | 9,150 |
luxwem1b | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>The energy released in the fusion of $$2 \mathrm{~kg}$$ of hydrogen deep in the sun is $$E_H$$ and the energy released in the fission of $$2 \mathrm{~kg}$$ of $${ }^{235} \mathrm{U}$$ is $$E_U$$. The ratio $$\frac{E_H}{E_U}$$ is approximately:
(Consider the fusion reaction as $$4_1^1H+2 \mathrm{e}^{-} \rightarrow{ }... | [{"identifier": "A", "content": "7.62"}, {"identifier": "B", "content": "25.6"}, {"identifier": "C", "content": "9.13"}, {"identifier": "D", "content": "15.04"}] | ["A"] | null | <p>To determine the ratio $$\frac{E_H}{E_U}$$, let's first consider the energy released in the fusion of hydrogen and the energy released in the fission of $$^{235}U$$.</p>
<p>For the fusion reaction: </p>
<p>The given reaction is:</p>
<p>$$4_1^1H + 2 \mathrm{e}^{-} \rightarrow {}_2^4 \mathrm{He} + 2 \mathrm{\nu} + ... | mcq | jee-main-2024-online-9th-april-evening-shift | 9,151 |
luyita3e | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>The energy equivalent of $$1 \mathrm{~g}$$ of substance is :</p> | [{"identifier": "A", "content": "$$5.6 \\times 10^{26} \\mathrm{~MeV}$$\n"}, {"identifier": "B", "content": "$$5.6 \\times 10^{12} \\mathrm{~MeV}$$\n"}, {"identifier": "C", "content": "$$5.6 \\mathrm{~eV}$$\n"}, {"identifier": "D", "content": "$$11.2 \\times 10^{24} \\mathrm{~MeV}$$"}] | ["A"] | null | <p>To determine the energy equivalent of a mass, we use Einstein's mass-energy equivalence principle given by the equation:</p>
<p>$$E = mc^2$$</p>
<p>where:</p>
<p>- $$E$$ is the energy</p>
<p>- $$m$$ is the mass</p>
<p>- $$c$$ is the speed of light in a vacuum, which is approximately $$3 \times 10^8 \mathrm{~m/s... | mcq | jee-main-2024-online-9th-april-morning-shift | 9,152 |
luz2u45k | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>A star has $$100 \%$$ helium composition. It starts to convert three $${ }^4 \mathrm{He}$$ into one $${ }^{12} \mathrm{C}$$ via triple alpha process as $${ }^4 \mathrm{He}+{ }^4 \mathrm{He}+{ }^4 \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+\mathrm{Q}$$. The mass of the star is $$2.0 \times 10^{32} \mathrm{~kg}$$ and ... | [] | null | 5 | <p>To determine the rate of converting $${ }^4 \mathrm{He}$$ to $${ }^{12} \mathrm{C}$$, we need to calculate the energy released per reaction and use the given power production of the star to find the rate of reactions. The relevant nuclear reaction is:</p>
<p>
<p>$${ }^4 \mathrm{He} + { }^4 \mathrm{He} + { }^4 \mat... | integer | jee-main-2024-online-9th-april-morning-shift | 9,153 |
lv3veetz | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>If $$M_0$$ is the mass of isotope $${ }_5^{12} B, M_p$$ and $$M_n$$ are the masses of proton and neutron, then nuclear binding energy of isotope is:</p> | [{"identifier": "A", "content": "$$(5 M_p+7 M_n-M_o) C^2$$\n"}, {"identifier": "B", "content": "$$(M_o-5 M_p-7 M_n) C^2$$\n"}, {"identifier": "C", "content": "$$(M_o-5 M_p) C^2$$\n"}, {"identifier": "D", "content": "$$(M_0-12 M_n) C^2$$"}] | ["A"] | null | <p>To determine the nuclear binding energy of the isotope $$_5^{12}B$$, we need to consider the mass defect concept. The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.</p>
<p>Let's calculate the mass defect first. The isotop... | mcq | jee-main-2024-online-8th-april-evening-shift | 9,156 |
lv3vefuq | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>In a hypothetical fission reaction</p>
<p>$${ }_{92} X^{236} \rightarrow{ }_{56} \mathrm{Y}^{141}+{ }_{36} Z^{92}+3 R$$</p>
<p>The identity of emitted particles (R) is :</p> | [{"identifier": "A", "content": "Proton"}, {"identifier": "B", "content": "Neutron"}, {"identifier": "C", "content": "Electron"}, {"identifier": "D", "content": "$$\\gamma$$-radiations"}] | ["B"] | null | <p>In the given hypothetical fission reaction:</p>
<p>$$^{236}_{92} X \rightarrow \, ^{141}_{56} Y + \, ^{92}_{36} Z + 3 R$$</p>
<p>We need to determine the identity of particles denoted by $ R $. Let's use the conservation of charge and mass number (nucleon number) to identify $ R $.</p>
<p>First, for the conservat... | mcq | jee-main-2024-online-8th-april-evening-shift | 9,157 |
lv7v3wj2 | physics | atoms-and-nuclei | nuclear-fission-and-fusion-and-binding-energy | <p>If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is
________ $$\times 10^{-2} \mathrm{~MeV}$$. (Given $$1 \mathrm{u}=931 \mathrm{~MeV} / \mathrm{c}^2$$, atomic mass of helium $$=4.002603 \mathrm{u}$$)</p> | [] | null | 727 | <p>To find the energy released when three helium nuclei combine to form a carbon nucleus, we first need to understand that this process is essentially nuclear fusion, forming a heavier nucleus from lighter ones. The mass defect in this fusion process is the key to calculating the energy released, according to Einstein'... | integer | jee-main-2024-online-5th-april-morning-shift | 9,159 |
gbwbvYz5vmma2a3j | physics | atoms-and-nuclei | nucleus-and-radioactivity | If $${N_0}$$ is the original mass of the substance of half-life period $${t_{1/2}} = 5$$ years, then the amount of substance left after $$15$$ years is | [{"identifier": "A", "content": "$${N_0}/8$$ "}, {"identifier": "B", "content": "$${N_0}/16$$"}, {"identifier": "C", "content": "$${N_0}/2$$"}, {"identifier": "D", "content": "$${N_0}/4$$"}] | ["A"] | null | After every half-life, the mass of the substance reduces to half its initial value.
<br><br>$${N_0}\mathop \to \limits^{5\,years} \,\,{{{N_0}} \over 2}\mathop \to \limits^{5\,years} {{{N_0}} \over {{2^2}}}\mathop \to \limits^{5\,years} {{{N_0}} \over 8}$$ | mcq | aieee-2002 | 9,160 |
pfUEG7Y8Eq3AWTy2 | physics | atoms-and-nuclei | nucleus-and-radioactivity | At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit
<br/>$$\eqalign{
& \left( i \right)\,\,\,\,\,\,\,electrons\,\,\,\,\,\,\,\,\,\,\,\,\left( {ii} \right)\,\,\,\,\,\,\,protons \cr
& \left( {iii} \right)\,\,\,H{e^{2 + }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,... | [{"identifier": "A", "content": "$$i, ii, iii$$ "}, {"identifier": "B", "content": "$$i, ii, iii, iv$$ "}, {"identifier": "C", "content": "$$iv$$ "}, {"identifier": "D", "content": "$$ii, iii$$ "}] | ["A"] | null | Charged particles are deflected in magnetic field. | mcq | aieee-2002 | 9,161 |
tC5L2rf3jjTIWp2q | physics | atoms-and-nuclei | nucleus-and-radioactivity | Which of the following cannot be emitted by radioactive substances during their decay ? | [{"identifier": "A", "content": "Protons "}, {"identifier": "B", "content": "Neutrinoes "}, {"identifier": "C", "content": "Helium nuclei "}, {"identifier": "D", "content": "Electrons "}] | ["A"] | null | The radioactive substances emit $$\alpha $$ -particles (Helium nucleus), $$\beta $$ -particles (electrons) and neutrinoes. | mcq | aieee-2003 | 9,165 |
N9HRUTVrw1p0QpSP | physics | atoms-and-nuclei | nucleus-and-radioactivity | A nucleus disintegrated into two nuclear parts which have their velocities in the ratio of $$2:1.$$ The ratio of their nuclear sizes will be | [{"identifier": "A", "content": "$${3^{{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}:1$$ "}, {"identifier": "B", "content": "$$1:{2^{1/3}}$$ "}, {"identifier": "C", "content": "$${2^{1/3}}:1$$ "}, {"identifier": "D", "content": "$$1:{3^{{\\raise0.5ex\\hbox... | ["B"] | null | From conservation of momentum $${m_1}{v_1} = {m_2}{v_2}$$
<br><br>$$ \Rightarrow \left( {{{{m_1}} \over {{m_2}}}} \right) = \left( {{{{v_2}} \over {{v_1}}}} \right)\,\,$$ given $$\,\,{{{v_1}} \over {v{}_2}} = 2$$
<br><br>$$ \Rightarrow {{{m_1}} \over {{m_2}}} = {1 \over 2}$$
<br><br>$$ \Rightarrow {{r_1^3} \over {r_2^3... | mcq | aieee-2004 | 9,167 |
7Yok3u8SgkYgKY5q | physics | atoms-and-nuclei | nucleus-and-radioactivity | If radius of the $$\matrix{
{27} \cr
{13} \cr
} $$ $$Al$$ nucleus is estimated to be $$3.6$$ fermi then the radius of $$\matrix{
{125} \cr
{52} \cr
} \,Te$$ nucleus is estimated to be nearly | [{"identifier": "A", "content": "$$8$$ fermi "}, {"identifier": "B", "content": "$$6$$ fermi "}, {"identifier": "C", "content": "$$5$$ fermi "}, {"identifier": "D", "content": "$$4$$ fermi "}] | ["B"] | null | <b>KEY CONCEPT :</b> $$R = {R_0}{\left( A \right)^{1/3}}$$
<br><br>Here A = Mass number
<br><br>$$\therefore$$ $${{{R_1}} \over {{R_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^{1/3}}$$
<br><br>$$ = {\left( {{{27} \over {125}}} \right)^{1/3}} = {3 \over 5}$$
<br><br>$${R_2} = {5 \over 3} \times 3.6 = 6$$ fermi | mcq | aieee-2005 | 9,168 |
gFbeq2tekhrcrxhu | physics | atoms-and-nuclei | nucleus-and-radioactivity | Starting with a sample of pure $${}^{66}Cu,{7 \over 8}$$ of it decays into $$Zn$$ in $$15$$ minutes. The corresponding half life is | [{"identifier": "A", "content": "$$15$$ minutes "}, {"identifier": "B", "content": "$$10$$ minutes "}, {"identifier": "C", "content": "$$7{1 \\over 2}$$ minutes "}, {"identifier": "D", "content": "$$5$$ minutes"}] | ["D"] | null | $${7 \over 8}$$ of $$Cu$$ decays in $$15$$ minutes.
<br><br>$$\therefore$$ $$Cu$$ undecayed $$ = N = 1 - {7 \over 8} = {1 \over 8} = {\left( {{1 \over 2}} \right)^3}$$
<br><br>$$\therefore$$ No. of half lifes $$=3$$
<br><br>$$n = {t \over T}$$ or $$3 = {{15} \over T}$$
<br><br>$$ \Rightarrow T = $$ half life period $$... | mcq | aieee-2005 | 9,169 |
WOUKKCA8AnBiof9A | physics | atoms-and-nuclei | nucleus-and-radioactivity | The intensity of gamma radiation from a given source is $$L$$. On passing through $$36$$ $$mm$$ of lead, it is reduced to $${{\rm I} \over 8}.$$ The thickness of lead which will reduce the intensity to $${{\rm I} \over 2}$$ will be | [{"identifier": "A", "content": "$$9mm$$ "}, {"identifier": "B", "content": "$$6mm$$ "}, {"identifier": "C", "content": "$$12mm$$ "}, {"identifier": "D", "content": "$$18mm$$ "}] | ["C"] | null | <b>KEY CONCEPT :</b> Intensity $$I = {I_0}.{e^{ - \mu d}},$$
<br><br>Applying logarithm on both sides,
<br><br>$$ - \mu d = \log \left( {{I \over {{I_0}}}} \right)$$
<br><br>$$ - \mu \times 36 = \log \left( {{{I/8} \over I}} \right).........\left( i \right)$$
<br><br>$$ - \mu \times d = \log \left( {{{I/2} \over I}} ... | mcq | aieee-2005 | 9,170 |
rRMlAiKhdCSM3BaD | physics | atoms-and-nuclei | nucleus-and-radioactivity | The $$'rad'$$ is the correct unit used to report the measurement of | [{"identifier": "A", "content": "the ability of a beam of gamma ray photons to produce ions in a target "}, {"identifier": "B", "content": "the energy delivered by radiation to a target "}, {"identifier": "C", "content": "the biological effect of radiation "}, {"identifier": "D", "content": "the rate of decay of radioa... | ["C"] | null | The risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue. Absorbed dose is expressed in rad. A rad is equal to $$100$$ $$ergs$$ of energy absorbed by $$1$$ gram of tissue. The more modern, internationally adopted unit is the gray (... | mcq | aieee-2006 | 9,171 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.