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Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.8 | NUM | Calculate the probability of tunneling in indium arsenide (InAs) if the applied electric field $(E)$ is $3 \times 10^{5} \mathrm{~V} / \mathrm{cm}$. Assume that the $\Pi_{\mathrm{c}}^{2}$ for electrons in InAs is $0.02 \Pi_{0}^{2}$. The band gap of InAs is 0.4 eV. | From Equation ($T \approx \exp \left(-\frac{4 \sqrt{2 m_{\mathrm{e}}^*} E_{\mathrm{g}}^{3 / 2}}{21 q \hbar E}\right)$), the probability of tunneling is given by
\[
T = \exp \left(
\frac{
-4 \sqrt{2 \left(0.02 \times 9.1 \times 10^{-31}~\mathrm{kg} \right)}
\left( 0.4~\mathrm{eV} \times 1.6 \times 10^{-19}~\mathrm{J/... | 2.82E-04 | * Threshold value of $10^{-6}$ for determining the importance of Zener tunneling. | single | medium | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.8 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.9 | NUM | Consider a voltage regulator supply connected to a $200 \Omega$ current-limiting resistance $\left(R_{\mathrm{L}}\right)$ and a reverse-biased Zener diode. The knee current ( $I_{\mathrm{ZK}}$) is 0.2 mA , and the maximum current ( $I_{\mathrm{ZM}}$ ) is 100 mA . Assume that the voltage across the Zener diode is always... | The job of this circuit (see Figure) is to provide a 10 V constant output even though the input voltage of the variable power supply will change. We need to find the range of voltages that can be controlled using this Zener diode and the $200 \Omega$ resistor.
If the applied voltage is too low, the Zener diode will no... | (10.04, 30) | (V, V) | No supplementary information needed. | multiple | hard | Materials: Semiconductors | Semiconductors | Electrical | Composites | Plastic | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.9 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 6.1 | NUM | A Schottky diode made from a Si and tungsten (W) junction has a saturation current density $(J_{\mathrm{S}})$ of $10^{-11} \mathrm{~A} / \mathrm{cm}^{2}$. (a) What is the current density if the applied forward bias $(V)$ is 0.3 V ? (b) If the cross-sectional area of this diode is $5 \times 10^{-4} \mathrm{~cm}^{2}$, wh... | 1. The current density under a forward bias of 0.3 V is calculated by assuming the exponential term is much larger than 1, which is valid for $V > \left(k_{\mathrm{B}} T / q\right)$. Therefore, the equation simplifies to:
$$
J=J_{S}\left[\exp \left(\frac{V}{\left(\frac{k_{\mathrm{B}} T}{q}\right)}\right)\right]=10^{-1... | (10.25, 5.12) | (A/cm^2, mA) | No supplementary information needed. | multiple | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 6.1 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 6.2 | NUM | An LED is driven by a maximum voltage supply of 8 V. For the driving voltage across an LED of $1.8-2.0 \mathrm{~V}$, what should be the series resistance $\left(R_{\mathrm{S}}\right)$? Assume that the LED current is 16 mA. | In this example, the maximum voltage for the circuit is 8 V. This voltage appears between the resistor $\left(R_{\mathrm{S}}\right)$ and the LED. We assume that the resistance of the LED itself is small $(\sim 5 \Omega)$ and can be ignored. Supplied voltage is divided to LED and series resistance. Therefore, forward cu... | 387.5 | Ω | No supplementary information needed. | single | medium | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 6.2 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 6.3 | FORMULA | Give the common emitter current gain $(\beta)$ in term of the common base current gain $(\alpha)$
| The emitter current $(I_{\mathrm{E}})$ is equal to the sum of collector and base currents:
$$
I_{\mathrm{E}}=I_{\mathrm{C}}+I_{\mathrm{B}}
$$
Divide both sides of this equation by $I_{\mathrm{C}}$ :
$$
I_{\mathrm{E}} / I_{\mathrm{C}}=1+\left(I_{\mathrm{B}} / I_{\mathrm{C}}\right)
$$
The common base current gain $\a... | $\beta=\frac{\alpha}{1-\alpha}$ | No supplementary information needed. | single | hard | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 6.3 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.2 | NUM | 1. Calculate the capacitance of a parallel-plate single-layer capacitor made using a $\mathrm{BaTiO}_{3}$ formulation of $k=2000$. Assume $d=3 \mathrm{~mm}$ and $A=5 \mathrm{~mm}^{2}$. What is the volumetric efficiency of this capacitor? 2. Calculate the capacitance of an MLC comprised of 60 dielectric layers connected... | 1. From Equation,
$$
\begin{aligned}
& C=\varepsilon_{0} \times k \times \frac{A}{d} \\
& C=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m} \times 2000 \times \frac{5 \times 10^{-6} \mathrm{~m}^{2}}{3 \times 10^{-3} \mathrm{~m}}
\end{aligned}
$$
Thus, the capacitance of a single-layer capacitor with a thickness of 3 mm... | (1.96 \times 10^{-3},7.080, 3600) | (nF/mm^3, nF/mm^3, ) | No supplementary information needed. | multiple | easy | Materials: Metals | Metals | Electrical | Composites | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.2 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.3 | NUM | IC fabrication technology uses silica $(\mathrm{SiO}_{2} / \mathrm{SiO}_{x})(k \sim 4)$ as a gate dielectric. The formula $\mathrm{SiO}_{x}$ indicates that the stoichiometry of the compound is not exactly known. The thickness of this $\mathrm{SiO}_{2}$ gate dielectric in state-of-the-art transistors is $\sim 2 \mathrm{... | 1. The capacitance per unit area of a $2-\mathrm{nm}$ film of $\mathrm{SiO}_{2}$ is given by $$\frac{C}{A}=\frac{\varepsilon_{0} \times k}{d}$$ $$\frac{C}{A}=\frac{8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m} \times 4}{2 \times 10^{-9} \mathrm{~m}}=1.77 \times 10^{-2} \mathrm{~F} / \mathrm{m}^{2}$$ Because $1 \mathrm{... | (17.7, 22.12) | (fF/μm², fF/μm²) | * An interfacial layer of SiO₂ forms between the HfO₂ and Si.
* The thickness of the interfacial SiO₂ layer is approximately 2 nm. | multiple | medium | Materials: Semiconductors | Semiconductors | Electrical | Composites | Finishing | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.3 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.4 | NUM | The atomic radius of an Ar atom is $1.15 \AA$.
1. What is the electronic polarizability $\left(\alpha_{\mathrm{e}}\right)$ of an Ar atom in units of $\mathrm{F} \cdot \mathrm{m}^{2}$ ?
2. What is the volume polarizability of an Ar atom in $\AA^{3}$ ?
3. What is the volume polarizability of an Ar atom in $\mathrm{cm}^{... | 1. The electronic polarizability $\left(\alpha_{e}\right)$ of Ar atoms is calculated from Equation ($\alpha_{\mathrm{e}}=4 \pi \varepsilon_0 R^3=3 \varepsilon_0 V_{\mathrm{a}}$) as follows:
$$\alpha_{\mathrm{e}}=4 \pi \varepsilon_{0} R^{3}=\left(4 \times \pi \times 8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)\... | (1.69 \times 10^{-40}, 1.52, 1.52 \times 10^{-24}, 1.32 \times 10^{-16}, 870284) | (F·m², ų, cm³, m, ) | * The value $Z=8$ used in the calculation of the displacement ($\delta$) in the formula $\delta=\frac{4 \pi \varepsilon_0 R^3 E}{Z q}$. | multiple | easy | Properties: Electrical | Electrical | Composites | Atomic Bonding | Corrosion | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.4 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.5 | NUM | If the natural frequency of oscillation $(\omega_{0})$ of the electron mass around the nucleus in a hydrogen $(\mathrm{H})$ atom is $4.5 \times 10^{16} \mathrm{rad} / \mathrm{s}$, what is the static electronic polarizability $(\alpha_{\mathrm{e}})$ of the H atom $(R=1.2 \AA)$ calculated by Bohr's model and the classica... | For Bohr's model,
$$
\alpha_{\mathrm{e}}=\frac{\left(1 \times 1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right)\left(4.5 \times 10^{16} \mathrm{rad} / \mathrm{s}\right)^{2}}=1.387 \times 10^{-41} \mathrm{~F} \cdot \mathrm{~m}^{2}
$$
The value can also be calculated using the cl... | (1.387 \times 10^{-41}, 1.92 \times 10^{-40}) | (F·m², F·m²) | No supplementary information needed. | multiple | easy | Materials: Semiconductors | Semiconductors | Thermal | Composites | Atomic Bonding | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.5 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.6 | NUM | Use the ion polarizabilities in Figure to estimate the dielectric constant of lithium silicate ($\mathrm{Li}_{2} \mathrm{SiO}_{3}$). The molar volume ($V_{\mathrm{m}}$) of $\mathrm{Li}_{2} \mathrm{SiO}_{3}$ is $59.01 \AA^{3}$. The experimental value of the dielectric constant ($\varepsilon_{\mathrm{r}, \text{exp}}$) fo... | From Figure, the ionic polarizability of $\mathrm{O}^{2-}$ ions is $2.01 \AA^{3}$. The ionic polarizabilities of lithium ($\mathrm{Li}^{+}$) and silicon ($\mathrm{Si}^{4+}$) ions are $1.20 \AA^{3}$ and $0.87 \AA^{3}$, respectively. Thus, the dielectric polarizability ($\alpha_{D}^{\top}$) of $\mathrm{Li}_{2} \mathrm{Si... | 6.82 | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Atomic Bonding | Corrosion | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.6 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.7 | NUM | The dielectric constant of MgO is 9.83, and its molar volume $(V_{\mathrm{m}})$ is $18.69 \AA^{3}$. The dielectric constant of $\mathrm{Al}_{2} \mathrm{O}_{3}$ is estimated to be 10.126, and its molar volume $(V_{\mathrm{m}})$ is $42.45 \AA^{3}$. What is the dielectric constant of $\mathrm{MgAl}_{2} \mathrm{O}_{4}$? Th... | We first calculate the total dielectric polarizability of MgO from its dielectric constant and volume using the modified form of the Clausius-Mossotti equation ($\alpha_{\mathrm{D}}^{\mathrm{T}}=\frac{3 V_{\mathrm{m}}}{4 \pi}\left(\frac{\varepsilon_{\mathrm{r}}-1}{\varepsilon_{\mathrm{r}}+2}\right)$) that Shannon used:... | 7.923 | No supplementary information needed. | single | medium | Materials: Metals | Metals | Electrical | Composites | Atomic Bonding | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.7 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.8 | NUM | The refractive index ($n$) of a silicate glass is 1.5. What is the high-frequency dielectric constant ($k_{\infty}$) of this glass? The dielectric constant of this glass at 1 kHz is $k_{\mathrm{S}}=7.6$. Based on the relaxation of the dielectric constant, approximately what fraction of the low-frequency dielectric cons... | Given the refractive index ($n$) of the silicate glass is 1.5, the high-frequency dielectric constant ($k_{\infty}$) can be calculated using the relationship $k_{\infty}=n^{2}=2.25$. The low-frequency dielectric constant is given as $k_{\mathrm{S}}=7.6$. The contribution of ionic polarization to the dielectric constant... | 0.704 | No supplementary information needed. | single | easy | Materials: Glasses | Glasses | Optical | Composites | Corrosion | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.8 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.9 | NUM | The static dielectric constant $(\varepsilon_{\mathrm{s}})$ of $\mathrm{H}_{2} \mathrm{O}$ is 78.4 at 298.15 K (Fernandez et al. 1995). It decreases to about $\varepsilon_{\mathrm{r}}(\omega)=$ 20 at a frequency of $f=40 \mathrm{GHz}$. If the high-frequency (optical) dielectric constant $(\varepsilon_{\infty})$ is 5, w... | We first convert the frequency $(f)$ into angular frequency $(\omega)$ by using
$$
\omega=2 \pi f
$$
Therefore, $\omega=2 \pi\left(40 \times 10^{9} \mathrm{~Hz}\right)=2.51327 \times 10^{11} \mathrm{rad} / \mathrm{s}$.
We assume that relaxation in the real part of the dielectric constant $(k^{\prime})$ of water follo... | 7.85 | picoseconds | No supplementary information needed. | single | easy | Materials: Metals | Metals | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.9 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.10 | NUM | What is the magnitude of the capacitive reactance $\left(X_{\mathrm{C}}\right)$ of a $200-\mathrm{pF}$ capacitor at a frequency of 10 MHz? | The magnitude of the capacitive reactance $\left(X_{\mathrm{C}}\right)$ for a capacitor in a parallel arrangement with a resistor, as described, is given by the formula:
$$
X_{\mathrm{C}}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}
$$
Given a capacitor of $200-\mathrm{pF}$ (where $1 \mathrm{pF} = 10^{-12} \mathrm{F}$) at ... | 79.5 | $\Omega$ | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.10 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 7.11 | NUM | A $10-\mathrm{V}$ variable-frequency AC supply is connected to a $200-\mu \mathrm{F}$ capacitor. What is the value of capacitive reactance for frequencies of (a) 0 Hz, (b) 60 Hz, and (c) 1 kHz? What is the value of the current (I) flowing through the circuit for each frequency? Give your answer as a tuple (capacitive r... | 1. When the frequency $(f)$ is zero, $X_{\mathrm{C}}=\infty$, and the value of current flowing through the circuit is zero. In other words, when a DC voltage is applied to the capacitor, the dielectric material just blocks this voltage.
2. When $f=60 \mathrm{~Hz}$, the value of the capacitive reactance $\left(X_{\mathr... | (\infty, 13.26, 0.796, 0, 0.754, 12.56) | (\Omega, \Omega, \Omega, A, A, A) | No supplementary information needed. | multiple | easy | Materials: Metals | Metals | Electrical | Composites | Plastic | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 7.11 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 8.1 | NUM | Both sides of an alumina plate are coated with a metal electrode. The surface area of the electrode is $5 \mathrm{~mm}^{2}$ and the relative dielectric constant and dielectric strength of the alumina are 9.1 and 17 volt $/ \mathrm{mm}$, respectively. What is the maximum amount of electric charge that can be stored on t... | 1. In Gauss's law (or the first Maxwell's equation),
$$\boldsymbol{\nabla} \cdot \boldsymbol{D}=\rho \text { and } \boldsymbol{D}=\varepsilon \boldsymbol{E}$$
Since the electric field equals the gradient of electric potential,
$$\boldsymbol{E}=-\nabla V$$
By combining these two equations,
$$\begin{gathered}
-\nabl... | 7.23E-12 | F | * The permittivity of alumina used in the solution is $9.62 \times \varepsilon_0$, which implies a relative dielectric constant of 9.62, whereas the problem description states the relative dielectric constant is 9.1. The value 9.62 is used in the calculation. | single | easy | Materials: Metals | Metals | Electrical | Structural Gradient | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 8.1 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 8.3 | NUM | Relative permittivity of water at $10^{8} \mathrm{~Hz}$ and $10^{11} \mathrm{~Hz}$ is $\sim 90$ and $\sim 6$, respectively. Estimate the relative contribution of dipolar polarization, ionic polarization, and electronic polarization to the permittivity of water. The refractive index of water is 1.33 at room temperature.... | 1. Since the valence electrons form bonds among atoms, covalent materials do not exhibit ionic polarization and dipolar polarization. Dielectric properties of strong covalent materials are determined by electronic polarization. Until the frequency of the incident wave reaches the natural resonance frequency of electron... | (84, 4.2, 1.8) | No supplementary information needed. | multiple | easy | Materials: Metals | Metals | Electrical | Composites | Atomic Bonding | Corrosion | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 8.3 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 8.4 | NUM | Estimate the plasma frequency \(\left(v_{p}\right)\) of lithium metal using the atomic density of lithium metal.
| The plasma frequency \(\left(v_{p}\right)\) of lithium metal can be estimated using the atomic density of lithium metal. The permittivity of the dielectric is given by
\[\varepsilon_{\mathrm{r}}=1+\frac{N_{\mathrm{d}} \alpha}{\varepsilon_{0}}=1+\frac{N_{\mathrm{d}} q x}{\varepsilon_{0} E}\]
where \(N_{\mathrm{d}}\) i... | 1.22E+16 | Hz | * Atomic density of lithium metal | single | medium | Materials: Metals | Metals | Electrical | Composites | Atomic Bonding | Corrosion | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 8.4 | ||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 8.5 | NUM | Green light with a wavelength of 530 nm travels in a waveguide. The refractive index of the core is 1.45 and that of the cladding is 1.43. What is the critical angle for total internal reflection at the core-cladding interface? | The critical angle $(\theta_{\mathrm{C}})$ for total internal reflection is determined by the condition:
$$
\sin \theta_{\mathrm{c}} = \frac{n_{\mathrm{f}}}{n_{\mathrm{i}}}
$$
Given that $n_{\mathrm{i}} = 1.45$ (core) and $n_{\mathrm{f}} = 1.43$ (cladding), the critical angle is calculated as:
$$
\theta_{\mathrm{c}}... | 80.5 | ${}^{\circ}$ | No supplementary information needed. | single | easy | Properties: Optical | Optical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 8.5 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 8.6 | NUM | 1. Sunlight is incident through a glass window. If there is no antireflection layer on the glass surface, how much radiation energy is transmitted through the glass for blue ray $(\lambda=450 \mathrm{~nm}, n=1.47)$ and infrared ray $(\lambda=3 \mu \mathrm{~m}, n=1.42)$? Assume that absorption is negligible. Give your a... | 1. When light is incident from air to the medium, the reflectance $(R)$ is given by $R=\left(\frac{\bar{n}-1}{\bar{n}+1}\right)^{2}=\frac{(n-1)^{2}+\kappa^{2}}{(n+1)^{2}+\kappa^{2}}$, where $\bar{n}=n+i\kappa$ is the complex refractive index. For light of $\lambda=450 \mathrm{~nm}$ and $3 \mu \mathrm{~m}$, with negligi... | (96.4, 97.0) | (%, %) | No supplementary information needed. | multiple | hard | Materials: Glasses | Glasses | Optical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 8.6 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 8.7 | NUM | If the AC conductivity ($\sigma$) and the real refractive index ($n$) of Cu are $3.1 \times 10^{2}(\Omega \cdot \mathrm{cm})^{-1}$ and 0.38 at $\lambda=1 \mu \mathrm{m}$, respectively, what are the extinction coefficient ($k$) and absorption coefficient ($\alpha$)? Give your answer as a tuple (k, $\alpha$). | 1. The extinction coefficient ($k$) and absorption coefficient ($\alpha$) can be calculated using the relations:
$$
\begin{gathered}
k=\frac{\sigma}{4 \pi \varepsilon_{0} n v} \\
\alpha=\frac{4 \pi v}{c} \frac{\sigma}{4 \pi \varepsilon_{0} n v}=\frac{\sigma}{\varepsilon_{0} c n}
\end{gathered}
$$
Given the AC conductiv... | (2.45, 3.1e7) | (,m^{-1}) | No supplementary information needed. | multiple | easy | Materials: Semiconductors | Semiconductors | Optical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 8.7 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 9.1 | NUM | Let us say that you want to install solar panels for your house in City A where the yearly solar radiation is 1,600 $\mathrm{kWh} / \mathrm{m}^{2}$. Each panel costs $\$ 550$, and the efficiency and active area of each panel is $16 \%$ and $2.5 \mathrm{~m}^{2}$.
1. In 2013, the average annual electricity consumption f... | 1. Total active area of solar cells for yearly production of $10,908 \mathrm{kWh}$ is
$$
10,908(\mathrm{kWh}) \div\left[0.16 \times 1600\left(\mathrm{kWh} / \mathrm{m}^{2}\right)\right]=42.9\left(\mathrm{~m}^{2}\right)
$$
Therefore, you need $\sim 17$ panels with an active area of $2.5 \mathrm{~m}^{2}$ to be self-suf... | (17, 11.458) | (m^2, year) | No supplementary information needed. | multiple | medium | Properties: Optical | Optical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 9.1 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 9.3 | NUM | You are trying to connect an external load to a solar cell. This solar cell exhibits the current density $(J)$ and power $(P)$ versus voltage relations in Figure. What is the resistance of the external load that allows you to pull out a maximum power from this solar cell? Active area (the area of the p-n junction semic... | In Figure, the output power $(P)$ is calculated by multiplying current and voltage at each point of the $J-V$ curve. A relation between $P$ and $V$ shows that the maximum power of the solar cell is found at $V_{\mathrm{m}} \sim 0.55$ volt and $J_{\mathrm{m}} \sim 5.5$ $\mathrm{mA} / \mathrm{cm}^{2}$. Since the active a... | 1 | Ω | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 9.3 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 9.4 | NUM | An indium gallium arsenide, (In,Ga)As, photodiode is irradiated by IR light with the wavelength $1 \mu \mathrm{m}$ and the power density $1.5 \mathrm{mW} / \mathrm{cm}^{2}$. Assume that the incident photon-to-current efficiency (IPCE or external quantum efficiency) of the photodiode is $90\%$ over IR light, and the lig... | Since the wavelength $\lambda$ is $1 \mu \mathrm{m}$, the energy of each photon $\left(E_{\mathrm{ph}}=h c / \lambda\right)$ is 1.24 eV. Also, the total incident power over the area of $2.5 \mathrm{cm}^{2}$ is $1.5 \mathrm{mW} / \mathrm{cm}^{2} \times 2.5 \mathrm{cm}^{2} = 3.75 \mathrm{mW}$. Therefore, the number of in... | 2.72 | mA | No supplementary information needed. | single | medium | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 9.4 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 10.3 | NUM | Question: A poled PZT type I piezoelectric ceramic disk (length $l_{0}=2 \mathrm{~mm}$ and poled in Direction 3) of a certain composition has $d_{33}=289 \mathrm{pC} / \mathrm{N}$, and its dielectric constant is 1300.
1. What is the value of the piezoelectric voltage constant or the $g_{33}$ coefficient?
2. What is th... | 1. The $g_{33}$ coefficient is calculated using the formula:
$$
g_{33}=\frac{d_{33}}{\varepsilon_{33}^{X}}
$$
Given $d_{33}=289 \mathrm{pC} / \mathrm{N}$ and $\varepsilon_{\mathrm{r} 33}^{X}=1300$, we first find $\varepsilon_{33}^{X}$:
$$
\varepsilon_{\mathrm{r} 33}^{X}=1300=\frac{\varepsilon}{8.85 \times 10^{-12} \... | (25.1 \times 10^{-3}, 5000, 7.225 \times 10^{-5}, 0.145) | (V·m/N, V, , µm) | No supplementary information needed. | multiple | easy | Materials: Ceramics | Ceramics | Electrical | Composites | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 10.3 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 10.4 | NUM | A poled PZT ceramic plate, 50 mm long, 5 mm wide, and 2 mm thick, is used as a micropositioner device. Assume that the dielectric constant $\left(\varepsilon_{\mathrm{r} 33}^{\mathrm{X}}\right)$ is 1200 and $g_{31}=10.5 \times 10^{-3} \mathrm{~V} \cdot \mathrm{~m} / \mathrm{N}$. 1. What will be the value of $d_{31}$? 2... | 1. We start with the relationship between $d_{31}$ and $g_{31}$ from Equation,
$$\frac{d_{31}}{g_{31}}=\varepsilon_{33}^{X}$$
Therefore,
$$d_{31}=g_{31} \times \varepsilon_{33}^{X}=10.5 \times 10^{-3} \mathrm{~V} \cdot \mathrm{~m} / \mathrm{N} \times 1200 \times 8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$$
Thus,... | ($139.9 \times 10^{-12}$, $0.348 \times 10^{-6}$) | (m/V, m) | No supplementary information needed. | multiple | easy | Materials: Ceramics | Ceramics | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 10.4 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 10.5 | NUM | A commercially available poled PVDF film, electroded using silver metallic ink, is $110 \mu \mathrm{~m}$ thick. A stress of $20,000 \mathrm{~Pa}$ is applied to this film over a 1 - in. ${ }^2$ area.
1. Assuming that the film has a rigid backing and deforms only in the thickness direction, what will be the open-circuit ... | 1. The stress is acting on a $1-\mathrm{in} .^2$ area, and because the film has a rigid backing, it can deform only along the thickness (Direction 3). Thus, we must use the $g_{33}$ coefficient.
From Equation ($E=-g \times X$), the electric field
$$
(E) \text { generated }=-\left(350 \times 10^{-3}\right) \frac{\mathr... | (0.77, 112.2) | (V, V) | * The length dimension of the cross-section subjected to force in part 2 (used as 2.54 cm or 1 inch). | multiple | easy | Materials: Metals | Metals | Mechanical | Composites | Diffusion & Kinetics | Finishing | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 10.5 | ||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 10.6 | NUM | A poled PZT cylinder with a diameter of 2 mm and a thickness of 5 mm is to be used as a spark igniter. What voltage is generated by applying a compressive force of 100 N to the circular cross section of this disk? Assume that the $d_{33}$ for this PZT is $289 \mathrm{pC} / \mathrm{N}$ and the dielectric constant in a f... | The electric field generated and the stress applied are related by Equations ($E=-g \times X$) and ($g_{33}=\frac{d_{33}}{\varepsilon_{33}^X}$). Therefore, for this ceramic, the $g_{33}$ will be given by $$g_{33}=\frac{d_{33}}{\varepsilon_{33}^{X} \varepsilon_{0}}$$ where $\varepsilon_{0}$ is the permittivity of free s... | 4058 | V | No supplementary information needed. | single | medium | Materials: Polymers | Polymers | Electrical | Composites | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 10.6 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 11.2 | NUM | From the magnetic moments of the ferrous ($\mathrm{Fe}^{2+}$) and ferric ($\mathrm{Fe}^{3+}$) ions, calculate the net magnetic moment per formula unit in ferrimagnetic iron oxide ($\mathrm{Fe}_{3} \mathrm{O}_{4}$). | The formula for ferrimagnetic iron oxide can also be written as $\mathrm{FeO}: \mathrm{Fe}_{2} \mathrm{O}_{3}$, highlighting the presence of divalent ($\mathrm{Fe}^{2+}$) and trivalent ($\mathrm{Fe}^{3+}$) iron ions. For one formula unit, the unpaired electrons are 4 (from $\mathrm{Fe}^{2+}$) and 10 (five each from two... | 4 | \mu_{\mathrm{B}} | Based on the analysis, the following specific necessary information is used in the Solution but is missing from the Problem Description:
* Number of unpaired electrons for the $\mathrm{Fe}^{2+}$ ion.
* Number of unpaired electrons for the $\mathrm{Fe}^{3+}$ ion.
* The distribution of $\mathrm{Fe}^{2+}$ and $\mat... | single | easy | Materials: Metals | Metals | Magnetic | Composites | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 11.2 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 11.3 | NUM | The applied magnetic field $(H)$ for a toroidal solenoid is given by $N \times I$, where $N$ is the number of turns per meter and $I$ is the current in amperes. (a) What is the flux density (B) created for a current of 0.5 A if the toroidal solenoid has an air core? (b) What is the flux density (B) if the core is fille... | 1. The average circumference of the solenoid winding is $l=75 \mathrm{~cm}$, and there are 1000 turns. Therefore, the value of $N$ (turns per meter) $=1000$ turns $/(0.75 \mathrm{~m})=1333.33$.
The magnetic field created is $H=N \times I=(1333.33) \times(0.5 \mathrm{~A})=666.66 \mathrm{~A} / \mathrm{m}$.
The flux dens... | (8.38 \times 10^{-4}, 0.838) | (T, T) | * The relative magnetic permeability of iron used in the solution is 1000, whereas the problem description states it is 900. | multiple | easy | Materials: Metals | Metals | Magnetic | Composites | Diffusion & Kinetics | Joining | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 11.3 | ||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 11.5 | NUM | (a) Calculate the saturation magnetization ($\mu_{0} M_{\mathrm{s}}$) for $\mathrm{Fe}_{3} \mathrm{O}_{4}$ if the lattice constant ($a_{0}$) of the larger unit cell, which consists of eight smaller unit cells, is $8.37 \AA$. | 1. In $\mathrm{Fe}_{3} \mathrm{O}_{4}$, the magnetic moments of ferric ions ($\mathrm{Fe}^{3+}$) cancel out because of the antiferromagnetic coupling of an equal number of $\mathrm{Fe}^{3+}$ ions located at both tetrahedral and octahedral sites. Thus, only the ferrous ions ($\mathrm{Fe}^{2+}$) contribute to the net mag... | 0.64 | T | * The magnetic moment of a $\mathrm{Fe}^{2+}$ ion in $\mathrm{Fe}_{3} \mathrm{O}_{4}$ is $4 \mu_{\mathrm{B}}$. | single | hard | Materials: Metals | Metals | Magnetic | Cellular | Crystal Structure | Shaping | Corrosion | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 11.5 | |
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 11.6 | NUM | The maximum $d_{33}$ magnetostriction coefficient for Terfenol-D is $57 \times 10^{-9} \mathrm{~m} / \mathrm{A}$ (du Trémolet de Lacheisserie et al. 2002). A bar made from this material is 10 cm long and is exposed to a magnetic field $(H)$ of $5 \mathrm{~A} / \mathrm{m}$. What will be the elongation produced in this b... | The $d_{33}$ coefficient represents the strain produced per magnetic field. Thus, in this case, the strain produced $\frac{\Delta l}{l}$ will be given by $$\lambda=57 \times 10^{-9}=\frac{\text { Strain }}{\text { Magnetic field }}=\frac{\Delta l / l}{5 \mathrm{~A} / \mathrm{m}}=\frac{\Delta l}{5 \mathrm{~A} / \mathrm{... | 28.5 | nm | No supplementary information needed. | single | easy | Materials: Elastomers | Elastomers | Magnetic | Composites | Corrosion | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 11.6 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 11.7 | NUM | The coercivity $\left(H_{\mathrm{C}}\right)$ of a sample of a material known as grain-oriented steel is $300 \mathrm{~A} / \mathrm{m}$. The saturation magnetization ( $\mu_0 M_{\mathrm{S}}$ ) of this material is 2.1 T . A hysteresis loop for this material was measured to determine the properties.
1. If the coercivity ... | 1. The coercivity is $300 \mathrm{~A} / \mathrm{m}$. To convert this value into Tesla, we multiply it by $\mu_{0} =4 \pi \times 10^{-7} \mathrm{~Wb} / \mathrm{m} \cdot \mathrm{A}$.
The coercivity value expressed in Tesla will be $\mu_{0} H_{\mathrm{C}}=(300 \mathrm{~A} / \mathrm{m})\left(4 \pi \times 10^{-7} \mathrm{~... | (3.78 \times 10^{-4}, 3.78, 21000, 0.88) | (T, Oe, G, T) | No supplementary information needed. | multiple | hard | Materials: Metals | Metals | Magnetic | Composites | Shaping | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 11.7 | |||
Fundamentals_of_Materials_Instructors_Example 2.2 | NUM | Silicon has three naturally occurring isotopes: $92.23 \%$ of ${ }^{28} \mathrm{Si}$, with an atomic weight of 27.9769 amu, $4.68 \%$ of ${ }^{29} \mathrm{Si}$, with an atomic weight of 28.9765 amu, and $3.09 \%$ of ${ }^{30} \mathrm{Si}$, with an atomic weight of 29.9738 amu. On the basis of these data, calculate the ... | The average atomic weight of silicon $\left(\bar{A}{\mathrm{Si}}\right)$ is computed by adding fraction-of-\frac{occurrence}{atomic} weight products for the three isotopes-i.e., using Equation ($\bar{A}_{\mathrm{M}}=\sum_i f_{i_{\mathrm{M}}} \mathrm{A}_{i_{\mathrm{M}}}$). (Remember: fraction of occurrence is equal to t... | 28.0854 | amu | single | easy | Materials: Semiconductors | Semiconductors | Composites | Fundamentals_of_Materials_Instructors | Example 2.2 | ||||||
Fundamentals_of_Materials_Instructors_Example 2.3 | NUM | Zinc has five naturally occurring isotopes: $48.63 \%$ of ${ }^{64} \mathrm{Zn}$ with an atomic weight of 63.929 amu; $27.90 \%$ of ${ }^{66} \mathrm{Zn}$ with an atomic weight of 65.926 amu; $4.10 \%$ of ${ }^{67} \mathrm{Zn}$ with an atomic weight of 66.927 amu; $18.75 \%$ of ${ }^{68} \mathrm{Zn}$ with an atomic wei... | The average atomic weight of zinc $\bar{A}{\mathrm{Zn}}$ is computed by adding fraction-of-occurrence-atomic weight products for the five isotopes-i.e., using Equation ($\bar{A}_{\mathrm{M}}=\sum_i f_{i_{\mathrm{M}}} \mathrm{A}_{i_{\mathrm{M}}}$). (Remember: fraction of occurrence is equal to the percent of occurrence ... | 65.400 | amu | single | easy | Materials: Metals | Metals | Composites | Fundamentals_of_Materials_Instructors | Example 2.3 | ||||||
Fundamentals_of_Materials_Instructors_Example 2.4 | NUM | Indium has two naturally occurring isotopes: ${ }^{113}$ In with an atomic weight of 112.904 amu, and ${ }^{115}$ In with an atomic weight of 114.904 amu. If the average atomic weight for In is 114.818 amu, calculate the fraction of occurrences of these two isotopes.
Give your answer as a tuple (${ }^{113}$, ${ }^{115... | The average atomic weight of indium $\left(\bar{A}_{\text {In }}\right)$ is computed by adding fraction-of-occurrence-atomic weight products for the two isotopes-i.e., using Equation ($\bar{A}_{\mathrm{M}}=\sum_i f_{i_{\mathrm{M}}} \mathrm{A}_{i_{\mathrm{M}}}$), or
$$
\bar{A}_{\mathrm{In}}=f_{113} A_{113}+f_{115} A_{1... | (0.043, 0.957) | multiple | easy | Materials: Metals | Metals | Composites | Fracture | Fundamentals_of_Materials_Instructors | Example 2.4 | ||||||
Fundamentals_of_Materials_Instructors_Example 2.5 | NUM | (a) How many grams are there in one amu of a material?
(b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there in a pound-mole of a substance?
Give your answer as a tuple (answer to (a), answer to (b)). | (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the \frac{amu}{atom}, $\mathrm{g} / \mathrm{mol}$, and atom $/ \mathrm{mol}$ relationships is all that is necessary, as
$$
\begin{gathered}
\# \mathrm{~g} / \mathrm{amu}=\left(\frac{1 \mathrm{~mol}}{6.022 \times 10^{23} \... | (1.66e-24, 2.73e26) | (g/amu, atoms/lb-mol) | multiple | easy | Structures: Composites | Composites | Fundamentals_of_Materials_Instructors | Example 2.5 | |||||||
Fundamentals_of_Materials_Instructors_Example 2.9 | FORMULA | Give the electron configurations for the following ions: $P^{3+}, P^{3-}, S n^{4+}, S e^{2-}, \Gamma$, and $N i^{2+}$.
Give your answer as a tuple (configuration of $P^{5+}$, configuration of $P^{3-}$, configuration of $Sn^{4+}$, configuration of $Se^{2-}$, configuration of $I^{-}$, $N i^{2+}$).
Refer to table 2.2 :... | The electron configurations for the ions are determined using table 2.2 (and figure 2.8 ).
$\mathrm{P}^{5+}$ : From table 2.2 , the electron configuration for an atom of phosphorus is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$. In order to become an ion with a plus five charge, it must lose five electrons - in this... | ($1 s^{2} 2 s^{2} 2 p^{6}$, $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$, $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10}$, $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}$, $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6}$, $1 s^{2} 2 ... | multiple | medium | Materials: Metals | Metals | Composites | Fundamentals_of_Materials_Instructors | Example 2.9 | |||||||
Fundamentals_of_Materials_Instructors_Example 2.15 | NUM | Calculate the force of attraction between a $\mathrm{Ca}^{2+}$ and an $O^{2-}$ ion whose centers are separated by a distance of 1.25 nm . | To solve this problem for the force of attraction between these two ions it is necessary to use Equation ($F_A=\frac{1}{4 \pi \varepsilon_0 r^2}\left(\left|Z_1\right| e\right)\left(\left|Z_2\right| e\right)$), which takes on the form of Equation ($\begin{aligned} F_A & =\frac{1}{4 \pi\left(8.85 \times 10^{-12} \mathrm{... | 5.91E-10 | N | single | easy | Fundamental: Diffusion & Kinetics | Composites | Diffusion & Kinetics | Fundamentals_of_Materials_Instructors | Example 2.15 | ||||||
Fundamentals_of_Materials_Instructors_Example 2.16 | NUM | The atomic radii of $\mathrm{Mg}^{2+}$ and $F^{-}$ions are 0.072 and 0.133 nm , respectively. Use k = 8.99×10^9 N·m²/C² (or equivalently k e² = 2.31×10^–28 N·m²).
(a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another).
(b) Wha... | (a) The force of attraction $F{A}$ is calculated using Equation ($\begin{aligned} F_A & =\frac{\left(2.31 \times 10^{-28} \mathrm{~N} \cdot \mathrm{~m}^2\right)\left(\left|Z_1\right|\right)\left(\left|Z_2\right|\right)}{r^2}\end{aligned}$) taking the interionic separation $r$ to be $r{0}$ the equilibrium separation dis... | (1.10e-8, -1.10e-8) | (N, N) | multiple | easy | Fundamental: Phase Diagram | Composites | Phase Diagram | Fundamentals_of_Materials_Instructors | Example 2.16 | ||||||
Fundamentals_of_Materials_Instructors_Example 2.17 | NUM | The force of attraction between a divalent cation and a divalent anion is $1.67 \times 10^{-8} \mathrm{~N}$. If the ionic radius of the cation is 0.080 nm , what is the anion radius? | To begin, let us rewrite Equation ($r_0=r_{\mathrm{K}^{+}}+r_{\mathrm{Br}^{-}}$) to read as follows:
$$
r{0}=r{\mathrm{C}}+r{\mathrm{A}}
$$
in which $r{\mathrm{C}}$ and $r{\mathrm{A}}$ represent, respectively, the radii of the cation and anion. Thus, this problem calls for us to determine the value of $r{\mathrm{A}}$... | 0.155 | nm | single | hard | Materials: Metals | Metals | Composites | Fundamentals_of_Materials_Instructors | Example 2.17 | ||||||
Fundamentals_of_Materials_Instructors_Example 2.18 | FORMULA | The net potential energy between two adjacent ions, $E{N}$, may be represented by the sum of Equations ($E_A=-\frac{A}{r}$) and ($E_R=\frac{B}{r^n}$); that is,
$$
E{N}=-\frac{A}{r}+\frac{B}{r^{n}}
$$
Calculate the bonding energy $E{0}$ in terms of the parameters $A, B$, and $n$ using the following procedure:
1. Diff... | Differentiation of Equation ($E_N=-\frac{A}{r}+\frac{B}{r^n}$) yields
$$
\begin{gathered}
\frac{d E{N}}{d r}=\frac{d\left(-\frac{A}{r}\right)}{d r}+\frac{d\left(\frac{B}{r^{n}}\right)}{d r} \\
=\frac{A}{r^{(1+1)}}-\frac{n B}{r^{(n+1)}}=0
\end{gathered}
$$
Now, solving for $r\left(=r{0}\right)$
$$
\frac{A}{r{0}^{2}}=... | $$
E{0}=-\frac{A}{\left(\frac{A}{n B}\right)^{1 /(1-n)}}+\frac{B}{\left(\frac{A}{n B}\right)^{n /(1-n)}}
$$ | single | hard | Materials: Metals | Metals | Phase Diagram | Fundamentals_of_Materials_Instructors | Example 2.18 | |||||||
Fundamentals_of_Materials_Instructors_Example 2.21 | FORMULA | The net potential energy $E{N}$ between two adjacent ions is sometimes represented by the expression
$$
E{N}=-\frac{C}{r}+D \exp \left(-\frac{r}{\rho}\right)
$$
in which $r$ is the interionic separation and $C, D$, and $\rho$ are constants whose values depend on the specific material.
(a) Derive an expression for the... | (a) Differentiating Equation ($E_N=-\frac{C}{r}+D \exp \left(-\frac{r}{\rho}\right)$) with respect to $r$ yields
$$
\begin{aligned}
\frac{d E}{d r}= & \frac{d\left(-\frac{C}{r}\right)}{d r}-\frac{d\left(D \exp \left(-\frac{r}{\rho}\right)\right)}{d r} \\
& =\frac{C}{r^{2}}-\frac{D \exp \left(-\frac{r}{\rho}\right)}{\r... | ($$E{0}=D\left(1-\frac{r{0}}{\rho}\right) \exp \left(-\frac{r{0}}{\rho}\right)$$, $$E{0}=\frac{C}{r{0}}\left(\frac{\rho}{r{0}}-1\right)$$) | multiple | medium | Materials: Metals | Metals | Composites | Atomic Bonding | Fundamentals_of_Materials_Instructors | Example 2.21 | ||||||
Fundamentals_of_Materials_Instructors_Example 2.25 | NUM | Compute the \%IC of the interatomic bond for each of the following compounds: $\mathrm{MgO}, \mathrm{GaP}, \mathrm{CsF}, \mathrm{CdS}$, and FeO.
Give your answer as a tuple (MgO, GaP, CsF, CdS, FeO). | The percent ionic character is a function of the electron negativities of the ions $X{\mathrm{A}}$ and $X{\mathrm{B}}$ according to Equation ($\% \mathrm{IC}=\left\{1-\exp \left[-(0.25)\left(X_{\mathrm{A}}-X_{\mathrm{B}}\right)^2\right]\right\} \times 100$). The electronegativities of the elements are found in figure ... | (73.4, 6.1, 93.4, 14.8, 51.4) | (%,%,%,%,%) | multiple | medium | Properties: Thermal | Thermal | Composites | Atomic Bonding | Fundamentals_of_Materials_Instructors | Example 2.25 | |||||
Fundamentals_of_Materials_Instructors_Example 2.26 | NUM | Calculate \%IC of the interatomic bonds for the intermetallic compound $A l_6 M n$. The electronegativities for Al and Mn are both 1.5. | The percent ionic character is a function of the electron negativities of the ions $X_{\mathrm{A}}$ and $X_{\mathrm{B}}$ according to Equation ($\% \mathrm{IC}=\left\{1-\exp \left[-(0.25)\left(X_{\mathrm{A}}-X_{\mathrm{B}}\right)^2\right]\right\} \times 100$). The electronegativities for Al and Mn are both 1.5 and. The... | 0 | % | single | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 2.26 | |||||
Fundamentals_of_Materials_Instructors_Example 3.2 | NUM | If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters. | Lead has an FCC crystal structure and, as given in the problem statement, an atomic radius of 0.1750 nm . The FCC unit cell volume may be computed from Equation ($V_C=16 R^3 \sqrt{2}$) as
$$
\begin{gathered}
V{C}=16 R^{3} \sqrt{2}=(16)\left(0.175 \times 10^{-9} \mathrm{~m}\right)^{3} \sqrt{2} \\
=1.213 \times 10^{-28}... | 1.213E-28 | m^3 | single | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.2 | |||||
Fundamentals_of_Materials_Instructors_Example 3.3 | FORMULA | Derive the relationship between the unit cell edge length (a) and the atomic radius (R) for a body-centered cubic (BCC) crystal structure. | Consider the BCC unit cell shown below
From the triangle NOP
$$
(\overline{N P})^{2}=a^{2}+a^{2}=2 a^{2}
$$
And then for triangle $N P Q$,
$$
(\overline{N Q})^{2}=(\overline{Q P})^{2}+(\overline{N P})^{2}
$$
But $\overline{N Q}=4 R, R$ being the atomic radius. Also, $\overline{Q P}=a$. Therefore,
$$
$(4 R)^{2}=a... | $$
a=\frac{4 R}{\sqrt{3}}
$$ | single | easy | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.3 | |||||||
Fundamentals_of_Materials_Instructors_Example 3.4 | NUM | For the HCP crystal structure, calculate the ideal c/a ratio. | A sketch of one-third of an HCP unit cell is shown below.
Consider the tetrahedron labeled as $J K L M$, which is reconstructed as follows:
The atom at point $M$ is midway between the top and bottom faces of the unit cell-that is $\bar{M} \bar{H}=\frac{c}{2}$. And, since atoms at points $J, K$, and $M$, all touch o... | 1.633 | single | easy | Materials: Semiconductors | Semiconductors | Thermal | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.4 | |||||
Fundamentals_of_Materials_Instructors_Example 3.6 | FORMULA | Calculate the atomic packing factor for HCP. | The APF is just the total sphere volume-unit cell volume ratio-i.e., $\left(V{S} / V{C}\right)$. For HCP, there are the equivalent of six spheres per unit cell, and thus
$$
V{S}=6\left(\frac{4 \pi R^{3}}{3}\right)=8 \pi R^{3}
$$
The unit cell volume $\left(V{C}\right)$ for the HCP unit cell was determined and given i... | 0.74 | single | medium | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.6 | |||||||
Fundamentals_of_Materials_Instructors_Example 3.7 | NUM | Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm , and an atomic weight of $95.94 \mathrm{~g} / \mathrm{mol}$. Compute its theoretical density. | This problem calls for a computation of the density of molybdenum. According to Equation ($\rho=\frac{n A}{V_C N_{\mathrm{A}}}$)
$$
\rho=\frac{n A{\mathrm{Mo}}}{V{C} N{\mathrm{A}}}
$$
For $\mathrm{BCC}, n=2$ \frac{atoms}{unit} cell. Furthermore, because $V{C}=a^{3}$, and $a=\frac{4 R}{\sqrt{3}}$, then
$$
V{C}=\left(... | 10.22 | g/cm^3 | single | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.7 | |||||
Fundamentals_of_Materials_Instructors_Example 3.8 | NUM | Strontium (Sr) has an FCC crystal structure, an atomic radius of 0.215 nm and an atomic weight of 87.62 $\mathrm{g} / \mathrm{mol}$. Calculate the theoretical density for Sr . | According to Equation ($\rho=\frac{n A}{V_C N_{\mathrm{A}}}$)
$$
\rho=\frac{n A{\mathrm{Sr}}}{V{C} N{\mathrm{A}}}
$$
For FCC, $n=4$ \frac{atoms}{unit} cell. Furthermore, because $V{C}=a^{3}$, and $a=2 R \sqrt{2}$, then
$$
V{C}=(2 R \sqrt{2})^{3}
$$
Thus, realizing that $A{\mathrm{Sr}}=87.62 \mathrm{~g} / \mathrm{mo... | 2.59 | g/cm^3 | single | easy | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.8 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.9 | NUM | Calculate the radius of a palladium (Pd) atom, given that Pd has an FCC crystal structure, a density of $12.0 \mathrm{~g} / \mathrm{cm}^{3}$, and an atomic weight of $106.4 \mathrm{~g} / \mathrm{mol}$. | We are asked to determine the radius of a palladium atom, given that Pd has an FCC crystal structure. For FCC, $n=4$ \frac{atoms}{unit} cell, and $V{C}=16 R^{3} \sqrt{2}$. Now, the density of Pd may be expressed using a form of Equation ($\rho=\frac{n A}{V_C N_{\mathrm{A}}}$) as follows:
$$
\begin{gathered}
\rho=\frac... | 0.138 | nm | single | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.9 | |||||
Fundamentals_of_Materials_Instructors_Example 3.10 | NUM | Calculate the radius of a tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 $\mathrm{g} / \mathrm{cm}^{3}$, and an atomic weight of $180.9 \mathrm{~g} / \mathrm{mol}$. | It is possible to compute the radius of a Ta atom using a rearranged form of Equation ($\rho=\frac{n A}{V_C N_{\mathrm{A}}}$). For BCC, $n=2$ \frac{atoms}{unit} cell. Furthermore, because $V{C}=a^{3}$ and $a=\frac{4 R}{\sqrt{3}}$ an expression for the BCC unit cell volume is as follows:
$$
V{C}=\left(\frac{4 R}{\sqrt{... | 0.143 | nm | single | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.10 | |||||
Fundamentals_of_Materials_Instructors_Example 3.11 | NUM | A hypothetical metal has the simple cubic crystal structure shown in figure 3.3 . If its atomic weight is 74.5 $\mathrm{g} / \mathrm{mol}$ and the atomic radius is 0.145 nm , compute its density. | For the simple cubic crystal structure ( figure 3.3 ), the value of $n$ in Equation ($\rho=\frac{n A}{V_C N_{\mathrm{A}}}$) is unity since there is only a single atom associated with each unit cell. Furthermore, for the unit cell edge length, $a=2 R$ ( figure 3.3 ); this means that the unit cell volume $V{C}=a^{3}=(2 R... | 5.07 | g/cm^3 | single | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.11 | |||||
Fundamentals_of_Materials_Instructors_Example 3.12 | NUM | Titanium (Ti) has an HCP crystal structure and a density of $4.51 \mathrm{~g} / \mathrm{cm}^{3}$. For HCP, $n=6$ \frac{atoms}{unit} cell, and the atomic weight for $\mathrm{Ti}, A{\mathrm{Ti}}=47.87 \mathrm{~g} / \mathrm{mol}$.
(a) What is the volume of its unit cell in cubic meters?
(b) If the c/a ratio is 1.58, compu... | (a) The volume of the Ti unit cell may be computed using a rearranged form of Equation ($\rho=\frac{n A}{V_C N_{\mathrm{A}}}$) as
$$
V{C}=\frac{n A{\mathrm{Ti}}}{\rho N{\mathrm{A}}}
$$
For HCP, $n=6$ \frac{atoms}{unit} cell, and the atomic weight for $\mathrm{Ti}, A{\mathrm{Ti}}=47.87 \mathrm{~g} / \mathrm{mol}$. Thu... | (1.058e-28, 0.468) | (m^3/unit cell, nm) | multiple | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.12 | |||||
Fundamentals_of_Materials_Instructors_Example 3.13 | NUM | Magnesium (Mg) has an HCP crystal structure and a density of $1.74 \mathrm{~g} / \mathrm{cm}^{3}$. For HCP, $n=6$ \frac{atoms}{unit} cell, and the atomic weight for $\mathrm{Mg}, A{\mathrm{Mg}}=24.31 \mathrm{~g} / \mathrm{mol}$.
(a) What is the volume of its unit cell in cubic centimeters?
(b) If the c/a ratio is 1.624... | (a) The volume of the Mg unit cell may be computed using a rearranged form of Equation ($\rho=\frac{n A}{V_C N_{\mathrm{A}}}$) as
$$
V{C}=\frac{n A{\mathrm{Mg}}}{\rho N{\mathrm{A}}}
$$
Now, for HCP, $n=6$ \frac{atoms}{unit} cell, and the atomic weight for $\mathrm{Mg}, A{\mathrm{Mg}}=24.31 \mathrm{~g} / \mathrm{mol}$... | (1.39e-22, 0.521) | (cm^3/unit cell, nm) | multiple | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.13 | |||||
Fundamentals_of_Materials_Instructors_Example 3.15 | FORMULA | Niobium ( Nb ) has an atomic radius of 0.1430 nm and a density of $8.57 \mathrm{~g} / \mathrm{cm}^{3}$. Determine whether it has an FCC or a BCC crystal structure. | In order to determine whether Nb has an FCC or a BCC crystal structure, we need to compute its density for each of the crystal structures. For FCC, $n=4$, and $a=2 R \sqrt{2}$. Also, from figure 2.8 , its atomic weight is $92.91 \mathrm{~g} / \mathrm{mol}$. Thus, for FCC
$$
\rho=\frac{n A{\mathrm{Nb}}}{V{C} N{\mathrm... | BCC | single | medium | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.15 | |||||||
Fundamentals_of_Materials_Instructors_Example 3.17 | NUM | The unit cell for uranium (U) has orthorhombic symmetry, with $a, b$, and c lattice parameters of 0.286 , 0.587 , and 0.495 nm , respectively. If its density, atomic weight, and atomic radius are $19.05 \mathrm{~g} / \mathrm{cm}^{3}, 238.03 \mathrm{~g} / \mathrm{mol}$, and 0.1385 nm , respectively, compute the atomic p... | In order to determine the APF for U , we need to compute both the unit cell volume $\left(V{C}\right)$ which is just the product of the three unit cell parameters, as well as the total sphere volume $\left(V{S}\right)$ which is just the product of the volume of a single sphere and the number of spheres in the unit cell... | 0.536 | multiple | medium | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.17 | |||||||
Fundamentals_of_Materials_Instructors_Example 3.18 | NUM | Indium (In) has a tetragonal unit cell for which the a and c lattice parameters are 0.459 and 0.495 nm , respectively.
(a) If the atomic packing factor and atomic radius are 0.693 and 0.1625 nm , respectively, determine the number of atoms in each unit cell.
(b) The atomic weight of indium is $114.82 \mathrm{~g} / \mat... | (a) For indium, and from the definition of the APF ($\mathrm{APF}=\frac{\text { volume of atoms in a unit cell }}{\text { total unit cell volume }}$) is
$$
\mathrm{APF}=\frac{V{X}}{V{C}}=\frac{n\left(\frac{4}{3} \pi R^{3}\right)}{a^{2} c}
$$
we may solve for the number of atoms per unit cell, $n$, as
$$
\begin{gathe... | (4.0, 7.31) | (atoms/unit cell, g/cm^3) | multiple | easy | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.18 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.19 | NUM | Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is 0.1143 nm , (a) determine the unit cell volume, and (b) calculate the theoretical density of Be and compare it with the literature value.
Give your answer as a tuple (the unit cell volume, t... | (a) We are asked to calculate the unit cell volume for Be. The volume of an HCP unit cell is provided by Equation as follows:
$$
V{C}=6 R^{2} c \sqrt{3}
$$
But, $c=1.568 a$, and $a=2 R$, or $c=(1.568)(2 R)=3.136 R$. Substitution of this expression for $c$ into the above equation leads to
$$
\begin{gathered}
V{C}=(6)... | (4.87e-23, 1.84) | (cm^3/unit cell, g/cm^3) | multiple | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.19 | |||||
Fundamentals_of_Materials_Instructors_Example 3.20 | NUM | Magnesium (Mg) has an HCP crystal structure, a c/a ratio of 1.624, and a density of $1.74 \mathrm{~g} / \mathrm{cm}^{3}$. Compute the atomic radius for Mg . | This problem calls for us to compute the atomic radius for Mg. In order to do this we must use Equation ($\rho=\frac{n A}{V_C N_{\mathrm{A}}}$), as well as the expression that relates the atomic radius to the unit cell volume for HCP-Equation ($\begin{aligned} V_C & =6 R^2 c \sqrt{3}\end{aligned}$)-that is
$$
V{C}=6 R... | 0.160 | nm | single | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.20 | |||||
Fundamentals_of_Materials_Instructors_Example 3.21 | NUM | Cobalt (Co) has an HCP crystal structure, an atomic radius of 0.1253 nm , and a c/a ratio of 1.623 . Compute the volume of the unit cell for Co. | This problem asks that we calculate the unit cell volume for Co , which has an HCP crystal structure. In order to do this, it is necessary to use Equation-an expression for the volume of an HCP unit cell in terms of the atomic radius $R$ and the lattice parameter - that is
$$
V{C}=6 R^{2} c \sqrt{3}
$$
The problem st... | 6.64E-02 | nm^3 | single | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.21 | |||||
Fundamentals_of_Materials_Instructors_Example 3.23 | NUM | What is the minimum cation-to-anion radius ratio for a coordination number of 4? | In this problem we are asked to show that the minimum cation-to-anion radius ratio for a coordination number of four is 0.225 . The sketch below shows the four anions surrounding a cation. If lines are drawn from the centers of the anions, then a tetrahedron is formed. The tetrahedron may be inscribed within a cube as
... | 0.225 | single | easy | Materials: Metals | Metals | Composites | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.23 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.24 | NUM | Calculate the minimum cation-to-anion radius ratio for a coordination number of 6. [Hint: Use the NaCl crystal structure ( figure 3.6 ), and assume that anions and cations are just touching along cube edges and across face diagonals.] | This problem asks us to show that the minimum cation-to-anion radius ratio for a coordination number of 6 is 0.414 (using the rock salt crystal structure). Below is shown one of the faces of the rock salt crystal structure in which anions and cations just touch along the edges, and also the face diagonals.
From trian... | 0.414 | single | medium | Materials: Metals | Metals | Micro/Nano-structure | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.24 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.25 | NUM | Calculate the minimum cation-to-anion radius ratio for a coordination number of 8. | This problem asks us to show that the minimum cation-to-anion radius ratio for a coordination number of 8 is 0.732 . From the following sketch of a cubic unit cell, it may be noted that the atom represented by the dark circle has eight nearest-neighbor atoms - denoted by the open circles at the corners of the unit cell... | 0.732 | single | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.25 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.28 | NUM | Compute the atomic packing factor for the rock salt crystal structure in which $r{C} / r{A}=0.414$. | This problem asks that we compute the atomic packing factor for the rock salt crystal structure when $r{\mathrm{C}} / r{\mathrm{A}}$ $=0.414$. The definition of the atomic packing (APF) is given in Equation as follows:
$$
\mathrm{APF}=\frac{V{S}}{V{C}}
$$
where $V{S}$ and $V{C}$ are, respectively sphere volume of ion... | 0.793 | single | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.28 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.29 | NUM | The unit cell for $\mathrm{Al}_{2} \mathrm{O}_{3}$ has hexagonal symmetry with lattice parameters $a=0.4759 \mathrm{~nm}$ and $c=1.2989 \mathrm{~nm}$. If the density of this material is $3.99 \mathrm{~g} / \mathrm{cm}^{3}$, calculate its atomic packing factor. For this computation use ionic radii listed in table 3.4 .... | Here
$\sum A_{\mathrm{Al}}=$ the sum of the atomic weights of all aluminum ions in $\mathrm{Al}_{2} \mathrm{O}_{3}=2 A_{\mathrm{Al}}=(2)(26.98 \mathrm{~g} / \mathrm{mol})$
$\sum A_{\mathrm{O}}=$ the sum of the atomic weights of all oxygen ions in $\mathrm{Al}_{2} \mathrm{O}_{3}=3 A_{\mathrm{O}}=(3)(16.00 \mathrm{~g} / ... | 0.842 | single | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.29 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.30 | NUM | Compute the atomic packing factor for cesium chloride using the ionic radii in table 3.4 and assuming that the ions touch along the cube diagonals.
Refer to table 3.4 :
table 3.4
lonic Radii for Several Cations and Anions for a Coordination Number of 6
\begin{tabular}{|l|l|l|l|}
\hline Cation & Ionic Radius (nm) ... | We are asked in this problem to compute the atomic packing factor for the CsCl crystal structure. This requires that we take the ratio of the sphere volume within the unit cell and the total unit cell volume. From figure 3.7 there is the equivalence of one Cs and one Cl ion per unit cell; the ionic radii of these two... | 0.684 | single | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.30 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.31 | NUM | Calculate the theoretical density of NiO, given that it has the rock salt crystal structure. The ionic radii of $\mathrm{Ni}^{2+}$ and $\mathrm{O}^{2-}$ are 0.069 nm and 0.140 nm.
Reference:
$$
\begin{aligned}
& \sum A{\mathrm{Ni}}=A{\mathrm{Ni}}=58.69 \mathrm{~g} / \mathrm{mol} \\
& \sum A{\mathrm{O}}=A{\mathrm{O}}=1... | This density of NiO may be computed using Equation ($\rho=\frac{n^{\prime}\left(\sum A_{\mathrm{C}}+\sum A_{\mathrm{A}}\right)}{V_C N_{\mathrm{A}}}$), which for this problem, is of the form
$$
\rho=\frac{n^{\prime}\left(\sum A{\mathrm{Ni}}+\sum A{\mathrm{O}}\right)}{V{C} N{\mathrm{A}}}
$$
But because the formula unit... | 6.80 | g/cm^3 | single | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.31 | |||||
Fundamentals_of_Materials_Instructors_Example 3.32 | NUM | Iron oxide (FeO) has the rock salt crystal structure and a density of $5.70 \mathrm{~g} / \mathrm{cm}^{3}$, determine the unit cell edge length.
Reference:
$$
\begin{aligned}
& \sum A{\mathrm{Fe}}=A{\mathrm{Fe}}=55.85 \mathrm{~g} / \mathrm{mol} \\
& \sum A{\mathrm{O}}=A{\mathrm{O}}=16.00 \mathrm{~g} / \mathrm{mol}
\en... | This part of the problem calls for us to determine the unit cell edge length for FeO . The density of FeO is $5.70 \mathrm{~g} / \mathrm{cm}^{3}$ and the crystal structure is rock salt. From Equation ($\rho=\frac{n^{\prime}\left(\sum A_{\mathrm{C}}+\sum A_{\mathrm{A}}\right)}{V_C N_{\mathrm{A}}}$) the density of Fe , i... | 0.437 | nm | single | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.32 | |||||
Fundamentals_of_Materials_Instructors_Example 3.33 | NUM | One crystalline form of silica $\left(\mathrm{SiO}{2}\right)$ has a cubic unit cell, and from x-ray diffraction data it is known that the cell edge length is 0.700 nm . If the measured density is $2.32 \mathrm{~g} / \mathrm{cm}^{3}$, how many $\mathrm{Si}^{4+}$ and $\mathrm{O}^{2-}$ ions are there per unit cell? (Round... | We are asked to determine the number of $\mathrm{Si}^{4+}$ and $\mathrm{O}^{2-}$ ions per unit cell for a crystalline form of silica $\left(\mathrm{SiO}{2}\right)$. For this material, $a=0.700 \mathrm{~nm}$ and $\rho=2.32 \mathrm{~g} / \mathrm{cm}^{3}$. Solving for $n^{\prime}$ from Equation ($\rho=\frac{n^{\prime}\lef... | (8, 16) | multiple | easy | Materials: Glasses | Glasses | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.33 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.34 | NUM | Using the ionic radii in table 3.4 , compute the theoretical density of CsCl. (Hint: Use a modification of $$
a=\frac{4 R}{\sqrt{3}}
$$.)
Refer to table 3.4 :
table 3.4
Ionic Radii for Several Cations and Anions for a Coordination Number of 6
\begin{tabular}{|l|l|l|l|}
\hline Cation & Ionic Radius (nm) & Anion & I... | We are asked to compute the density of CsCl . To solve this problem it is necessary to use Equation-namely
$$
\rho=\frac{n^{\prime}\left(\sum \mathcal{A}{\mathrm{Cs}}+\sum \mathcal{A}{\mathrm{Cl}}\right)}{V{C} N{\mathrm{A}}}
$$
But because the formula unit is "CsCl" then
$$
\begin{aligned}
& \sum \mathcal{A}{\mathrm... | 4.21 | g/cm^3 | single | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.34 | |||||
Fundamentals_of_Materials_Instructors_Example 3.35 | NUM | From the data in table 3.4 , compute the theoretical density of $\mathrm{CaF}{2}$, which has the fluorite structure.
Reference:
"Table 3.4
Ionic Radii for Several Cations and Anions for a Coordination Number of 6
\begin{tabular}{|l|l|l|l|}
\hline Cation & Ionic Radius (nm) & Anion & Ionic Radius (nm) \\
\hline $\math... | This problem asks that we compute the density of $\mathrm{CaF}{2}$. In order to solve this problem it is necessary to use equation (\rho=\frac{n^{\prime}\left(\sum A_{\mathrm{C}}+\sum A_{\mathrm{A}}\right)}{V_C N_{\mathrm{A}}}), which for our problem takes the form:
$$
\rho=\frac{n^{\prime}\left(\sum A{\mathrm{Ca}}+\s... | 3.32 | g/cm^3 | single | medium | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.35 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.37 | NUM | The unit cell for $\mathrm{Fe}{3} \mathrm{O}{4}\left(\mathrm{FeO}-\mathrm{Fe}{2} \mathrm{O}{3}\right)$ has cubic symmetry with a unit cell edge length of 0.839 nm . If the density of this material is $5.24 \mathrm{~g} / \mathrm{cm}^{3}$, compute its atomic packing factor. For this computation, you will need to use the ... | This problem asks us to compute the atomic packing factor for $\mathrm{Fe}{3} \mathrm{O}{4}$ given its density and unit cell edge length. The APF is determined using equation (\mathrm{APF}=\frac{\text { volume of atoms in a unit cell }}{\text { total unit cell volume }}). Computation of the sphere volume is possible by... | 0.686 | single | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.37 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.39 | NUM | Determine the angle between covalent bonds in an $\mathrm{SiO}{4}^{4-}$ tetrahedron. | Below is shown a $\mathrm{SiO}{4}^{4-}$ tetrahedron situated within a cube; oxygen atoms are represented by open circles, the Si atom by the dark circle.
Now if we extend the base diagonal (having a length of $y$ ) from one corner to the other, then
$$
$(2 y)^{2}=a^{2}+a^{2}=2 a^{2}$
$$
where $a$ is the unit cell e... | 109.48 | ° | single | medium | Materials: Semiconductors | Semiconductors | Cellular | Atomic Bonding | Fundamentals_of_Materials_Instructors | Example 3.39 | |||||
Fundamentals_of_Materials_Instructors_Example 3.40 | NUM | Compute the theoretical density of diamond, given that the $\mathrm{C}-\mathrm{C}$ distance and bond angle are 0.154 nm and $109.5^{\circ}$, respectively. | This problem asks that we compute the theoretical density of diamond given that the $\mathrm{C}-\mathrm{C}$ distance and bond angle are 0.154 nm and $109.5^{\circ}$, respectively. To compute the density of diamond it is necessary that we use equation (\rho=\frac{n^{\prime}\left(\sum A_{\mathrm{C}}+\sum A_{\mathrm{A}}\r... | 3.54 | g/cm^3 | single | hard | Materials: Semiconductors | Semiconductors | Cellular | Crystal Structure | Shaping | Fundamentals_of_Materials_Instructors | Example 3.40 | ||||
Fundamentals_of_Materials_Instructors_Example 3.41 | NUM | Compute the theoretical density of ZnS , given that the $\mathrm{Zn}-\mathrm{S}$ distance and bond angle are 0.234 nm and $109.5^{\circ}$, respectively. | This problem asks that we compute the theoretical density of ZnS given that the $\mathrm{Zn}-\mathrm{S}$ distance and bond angle are 0.234 nm and $109.5^{\circ}$, respectively. To compute the density of zinc sulfide it is necessary that we use equation (\rho=\frac{n^{\prime}\left(\sum A_{\mathrm{C}}+\sum A_{\mathrm{A}}... | 4.11 | g/cm^3 | single | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Shaping | Fundamentals_of_Materials_Instructors | Example 3.41 | ||||
Fundamentals_of_Materials_Instructors_Example 3.42 | NUM | Compute the atomic packing factor for the diamond cubic crystal structure ( figure 3.17 ). Assume that bonding atoms touch one another, that the angle between adjacent bonds is $109.5^{\circ}$, and that each atom internal to the unit cell is positioned a/4 of the distance away from the two nearest cell faces (a is the ... | We are asked in this problem to compute the atomic packing factor for the diamond cubic crystal structure, given that the angle between adjacent bonds is $109.5^{\circ}$. Let us consider he drawing below, which represents the diamond cubic unit cell with those carbon atoms that bond to one another in one-quarter of the... | 0.340 | single | hard | Materials: Semiconductors | Semiconductors | Cellular | Crystal Structure | Shaping | Fundamentals_of_Materials_Instructors | Example 3.42 | |||||
Fundamentals_of_Materials_Instructors_Example 3.43 | NUM | Iron (Fe) undergoes an allotropic transformation at $912^{\circ} \mathrm{C}$ : upon heating from a $B C C$ ( $\alpha$ phase) to an FCC ( $\gamma$ phase). Accompanying this transformation is a change in the atomic radius of Fe -from $R{B C C}=0.12584$ nm to $R{F C C}=0.12894 \mathrm{~nm}$-and, in addition a change in de... | To solve this problem let us first compute the density of each phase using equation (\rho=\frac{n A}{V_C N_{\mathrm{A}}}), and then determine the volumes per unit mass (the reciprocals of densities). From these values it is possible to calculate the percent volume change.
The density of each phase may be computed usin... | 1.19 | % | single | medium | Materials: Metals | Metals | Thermal | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.43 | ||||
Fundamentals_of_Materials_Instructors_Example 3.44 | NUM | The accompanying figure shows a unit cell for a hypothetical metal. Calculate the density of the material, given that its atomic weight is $141 \mathrm{~g} / \mathrm{mol}$. | The unit cell shown in the problem statement belongs to the tetragonal crystal system since $a=b=0.35$ $\mathrm{nm}, c=0.45 \mathrm{~nm}$, and $\alpha=\beta=\gamma=90^{\circ}$.
The crystal structure would be called body-centered tetragonal.
As with BCC, $n=2$ \frac{atoms}{unit} cell. Also, for this unit cell
$$
\begin... | 8.49 | g/cm^3 | single | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.44 | |||||
Fundamentals_of_Materials_Instructors_Example 3.59 | FORMULA | (a) What are the direction indices for a vector that passes from point $\frac{1}{4} 0 \frac{1}{2}$ to point $\frac{3}{4} \frac{1}{2} \frac{1}{2}$ in a cubic unit cell?
(b) Repeat part (a) for a monoclinic unit cell.
Give your answer as a tuple (answer to (a), answer to (b)) | (a) Point coordinate indices for the vector tail, $\frac{1}{4} 0 \frac{1}{2}$, means that
$$
q=\frac{1}{4} \quad r=0 \quad s=\frac{1}{2}
$$
or that, using equation (q a=\text { lattice position referenced to the } x \text { axis }) through equation (s c=\text { lattice position referenced to the } z \text { axis }), ... | ([1 1 0], [1 1 0]) | multiple | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.59 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.60 | FORMULA | (a) What are the direction indices for a vector that passes from point $\frac{1}{3} \frac{1}{2} 0$ to point $\frac{2}{3} \frac{3}{4} \frac{1}{2}$ in a tetragonal unit cell?
(b) Repeat part (a) for a rhombohedral unit cell.
Give your answer as a tuple (answer to (a), answer to (b)). | (a) Point coordinate indices for the vector tail, $\frac{1}{3} \frac{1}{2} 0$, means that
$$
q=\frac{1}{3} \quad r=\frac{1}{2} \quad s=0
$$
or that. using equation (q a=\text { lattice position referenced to the } x \text { axis }) through equation (s c=\text { lattice position referenced to the } z \text { axis }), ... | ([4 3 6], [4 3 6]) | multiple | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.60 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.62 | FORMULA | Convert the [110] and [00 $\overline{1}]$ directions into the four-index Miller-Bravais scheme for hexagonal unit cells. Give your answer as a tuple: (conversion of [110], conversion of [00 $\overline{1}$]) | We are asked to convert [110] and [00 $\overline{1}]$ directions into the four-index Miller-Bravais scheme for hexagonal unit cells. For [110]
$$
\begin{aligned}
& U=1 \\
& V=1 \\
$& W=0$
\end{aligned}
$$
From equation (u=\frac{1}{3}(2 U-V)) through equation (w=W)
$$
\begin{gathered}
u=\frac{1}{3}(2 U-V)=\frac{1}{3}... | ([1 1 -2 0], [0 0 0 -1]) | single | easy | Fundamental: Crystal Structure | Composites | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.62 | |||||||
Fundamentals_of_Materials_Instructors_Example 3.66 | FORMULA | What are the indices for the two planes drawn in the following sketch?
Give your answer as a tuple (index of plane 1, index of plane 2). | In order to solve for the $h, k$, and $l$ indices for these two crystallographic planes it is necessary to use equations (h=\frac{n a}{A}), (k=\frac{n b}{B}) and (l=\frac{n c}{C}).
For Plane 1, the intercepts with the $x, y$, and $z$ axes are $\frac{a}{2}(0.2 \mathrm{~nm}), b(0.4 \mathrm{~nm})$, and $c(0.2 \mathrm{~nm... | ((2 1 1), (0 -1 0)) | multiple | medium | Materials: Metals | Metals | Composites | Diffusion & Kinetics | Fundamentals_of_Materials_Instructors | Example 3.66 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.68 | FORMULA | Determine the Miller indices for the planes shown in the following unit cell:
Give your answer as a tuple (the Miller indices of plane A, plane B). | For plane A , the first thing we need to do is determine the intercepts of this plane with the $x, y$, and $z$ axes. If we extend the plane back into the plane of the page, it will intersect the $z$ axis at $-c$. Furthermore, intersections with the $x$ and $y$ axes are, respectively, $a$ and $b$. The is, values of the ... | ($(11\bar{1})$, (230)) | multiple | easy | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.68 | |||||||
Fundamentals_of_Materials_Instructors_Example 3.69 | FORMULA | Determine the Miller indices for the planes shown in the following unit cell.
Give your answer as a tuple (the Miler indices of plane A, plane B). | For plane A , we will move the origin of the coordinate system one unit cell distance to the right along the $y$ axis. Referenced to this new origin, the plane's intersections with with the $x$ and $y$ axes are $a / 2,-b / 2$; since it is parallel to the $z$ axis, the intersection is taken as and $\infty$ - these three... | ($(1\bar{1}0), (122)) | multiple | easy | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.69 | |||||||
Fundamentals_of_Materials_Instructors_Example 3.70 | FORMULA | Determine the Miller indices for the planes shown in the following unit cell.
Give your answer as a tuple (the Miller indices for plane A, plane B). | Since Plane A passes through the origin of the coordinate system as shown, we will move the origin of the coordinate system one unit cell distance vertically along the $z$ axis. Referenced to this new origin, intercepts with the $x, y$, and $z$ axes are, respectively, $\frac{a}{2}, b$, and $-c$. If we assume that the v... | ($(21\bar{1})$, $(02\bar{1})$) | multiple | easy | Fundamental: Crystal Structure | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.70 | |||||||
Fundamentals_of_Materials_Instructors_Example 3.76 | NUM | The figure below shows a unit cell with atoms at all eight corners and at the centers of all six faces. If the density of the metal is 18.91 g cm⁻³, what is its atomic weight? | The unit cells constructed below show the three crystallographic planes that were provided in the problem statement.
This unit cell belongs to the orthorhombic crystal system since $a=0.25 \mathrm{~nm}, b=0.30 \mathrm{~nm}, c=0.20 \mathrm{~nm}$, and $\alpha=\beta=\gamma=90^{\circ}$. This crystal structure would be call... | 42.7 | g/mol | single | medium | Materials: Metals | Metals | Thermal | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.76 | ||||
Fundamentals_of_Materials_Instructors_Example 3.77 | FORMULA | Convert the (111) and (0 -1 2) planes into the four-index Miller-Bravais scheme for hexagonal unit cells.
Give your answer as a tuple (converted (111), converted (012)). | This problem asks that we convert (111) and $(0 \overline{1} 2)$ planes into the four-index Miller-Bravais scheme, (hkil), for hexagonal cells. For (111), $h=1, k=1$, and $l=1$, and, from Equation 3.15, the value of $i$ is equal to
$$
i=-(h+k)=-(1+1)=-2
$$
Therefore, the (111) plane becomes $(11 \overline{2} 1)$.
No... | ((1 1 -2 1), (0 -1 1 2)) | multiple | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.77 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.80 | FORMULA | (a) Derive linear density expressions for FCC [100] and [111] directions in terms of the atomic radius $R$.
(b) Compute and compare linear density values for these same two directions for copper $(\mathrm{Cu})$.
The atomic radius for copper is 0.128 nm .
Give your answer as a tuple (linear density expressions for FCC ... | (a) In the figure below is shown a [100] direction within an FCC unit cell.
For this [100] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalent of 1 atom that is centered on the direction vector. The length of this direction vector is just the unit cell edge length, $2 ... | ($\frac{1}{2 R \sqrt{2}}$, $\frac{1}{2 R \sqrt{6}}$, 2.76e9, 1.59e9) | (,,$m^{-1}$,$m^{-1}$) | multiple | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.80 | |||||
Fundamentals_of_Materials_Instructors_Example 3.81 | FORMULA | (a) Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius $R$.
(b) Compute and compare linear density values for these same two directions for iron (Fe).
The atomic radius for iron is 0.124 nm .
Give your answer as a tuple (LD expression for BCC [110], LD expression for BCC... | (a) In the figure below is shown a [110] direction within a BCC unit cell.
For this [110] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of this direction vector is denoted by $x$ in this figure, whic... | ($\frac{\sqrt{3}}{4R\sqrt{2}}$, $\frac{1}{2 R}$, 2.47e9, 4.03e9) | (,,$m^{-1}$,$m^{-1}$) | multiple | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.81 | |||||
Fundamentals_of_Materials_Instructors_Example 3.82 | FORMULA | (a) Derive expressions for the planar density of the {100} and {111} planes in an FCC crystal in terms of the atomic radius R.
(b) Using R = 0.143 nm for aluminum, compute the numerical values of these two planar densities for Al.
The atomic radius for aluminum is 0.143 nm .
Give your answer as a tuple: (PD₁₀₀ expr... | (a) In the figure below is shown a (100) plane for an FCC unit cell.
For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms associated with this ... | ($\frac{1}{4R^2}$, $\frac{1}{2 R^{2} \sqrt{3}}$, 12.23, 14.12) | (,, $nm^{-2}$, $nm^{-2}$) | multiple | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Plastic | Fundamentals_of_Materials_Instructors | Example 3.82 | ||||
Fundamentals_of_Materials_Instructors_Example 3.83 | FORMULA | (a) Derive planar density expressions for BCC (100) and (110) planes in terms of the atomic radius $R$.
(b) Compute and compare planar density values for these same two planes for molybdenum (Mo).
the atomic radius for molybdenum is 0.136 nm .
Give your answer as a tuple (planar density expression for BCC [100], planar... | (a) A BCC unit cell within which is drawn a (100) plane is shown below.
For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells. Thus, there is the equivalence of 1 atom associated with this BCC (100) plane. The planar section represented in the a... | ($\frac{3}{16 R^{2}}$, $\frac{3}{8 R^{2} \sqrt{2}}$,1.014e19, 1.434e19) | (,,$m^{-2}$,$m^{-2}$) | multiple | hard | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.83 | |||||
Fundamentals_of_Materials_Instructors_Example 3.84 | FORMULA | (a) Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius $R$.
(b) Compute the planar density value for this same plane for titanium (Ti).
The atomic radius for titanium is 0.145 nm .
Give your answer as a tuple (planar density expression, planar density value for Ti) | (a) A (0001) plane for an HCP unit cell is show below.
Each of the 6 perimeter atoms in this plane is shared with three other unit cells, whereas the center atom is shared with no other unit cells; this gives rise to three equivalent atoms belonging to this plane.
In terms of the atomic radius $R$, the area of each ... | ($\frac{1}{2 R^{2} \sqrt{3}}$, 1.373e19) | (, $m^{-2}$) | multiple | medium | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.84 | |||||
Fundamentals_of_Materials_Instructors_Example 3.86 | NUM | The corundum crystal structure, found for $\mathrm{Al}{2} \mathrm{O}{3}$, consists of an HCP arrangement of $\mathrm{O}^{2-}$ ions; the $\mathrm{Al}^{3+}$ ions occupy octahedral positions.
What fraction of the available octahedral positions are filled with $\mathrm{Al}^{3+}$ ions? | This question is concerned with the corundum crystal structure in terms of close-packed planes of anions.
For this crystal structure, two-thirds of the octahedral positions will be filled with $\mathrm{Al}^{3+}$ ions since there is one octahedral site per $\mathrm{O}^{2-}$ ion, and the ratio of $\mathrm{Al}^{3+}$ to $\... | 2/3 | single | easy | Materials: Semiconductors | Semiconductors | Micro/Nano-structure | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.86 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.87 | NUM | Beryllium oxide ( BeO ) may form a crystal structure that consists of an HCP arrangement of $\mathrm{O}^{2-}$ ions. If the ionic radius of $\mathrm{Be}^{2+}$ is 0.035 nm , r_{\mathrm{O}^{2-}} = 0.140mn, then what fraction of these available interstitial sites will be occupied by $\mathrm{Be}^{2+}$ ions? | We are now asked what fraction of these available interstitial sites are occupied by $\mathrm{Be}^{2+}$ ions. Since there are two tetrahedral sites per $\mathrm{O}^{2-}$ ion, and the ratio of $\mathrm{Be}^{2+}$ to $\mathrm{O}^{2-}$ is $1: 1$, one-half of these sites are occupied with $\mathrm{Be}^{2+}$ ions. | 1/2 | single | easy | Materials: Ceramics | Ceramics | Micro/Nano-structure | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.87 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.88 | NUM | Iron titanate, $\mathrm{FeTiO}{3}$, forms in the ilmenite crystal structure that consists of an HCP arrangement of $\mathrm{O}^{2-}$ ions.
(a) What fraction of the total tetrahedral sites will be occupied?
(b) What fraction of the total octahedral sites will be occupied?
Give your answer as a tuple (fraction of the to... | (a) Since both $\mathrm{Fe}^{2+}$ and $\mathrm{Ti}^{4+}$ ions occupy octahedral sites, no tetrahedral sites will be occupied.
(b) For every $\mathrm{FeTiO}{3}$ formula unit, there are three $\mathrm{O}^{2-}$ ions, and, therefore, three octahedral sites; since there is one ion each of $\mathrm{Fe}^{2+}$ and $\mathrm{Ti}... | (0, 2/3) | multiple | easy | Materials: Metals | Metals | Micro/Nano-structure | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.88 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.90 | FORMULA | The interplanar spacing $d{h k l}$ for planes in a unit cell having orthorhombic geometry is given by
$$
\frac{1}{d{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{b^{2}}+\frac{l^{2}}{c^{2}}
$$
where $a, b$, and $c$ are the lattice parameters.
(a) To what equation does this expression reduce for crystals having cubic sy... | (a) For the crystals having cubic symmetry, $a=b=c$. Making this substitution into the above equation leads to
$$
\begin{aligned}
\frac{1}{d{h k l}^{2}} & =\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{a^{2}} \\
& =\frac{h^{2}+k^{2}+l^{2}}{a^{2}}
\end{aligned}
$$
(b) For crystals having tetragonal symmetry, $a... | ($\frac{1}{d{h k l}^{2}}=\frac{h^{2}+k^{2}+l^{2}}{a^{2}}$, $\frac{1}{d{h k l}^{2}}=\frac{h^{2}+k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}}$) | multiple | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.90 | ||||||
Fundamentals_of_Materials_Instructors_Example 3.91 | NUM | Using the data for aluminum in table 3.1 , compute the interplanar spacing for the (110) set of planes.
"Table 3.1
Atomic Radii and Crystal Structures for 16 Metals
\begin{tabular}{|l|l|l|l|l|l|}
\hline Metal & Crystal Structure ${ }^a$ & Atomic Radius ${ }^{\text {b }}$ (nm) & Metal & Crystal Structure & Atomic Rad... | From the table 3.1 , aluminum has an FCC crystal structure and an atomic radius of 0.1431 nm . Using equation (a=2 R \sqrt{2}), the lattice parameter $a$ may be computed as
$$
a=2 R \sqrt{2}=(2)(0.1431 \mathrm{~nm}) \sqrt{2}=0.4045 \mathrm{~nm}
$$
Now, since $h=1, k=1$, and $l=0$, the interplanar spacing $d{110}$ is... | 0.2860 | nm | single | easy | Materials: Metals | Metals | Micro/Nano-structure | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.91 | |||||
Fundamentals_of_Materials_Instructors_Example 3.92 | NUM | Using the data for $\alpha$-iron in table 3.1 , compute the interplanar spacings for the (111) and (211) sets of planes.
Give your name as an answer ((111), (211)).
"Table 3.1
Atomic Radii and Crystal Structures for 16 Metals
\begin{tabular}{|l|l|l|l|l|l|}
\hline Metal & Crystal Structure ${ }^a$ & Atomic Radius ${ ... | From the table, $\alpha$-iron has a BCC crystal structure and an atomic radius of 0.1241 nm . Using equation (a=\frac{4 R}{\sqrt{3}}) the lattice parameter, $a$, may be computed as follows:
$$
a=\frac{4 R}{\sqrt{3}}=\frac{(4)(0.1241 \mathrm{~nm})}{\sqrt{3}}=0.2866 \mathrm{~nm}
$$
Now, since $h=k=l$, the $d{111}$ inte... | (0.1655, 0.1170) | (nm, nm) | multiple | easy | Materials: Metals | Metals | Micro/Nano-structure | Crystal Structure | Fundamentals_of_Materials_Instructors | Example 3.92 |
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