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Introduction_to_Materials_Science_for_Engineers_Shackelford_12.45 | NUM | Calculate the modulus of elasticity of a metal-matrix composite under isostrain conditions. Assume an aluminum matrix is reinforced by $60 \mathrm{vol} \% \mathrm{SiC}$ fibers. Give your answer as the value of modulus_of_elasticity. | 2.86E+05 | MPa | single | easy | Materials: Metals | Metals | Mechanical | Composites | Elastic | Introduction_to_Materials_Science_for_Engineers_Shackelford | 12.45 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_12.49 | NUM | Calculate the composite modulus for polyester reinforced with $10 \mathrm{vol} \% \mathrm{Al}_2 \mathrm{O}_3$ whiskers under isostrain conditions. (See Tables for appropriate moduli.) Give your answer as the value of composite_modulus.
Refer to Tables:
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline \multicolumn{8}{|c|}{Mec... | 4.92E+04 | MPa | single | easy | Materials: Polymers | Polymers | Mechanical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 12.49 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_12.51 | NUM | Calculate the elastic modulus and thermal conductivity perpendicular to continuous, reinforcing fibers in an E-glass ( $60 \mathrm{vol} \%$ )/polyester composite. Give your answer as the value of thermal_conductivity. | 0.34 | W/(m*K) | multiple | easy | Materials: Glasses | Glasses | Thermal | Composites | Elastic | Introduction_to_Materials_Science_for_Engineers_Shackelford | 12.51 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_12.55 | NUM | Calculate the error in assuming the isostrain modulus of an epoxy reinforced with $67 \mathrm{vol} \% \mathrm{C}$ fibers is given by Equation($E_c=\nu_m E_m+\nu_f E_f$). (Note Table for experimental data.) Give your answer as the value of error.
(Table:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline \multicolumn{7}{|l|}{Mecha... | 4.1 to 16 | % | single | hard | Materials: Metals | Metals | Mechanical | Composites | Atomic Bonding | Introduction_to_Materials_Science_for_Engineers_Shackelford | 12.55 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_12.59 | NUM | Calculate the specific strength for the $W$ fibers ( $50 \mathrm{vol} \%$ )/copper composite listed in Table. Give your answer as the value of specific_strength.
(Table:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline \multicolumn{7}{|l|}{Mechanical Properties of Common Composite Systems} \\
\hline Class & $$
\begin{gathered}
... | 7.96E+06 | mm | single | hard | Materials: Metals | Metals | Mechanical | Composites | Atomic Bonding | Introduction_to_Materials_Science_for_Engineers_Shackelford | 12.59 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_12.63 | NUM | The Feature Box in this chapter introduced spider silk as an impressive natural material with a specific strength greater than structural steel. Given a tensile strength of a particular spider silk as $1.2 \times 10^3 \mathrm{MPa}$ with a density of $1.30 \mathrm{Mg} / \mathrm{m}^3$, calculate its specific strength and... | 9.42E+07 | mm | single | hard | Materials: Metals | Metals | Mechanical | Composites | Shaping | Introduction_to_Materials_Science_for_Engineers_Shackelford | 12.63 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_12.67 | NUM | Given data on modulus of elasticity (in tension) and tensile strength for various thermoplastic polymers in Table , select the polymers that would meet the following design specifications for a mechanical gear application:
modulus of elasticity, $E$ :
$$
2,000 \mathrm{MPa}<E<3,000 \mathrm{MPa}
$$
and
tensile streng... | 0.341 | MPa | single | hard | Materials: Polymers | Polymers | Mechanical | Composites | Elastic | Introduction_to_Materials_Science_for_Engineers_Shackelford | 12.67 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.1 | NUM | (a) Assume that the circuit in Figure contains, as a sample, a cylindrical steel bar 1 cm diameter $\times 10 \mathrm{~cm}$ long with a conductivity of $7.00 \times 10^6 \Omega^{-1} \cdot \mathrm{~m}^{-1}$. What would be the current in this bar due to a voltage of 10 mV ? (b) Repeat part (a) for a bar of high-purity si... | (55.0, 3.14e-9, 7.85e-19) | (A, A, A) | multiple | medium | Materials: Metals | Metals | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.1 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.3 | NUM | A semiconductor wafer is 0.6 mm thick. A potential of 100 mV is applied across this thickness. (a) What is the electron drift velocity if their mobility is $0.2 \mathrm{~m}^2 /(\mathrm{V} \cdot \mathrm{s})$ ? (b) How much time is required for an electron to move across this thickness? Give your answer as a tuple: (elec... | (33.3, 18.0) | (m/s, μs) | multiple | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.3 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.5 | NUM | A strip of aluminum metallization on a solid-state device is 1 mm long with a thickness of $1 \mu \mathrm{~m}$ and a width of $6 \mu \mathrm{~m}$. What is the resistance of this strip? Give your answer as the value of resistance. | 4.71 | Ω | single | easy | Materials: Metals | Metals | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.5 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.7 | NUM | A structural design involves a steel wire 2 mm in diameter that will carry an electrical current. If the resistance of the wire must be less than $25 \Omega$, calculate the maximum length of the wire, given the data in Table. Give your answer as the value of maximum_length.
(Table:
\begin{tabular}{llc}
\hline Electrica... | 734 | m | single | easy | Materials: Metals | Metals | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.7 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.9 | NUM | At what temperature will the $5.60-\mathrm{eV}$ energy level for electrons in silver be $25 \%$ filled? (The Fermi level for silver is 5.48 eV .) Give your answer as the value of temperature. | 994 | °C | single | easy | Materials: Metals | Metals | Thermal | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.9 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.11 | NUM | What is the probability of an electron's being promoted to the conduction band in indium antimonide, InSb , at (a) $25^{\circ} \mathrm{C}$ and (b) $50^{\circ} \mathrm{C}$ ? Give your answer as a tuple: (probability_25C, probability_50C). | (0.0353, 0.0451) | multiple | medium | Structures: Composites | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.11 | |||||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.13 | NUM | Gallium forms semiconducting compounds with various group VA elements. The band gap systematically drops with increasing atomic number of the VA elements. For example, the band gaps for the III-V semiconductors $\mathrm{GaP}, \mathrm{GaAs}$, and GaSb are 2.25 eV , 1.47 eV , and 0.68 eV , respectively. Calculate the pro... | (3.75e-13, 1.79e-6) | multiple | medium | Materials: Semiconductors | Semiconductors | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.13 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.15 | NUM | A strip of copper metallization on a solid-state device is 1 mm long with a thickness of $1 \mu \mathrm{~m}$ and a width of $6 \mu \mathrm{~m}$. If a voltage of 0.1 V is applied along the long dimension, what is the resulting current? Give your answer as the value of current. | 34.8 | mV | single | easy | Materials: Metals | Metals | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.15 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.21 | NUM | An important application of metal conductors in the field of materials processing is in the form of metal wire for resistance-heated furnace elements. Some of the alloys used as thermocouples also serve as furnace elements. For example, consider the use of a $1-\mathrm{mm}$-diameter chromel wire to produce a $1-\mathrm... | 8.80 | m | single | medium | Materials: Metals | Metals | Electrical | Composites | Diffusion & Kinetics | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.21 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.27 | NUM | If progress in increasing $T_c$ for superconductors had continued at the linear rate followed through 1975, by what year would a $T_c$ of 95 K be achieved? Give your answer as the value of year. | 2210 | single | easy | Materials: Metals | Metals | Electrical | Diffusion & Kinetics | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.27 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.31 | NUM | Calculate the charge density on a 3 mm -thick capacitor made of $99.5 \% \mathrm{Al}_2 \mathrm{O}_3$ under an applied voltage of 1 kV . Give your answer as the value of charge_density. | 2.89E-05 | C/m^2 | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.31 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.33 | NUM | Calculate the charge density on a capacitor made of cordierite at its breakdown dielectric strength of $3 \mathrm{kV} / \mathrm{mm}$. The dielectric constant is 4.5 . Give your answer as the value of charge_density. | 1.2E-04 | C/m^2 | single | easy | Materials: Ceramics | Ceramics | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.33 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.39 | NUM | If the elastic modulus of $\mathrm{BaTiO}_3$ in the $c$-direction is $109 \times 10^3 \mathrm{MPa}$, what stress is necessary to reduce its polarization by $0.1 \%$ Give your answer as the value of stress. | 109 | MPa | single | easy | Properties: Mechanical | Mechanical | Composites | Elastic | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.39 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.43 | NUM | Calculate the fraction of Ge atoms that provides a conduction electron at room temperature. Give your answer as the value of fraction. | 5.2E-10 | single | easy | Properties: Thermal | Thermal | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.43 | |||||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.45 | NUM | What fraction of the conductivity at room temperature for (a) germanium and (b) CdS is contributed by (i) electrons and (ii) electron holes? Give your answer as a tuple: (germanium_electrons, germanium_holes, CdS_electrons, CdS_holes). | (0.657, 0.343, 0.950, 0.050) | multiple | medium | Materials: Semiconductors | Semiconductors | Thermal | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.45 | ||||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.47 | NUM | Using the data in Table, calculate the roomtemperature conductivity of intrinsic InSb. Give your answer as the value of conductivity.
(Table:
\begin{tabular}{|l|l|l|l|l|}
\hline \multicolumn{5}{|c|}{Properties of Some Common Semiconductors at Room Temperature ( 300 K )} \\
\hline Material & Energy gap, $E_g(\mathrm{eV}... | 1.74E+04 | Ω^-1·m^-1 | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.47 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.51 | NUM | There is a slight temperature dependence for the band gap of a semiconductor. For silicon, this dependence can be expressed as $$ E_g(T)=1.152 \mathrm{eV}-\frac{A T^2}{T+B} $$ where $A=4.73 \times 10^{-4} \mathrm{eV} / \mathrm{K}, B=636 \mathrm{~K}$, and $T$ is in Kelvin. What is the percentage error in taking the band... | 4.77 | % | single | medium | Materials: Semiconductors | Semiconductors | Thermal | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.51 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.53 | NUM | An $n$-type semiconductor consists of 100 ppb of P doping, by weight, in silicon. What is (a) the mole percentage $P$ and (b) the atomic density of $P$ atoms? Compare your answer in part (b) with the maximum solid solubility level given in Table. Give your answer as a tuple: (mole_percentage_P, atomic_density_P_atoms).... | (9.07e-6, 4.54e21) | (mol %, atoms/m^3) | multiple | hard | Materials: Semiconductors | Semiconductors | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.53 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.65 | NUM | In designing a solid-state device using B-doped Si, it is important that the conductivity not increase more than $10 \%$ (relative to the value at room temperature) during the operating lifetime. For this factor alone, what is the maximum operating temperature to be specified for this design? Give your answer as the va... | 317 | K | single | medium | Materials: Metals | Metals | Thermal | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.65 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.67 | NUM | (a) It was pointed out that the temperature sensitivity of conductivity in semiconductors makes them superior to traditional thermocouples for certain high-precision temperature measurements. Such devices are referred to as thermistors. As a simple example, consider a wire 0.5 mm in diameter $\times 10 \mathrm{~mm}$ lo... | (1.11e-10, 9.31e-7, 289) | (K, K, K) | multiple | hard | Materials: Metals | Metals | Electrical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.67 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.73 | NUM | Starting from an ambient temperature of 300 K , what temperature increase is necessary to double the conductivity of intrinsic GaAs? Give your answer as the value of temperature_increase. | 7.5 | K | single | hard | Materials: Semiconductors | Semiconductors | Thermal | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.73 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.75 | NUM | What temperature increase (relative to room temperature) is necessary to increase the conductivity of intrinsic GaAs by $1 \%$ ? Give your answer as the value of temperature_increase. | 0.1 | °C | single | hard | Materials: Semiconductors | Semiconductors | Thermal | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.75 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.77 | NUM | In intrinsic semiconductor GaAs , what fraction of the current is carried by electrons, and what fraction is carried by holes? Give your answer as a tuple: (electron_fraction, hole_fraction). | (0.973, 0.027) | multiple | medium | Materials: Semiconductors | Semiconductors | Electrical | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.77 | ||||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.79 | NUM | When the aluminum impurity level in a silicon bar has reached 1 ppb , what would have been the purity of the liquid on the previous pass? | 2.76 | ppm | single | easy | Materials: Metals | Metals | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.79 | ||||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.83 | NUM | The high-frequency operation of solid-state devices can be limited by the transit time of an electron across the gate between the source and drain of an FET. For a device to operate at 1 gigahertz $\left(10^9 \mathrm{~s}^{-1}\right)$, a transit time of $10^{-y} \mathrm{~s}$ is required. (a) What electron velocity is re... | (1e3, 7.14e3, 5.14) | (m/s, V/m, gigahertz) | multiple | hard | Materials: Semiconductors | Semiconductors | Electrical | Composites | Diffusion & Kinetics | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.83 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.89 | NUM | The first Intel microprocessor (4004) contained 2,300 transistors in 1971. Use Moore's Law to estimate the number of transistors that might have been contained on the Pentium III when it was introduced in 1999 and compare your answer to the actual number ( 32 million). Give your answer as the value of estimated_transis... | 3.8E+07 | single | medium | Materials: Semiconductors | Semiconductors | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.89 | ||||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_13.91 | NUM | Calculate the conductivity at $20^{\circ} \mathrm{C}$ parallel to the W filaments in the Cu -matrix composite in Table. Give your answer as the value of conductivity.
(Table:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline \multicolumn{7}{|l|}{Mechanical Properties of Common Composite Systems} \\
\hline Class & $$
\begin{gathe... | 3.81E+07 | Ω^-1·m^-1 | single | easy | Materials: Metals | Metals | Mechanical | Composites | Atomic Bonding | Introduction_to_Materials_Science_for_Engineers_Shackelford | 13.91 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.1 | NUM | A ruby laser produces a beam of monochromatic photons with a wavelength of 694.3 nm . Calculate the corresponding (a) photon frequency and (b) photon energy. Give your answer as a tuple: (photon_frequency, photon_energy). | (3.318e14, 1.786) | (s^-1, eV) | multiple | easy | Structures: Composites | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.1 | ||||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.3 | NUM | By what percentage is the critical angle of incidence different for a lead oxide-containing "crystal" glass (with $n=1.7$ ) compared with plain silica glass? Give your answer as the value of percentage_difference. | 16.9 | % | single | medium | Materials: Glasses | Glasses | Optical | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.3 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.5 | NUM | (a) Consider a translucent orthoclase ceramic with a thin orthoclase glass coating (glaze). What is the maximum angle of incidence at the ceramic-glaze interface to ensure that an observer can see any visible light transmitted through the product (into an air atmosphere)? (Consider only specular transmission through th... | (40.98, 41.47) | (°, °) | multiple | hard | Materials: Ceramics | Ceramics | Optical | Composites | Finishing | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.5 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.7 | NUM | Silica glass is frequently and incorrectly referred to as quartz. This error is the result of shortening the traditional term fused quartz, which described the original technique of making silica glass by melting quartz powder. What is the percentage error in calculating the reflectance of silica glass by using the ind... | 34.0 | % | single | easy | Materials: Glasses | Glasses | Optical | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.7 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.9 | NUM | Calculate the critical angle of incidence for the airnylon 66 interface. Give your answer as the value of critical_angle. | 40.8 | ° | single | easy | Materials: Polymers | Polymers | Optical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.9 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.11 | NUM | We shall find that many materials can be damaged by various forms of radiation. Polymers are often susceptible to ultraviolet-light damage. Calculate the wavelength of ultraviolet light necessary to break the $\mathrm{C}-\mathrm{C}$ single bond. (Bond energies are given in Table.) Give your answer as the value of wavel... | 323 | nm | single | easy | Materials: Polymers | Polymers | Composites | Atomic Bonding | Fracture | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.11 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.15 | NUM | By what factor is the energy of a photon of red visible light ( $\lambda=700 \mathrm{~nm}$ ) greater than that of a photon of infrared light with $\lambda=5 \mu \mathrm{~m}$ ? Give your answer as the value of energy_factor. | 7.14 | single | medium | Structures: Composites | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.15 | |||||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.19 | NUM | For a single-mode fiber with a core index of refraction of $n=1.460$, how long will it take for a single light pulse to travel a length of 1 kilometer along a cable-television line? Give your answer as the value of travel_time. | 4.87 | μs | single | easy | Properties: Optical | Optical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.19 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.21 | NUM | For what range of visible light wavelengths would ZnTe , with a band gap of 2.26 eV , be a photoconductor? Give your answer as a tuple: (min_wavelength, max_wavelength). | (400, 549) | (nm, nm) | multiple | hard | Materials: Semiconductors | Semiconductors | Electrical | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.21 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.23 | NUM | Calculate the induction and magnetization of a diamagnetic material (with $\mu_r=0.99995$ ) under an applied field strength of $2.0 \times 10^5$ amperes $/ \mathrm{m}$. Give your answer as a tuple: (induction, magnetization). | (0.251, 10) | (Wb/m^2, A/m) | multiple | easy | Properties: Magnetic | Magnetic | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.23 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.27 | NUM | The following data are obtained for a metal subjected to a magnetic field:
\begin{tabular}{cc} \hline $\boldsymbol{H}($ amperes $/ \mathbf{m})$ & $\boldsymbol{B}\left(\right.$ weber $\left./ \mathbf{m}^{\mathbf{2}}\right)$ \\ \hline 0 & 0 \\ $4 \times 10^5$ & 0.50263 \\ \hline \end{tabular}
Calculate the relative per... | 0.99995 | single | easy | Materials: Metals | Metals | Magnetic | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.27 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.29 | NUM | The following data are obtained for an armco iron alloy during the generation of steady-state ferromagnetic hysteresis loop: \begin{tabular}{cc} \hline $\boldsymbol{H}($ amperes $/ \mathrm{m})$ & $\boldsymbol{B}\left(\right.$ weber $\left./ \mathrm{m}^{\mathbf{2}}\right)$ \\ \hline 56 & 0.50 \\ 30 & 0.46 \\ 10 & 0.40 \... | (0.36, -25) | (Wb/m^2, A/m) | multiple | easy | Materials: Metals | Metals | Magnetic | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.29 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.31 | NUM | The following data are obtained for a nickel-iron alloy during the generation of a steady-state ferromagnetic hysteresis loop: \begin{tabular}{cc} \hline $\boldsymbol{H}$ (amperes/m) & $\boldsymbol{B}\left(\right.$ weber $\left./ \mathbf{m}^{\mathbf{2}}\right)$ \\ \hline 50 & 0.95 \\ 25 & 0.94 \\ 0 & 0.92 \\ -10 & 0.90... | (0.92, -18) | (Wb/m^2, A/m) | multiple | medium | Materials: Metals | Metals | Magnetic | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.31 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.35 | NUM | (a) Calculate the magnetic moment of a unit cell of manganese ferrite. (b) Calculate the corresponding saturation magnetization, given a lattice parameter of 0.850 nm . Give your answer as a tuple: (magnetic_moment, saturation_magnetization). | (40, 6.04e5) | (\mu_{\mathrm{B}}, A/m) | multiple | hard | Materials: Metals | Metals | Magnetic | Cellular | Crystal Structure | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.35 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.47 | NUM | The hysteresis loss for soft magnets is generally given in units of $\mathrm{W} / \mathrm{m}^3$. Calculate the loss in these units for the $\mathrm{Fe}-\mathrm{B}$ amorphous metal in Table at a frequency of 60 Hz . Give your answer as the value of hysteresis_loss.
(Table:
\begin{tabular}{|c|c|c|c|}
\hline \multicolumn{... | 1.50 | kW/m^3 | single | easy | Materials: Metals | Metals | Magnetic | Composites | Crystal Structure | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.47 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.55 | NUM | The plot of $H_c$ versus $T$ for a metallic compound, such as $\mathrm{Nb}_3 \mathrm{Ge}$ in Figure, can be approximated by the equation for a parabola, namely $$ H_c=H_0\left[1-\left(\frac{T^2}{T_c^2}\right)\right] $$ where $H_0$ is the critical field at 0 K . Given that $H_c$ for $\mathrm{Nb}_3 \mathrm{Ge}$ is $22 \t... | (38e4, 14) | (A-turns/m, %) | multiple | easy | Materials: Metals | Metals | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.55 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_14.57 | NUM | A commercial ceramic magnet, Ferroxcube A, has a hysteresis loss per cycle of $40 \mathrm{~J} / \mathrm{m}^3$. Assuming the hysteresis loop is traversed at a frequency of 60 Hz , calculate the power loss for this magnet. | 2.4 | kW/m^3 | single | easy | Materials: Ceramics | Ceramics | Magnetic | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 14.57 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.13 | NUM | A Boeing 777 aircraft has the following breakdown of materials usage: Aluminum alloys - $70 \mathrm{wt} \%$ Composites - $11 \mathrm{wt} \%$ Steel - 11 wt \% Titanium alloys - $7 \mathrm{wt} \%$ Other - $1 \mathrm{wt} \%$. Take the average density of an aluminum alloy to be $2.80 \mathrm{Mg} / \mathrm{m}^3$, the averag... | 2.80 | Mg/m^3 | single | medium | Materials: Metals | Metals | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.13 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.19 | NUM | Calculate the band gap, $E_g$, that corresponds to the LEDs in Table made of (a) GaP and (b) GaAsP. Give your answer as a tuple: (band_gap_GaP, band_gap_GaAsP).
Refer to Table:
\begin{tabular}{lcl}
\hline \multicolumn{2}{l}{\begin{tabular}{l}
Common Light-Emitting Diode Compounds \\
and Wavelengths ${ }^{\mathrm{a}}$
... | (2.19, 1.91 to 2.10) | (eV, eV) | multiple | medium | Materials: Semiconductors | Semiconductors | Optical | Composites | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.19 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.21 | NUM | The densities for three iron oxides are FeO $\left(5.70 \mathrm{Mg} / \mathrm{m}^3\right), \mathrm{Fe}_3 \mathrm{O}_4\left(5.18 \mathrm{Mg} / \mathrm{m}^3\right)$, and $\mathrm{Fe}_2 \mathrm{O}_3$ $\left(5.24 \mathrm{Mg} / \mathrm{m}^3\right)$. Calculate the Pilling-Bedworth ratio for iron relative to each type of oxid... | (1.78, 2.10, 2.15) | multiple | medium | Materials: Metals | Metals | Surface Texture | Finishing | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.21 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.25 | NUM | In an ionic-concentration corrosion cell involving nickel (forming $\mathrm{Ni}^{2+}$ ), an electrical current of 5 mA is measured. How many Ni atoms per second are oxidized at the anode? Give your answer as the value of Ni_atoms_per_second. | 1.56E+16 | s^-1 | single | medium | Materials: Metals | Metals | Electrical | Cellular | Shaping | Corrosion | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.25 | ||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.27 | NUM | In an ionic-concentration corrosion cell involving chromium, which forms a trivalent ion $\left(\mathrm{Cr}^{3+}\right)$, an electrical current of 10 mA is measured. How many atoms per second are oxidized at the anode? Give your answer as the value of atoms_per_second. | 2.08E+16 | s^-1 | single | medium | Materials: Metals | Metals | Electrical | Cellular | Corrosion | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.27 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.29 | NUM | In a simple galvanic cell consisting of Co and Cr electrodes immersed in 1-molar ionic solutions, calculate the cell potential. | 0.467 | V | single | easy | Materials: Metals | Metals | Cellular | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.29 | |||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.33 | NUM | A copper-nickel ( $35 \mathrm{wt} \%-65 \mathrm{wt} \%$ ) alloy is corroded in an oxygen-concentration cell using boiling water. What volume of oxygen gas (at 1 atm ) must be consumed at the cathode to corrode 10 g of the alloy? (Assume only divalent ions are produced.) | 2.54E-03 | m^3 | single | medium | Materials: Metals | Metals | Cellular | Corrosion | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.33 | ||||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.35 | NUM | For the rusting mechanism illustrated in Figure, calculate the volume of $\mathrm{O}_2$ gas consumed (at STP) in the production of 100 gm of rust $\left[\mathrm{Fe}(\mathrm{OH})_3\right]$. | 0.0157 | m^3 | single | easy | Materials: Metals | Metals | Composites | Diffusion & Kinetics | Corrosion | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.35 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.37 | NUM | In designing the hull of a new fishing vessel to ensure corrosion protection, you find that a sacrificial anode of zinc provides an average corrosion current of 2 A over the period of 1 year. What mass of zinc is required to give this protection? | 21.8 | kg | single | easy | Materials: Metals | Metals | Electrical | Cellular | Corrosion | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.37 | |||||
Introduction_to_Materials_Science_for_Engineers_Shackelford_15.39 | NUM | The maximum corrosion-current density in a galvanized steel sheet used in the design of the new engineering laboratories on campus is found to be $5 \mathrm{~mA} / \mathrm{m}^2$. What thickness of the zinc layer is necessary to ensure at least (a) 1 year and (b) 5 years of rust resistance? Give your answer as a tuple: ... | (7.50, 37.5) | (μm, μm) | multiple | hard | Materials: Metals | Metals | Electrical | Composites | Corrosion | Introduction_to_Materials_Science_for_Engineers_Shackelford | 15.39 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 1.1 | NUM | Calculate the maximum possible packing fractions for the (a) simple cubic (SC), (b) face-centered cubic (FCC), and (c) body-centered cubic (BCC) structures. Assume that all atoms have a radius $r$ and the unit-cell parameter is $a$.
Give your answer as a tuple (SC packing fraction, FCC packing fraction, BCC packing f... | a:
description: In the SC structure, atoms touch along the cube edges.
relation: $a = 2r$
packing_fraction: $\frac{\pi}{6} \approx 0.52$
void_space: 48%
b:
description: In the FCC structure, atoms touch along the face diagonals.
relation: $a = 2\sqrt{2}r$
packing_fraction: $\frac{\pi}{3\sqrt{2}} \approx 0... | (0.52, 0.74, 0.68) | No supplementary information needed. | multiple | hard | Fundamental: Crystal Structure | Cellular | Crystal Structure | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 1.1 | ||||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 1.2 | FORMULA | What are the Miller indices for any one of the close-packed (CP) directions in FCC and BCC structures? Give your answer as a tuple (FCC, BCC). | In an FCC structure, the atoms touch along the face diagonal. Therefore, the face diagonals are the CP directions. One face diagonal (marked as OA) is shown in Figure. We follow the procedure described earlier for obtaining the Miller indices of direction:
1. The coordinates of the 'head' point A are $1,1,0$.
2. The c... | ([110], [111]) | No supplementary information needed. | multiple | medium | Materials: Metals | Metals | Crystal Structure | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 1.2 | ||||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 1.3 | FORMULA | What are the Miller indices for the plane P shown in Figure? | We see that this plane, shown in Figure, intersects the $x$ - and $y$-axes at a length of " $1 x$ " lattice parameter. The plane is parallel to the $z$-axis, that is, it does not intersect the $z$-axis at all; hence, this intercept is $\infty$. We follow the directions for establishing the Miller indices of a plane as
... | -110 | No supplementary information needed. | single | easy | Fundamental: Crystal Structure | Composites | Crystal Structure | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 1.3 | ||||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 1.4 | NUM | X-rays of a single wavelength $(\lambda)$ were used to analyze a glittering sample suspected to be Au. The X-ray diffraction (XRD) analysis, by which the lattice constant of a unit cell is determined, showed that the (400) planes in this sample were separated by a distance of $0.717 \AA$. What is the lattice constant f... | Given $d_{400} = 0.717 \AA$, we use the equation for interplanar spacing in a cubic system:
$$
d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}
$$
For the (400) planes, this simplifies to:
$$
d_{400} = \frac{a}{\sqrt{4^2 + 0^2 + 0^2}} = \frac{a}{4} = 0.717 \AA
$$
Therefore, the lattice constant for this sample is $a = (0... | 2.868 | Å | No supplementary information needed. | single | easy | Properties: Thermal | Thermal | Cellular | Crystal Structure | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 1.4 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 1.5 | NUM | NaCl and potassium chloride $(\mathrm{KCl})$ have the same stoichiometry. The radius of the potassium ion $(\mathrm{K}^{+})$ is 0.133 nm. The radius of the chlorine ion $(\mathrm{Cl}^{-})$ is 0.181 nm. What will be the expected coordination number (CN) for $\mathrm{K}^{+}$ ions? | The $r_{\text{cation}} / r_{\text{anion}}$ for KCl is $0.133 / 0.181=0.735$. This suggests that the CN for $\mathrm{K}^{+}$ ions will be 8. | 8 | * The relationship between the cation-to-anion radius ratio and the corresponding coordination number. Specifically, the radius ratio range that corresponds to a coordination number of 8. | single | easy | Properties: Thermal | Thermal | Composites | Crystal Structure | Fracture | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 1.5 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 1.6 | NUM | The radius of Si in a covalent structure is $1.176 \AA$. What is the lattice constant of Si? What is the theoretical density of Si if its atomic mass is 28.1? Using Avogadro's number ($6.023 \times 10^{23}$ atoms per mole). Give your answer as a tuple (lattice constant, theoretical density). | In the diamond cubic (DC) structure exhibited by Si, the body diagonal can be analyzed to find the lattice constant. The body diagonal is $\sqrt{3} a$, where $a$ is the length of the unit cell. Given the radius $r$ of the atoms in the DC structure, the relationship $\sqrt{3} \times a = 8 r$ holds. For Si, substituting ... | (5.4317, 2.33) | (\AA, g/cm^3) | No supplementary information needed. | multiple | easy | Materials: Metals | Metals | Cellular | Crystal Structure | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 1.6 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 1.7 | NUM | If the lattice constant (a) of InP is about $5.8687 \AA$, what is its theoretical density? The atomic masses of indium (In) and phosphorus (P) are 114.81 and 31, respectively. The structure of InP is similar to the zinc blende structure, where there are four In atoms and four P atoms inside the InP unit cell. | To calculate the theoretical density of InP, we use the formula:
Density of indium phosphide $=\frac{\text { Mass of four In atoms }+ \text { mass of four } \mathrm{P} \text { atoms }}{\text { Volume of the unit cell }}$
Where the volume of the unit cell is $(5.8687 \times 10^{-8} \mathrm{~cm})^{3}$. The mass of fo... | 4.79 | g/cm^3 | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Cellular | Crystal Structure | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 1.7 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.1 | NUM | The conductivity of an Ag sample is listed as $62.9 \times 10^{4} \mathrm{~S} / \mathrm{cm}$. What is the resistivity in $\mu \Omega \cdot \mathrm{cm}$? | 1. The Cu sample resistivity in $\Omega \cdot \mathrm{cm} = 1.673 \mu \Omega \cdot \mathrm{cm} \times 10^{-6} \Omega / \mu \Omega = 1.673 \times 10^{-6} \Omega \cdot \mathrm{cm}$. The Cu sample resistivity in $\mathrm{n} \Omega \cdot \mathrm{m} = 1.673 \times 10^{-6} \Omega \cdot \mathrm{cm} \times 10^{9} \mathrm{n} \O... | 1.589 | $ \mu \Omega \cdot \mathrm{cm}$ | No supplementary information needed. | single | easy | Properties: Electrical | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.1 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.2 | NUM | 1. An antenna is made by etching solid Cu with $\sigma=5.8 \times 10^{7} \mathrm{~S} / \mathrm{m}$. What is its electrical resistance (R)? (Assume that the antenna is in the form of a strip that is 20 cm long, 5 mm wide, and $25 \mu \mathrm{~m}$ thick.)
2. What will be the resistance of a similar antenna made by screen... | 1. Assuming the current flows along the direction of the length, the cross-sectional area will be $5 \mathrm{~mm} \times 25 \mu \mathrm{~m}$ or $0.5 \mathrm{~cm} \times 25 \mu \mathrm{~m} \times 10^{-4} \mathrm{~cm} / \mu \mathrm{m}=125 \times 10^{-5} \mathrm{~cm}^{2}$. Using the formula $R=\rho \frac{L}{A}$, the resis... | (0.02758, 1.0) | ($\Omega$,$\Omega$) | No supplementary information needed. | multiple | easy | Materials: Metals | Metals | Electrical | Micro/Nano-structure | Finishing | Plastic | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.2 | ||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.3 | NUM | Calculate the resistance of 1000 m of American wire gauge (AWG) #18 Cu wire (see Table). Assume $\sigma_{\mathrm{Cu}}=5.8 \times 10^{7} \mathrm{~S} / \mathrm{m}$.
Refer to Table:
American Wire Gauge Diameter Conversion
\begin{tabular}{|l|l|l|}
\hline American Wire Gauge Number & Conductor Diameter (inches) & Conductor... | From Table, for AWG #18, the wire diameter (d) is 0.040303 in (1.023696 mm).
Thus, the area of cross section $=\frac{\pi d^{2}}{4}=8.2306 \times 10^{-7} \mathrm{~m}^{2}$
Therefore, the resistance of the 1000-meter-long wire will be as follows (according to Equation $R=\rho \frac{L}{A}$ where $\rho$ is the resistivity... | 20.9478 | $\Omega$ | No supplementary information needed. | single | easy | Properties: Electrical | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.3 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.6 | NUM | Calculate the concentration of conduction electrons in (a) Al and (b) Si. Assume that the densities of Al and Si are $2.7 \mathrm{~g} / \mathrm{cm}^{3}$ and $2.33 \mathrm{~g} / \mathrm{cm}^{3}$, respectively. The atomic masses for Al and Si are $\sim 27$ and $28$, respectively. Assuming that the electron mobility value... | 1. For Al, which has a valence of +3, we assume each Al atom donates three conduction electrons. The atomic mass of Al is $27 \mathrm{~g}$, meaning the mass of $6.023 \times 10^{23}$ atoms (Avogadro's number) is $27 \mathrm{~g}$. A volume of $1 \mathrm{~cm}^{3}$ of Al weighs $2.7 \mathrm{~g}$. The number of atoms in th... | (1.807e23, 2.0e23) | (electrons/cm^3, electrons/cm^3) | * Valence of Aluminum (Al) is 3.
* Valence of Silicon (Si) is 4. | multiple | hard | Materials: Metals | Metals | Electrical | Composites | Atomic Bonding | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.6 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.8 | NUM | Given that the mobility of electrons in copper is $32 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}$ and conductivity σ=5.97×10^5 S/cm. What is the number of conduction electrons contributed per Cu atom?
Reference: $\begin{gathered}\rho=8.96 \mathrm{~g} / \mathrm{cm}^3=8960 \mathrm{~kg} / \mathrm{m}^3, \quad M=63.546... | We are given the conductivity and mobility of electrons in copper as:
\[
\sigma = 5.97 \times 10^5\ \mathrm{S/cm}, \quad \mu = 32\ \mathrm{cm^2/V \cdot s}
\]
Using the relation:
\[
\sigma = n e \mu \quad \Rightarrow \quad n = \frac{\sigma}{e \mu}
\]
Substituting values (and converting mobility to SI units, \(\mu = 32... | 1.38 | No supplementary information needed. | single | medium | Materials: Metals | Metals | Thermal | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.8 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.9 | NUM | A particular single crystal of a semiconducting Si sample, containing a small but deliberately added level of phosphorus (P), provides conductivity with electrons as the majority carriers. Assume that the concentration of P added is $10^{18}$ atoms $/ \mathrm{cm}^{3}$ and that each P atom provides one electron. Ignore ... | Because each P atom contributes only one conduction electron, the concentration of P atoms and that of the conduction electrons donated by P atoms are equal; therefore, $n=10^{18}$ electrons $/ \mathrm{cm}^{3}$. From Equation ($\sigma=\mu_{\mathrm{n}} \times n \times q$),
$$
\sigma=\left(10^{18} \text { electrons } / ... | 8.9E-03 | $\Omega \cdot \mathrm{cm}$ | No supplementary information needed. | single | medium | Materials: Semiconductors | Semiconductors | Electrical | Micro/Nano-structure | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.9 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.10 | NUM | A resistance-temperature detector (RTD) utilizes a metal wire, such as platinum (Pt), and is capable of operating within a temperature range of $-200^{\circ} \mathrm{C}$ to $+700^{\circ} \mathrm{C}$. The principle behind its operation is the change in electrical resistance with temperature, allowing temperature measure... | part_a:
calculation: The relationship between resistance and temperature is given by the formula: $$\begin{aligned}\alpha_{R} & =\frac{1}{\rho_{0}}\left(\frac{\rho-\rho_{0}}{T-T_{0}}\right) \\\therefore 0.00393 & =\frac{1}{100}\left(\frac{200-100}{T-0}\right) \\\therefore T & =\frac{200-100}{0.393}=254.45^{\circ} \ma... | (254.45, 259.7) | (${}^{\circ}C$, ${}^{\circ}C$) | No supplementary information needed. | multiple | easy | Materials: Metals | Metals | Electrical | Composites | Diffusion & Kinetics | Corrosion | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.10 | ||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.11 | NUM | A Cu bus bar, or a conductor used in power transmission, is 300 ft long and has a cross section of $1 / 4 \times 4$ in. (a) What is the resistance of this bar at room temperature? (b) Calculate the Joule losses in kilowatts if the current is 1000 A. Assume that the current is DC. (c) What is the energy lost (in kW $\cd... | 1. The resistance of the Cu bus bar at room temperature is calculated as follows:
$$
\begin{aligned}
R_{\text{Cu bus bar}}=\rho \frac{L}{A} & =\frac{300 \mathrm{ft} . \times 12 \mathrm{in} . / \mathrm{ft} . \times 2.54 \mathrm{~cm} / \mathrm{in} .}{\left(\frac{1}{4} \times 4\right) \mathrm{in} .^{2} \times\left(2.54 \... | (2.44366 \times 10^{-3}, 2.443, 58.6, 2566) | (Ω, kW, kW·h, $) | No supplementary information needed. | multiple | easy | Properties: Electrical | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.11 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.12 | NUM | What is the resistivity of an Au alloy containing 1.5 weight $\% \mathrm{Cu}$? Assume that Nordheim's coefficient (C) for Cu dissolved in Au is $450 \mathrm{n} \Omega \cdot \mathrm{m}$.
Reference: The resistivity of Au is $2.35 \mu \Omega \cdot \mathrm{~cm}$ or $23.5 \mathrm{n} \Omega \cdot \mathrm{m}$. $M_{\mathrm{Au... | The resistivity of Au is $2.35 \mu \Omega \cdot \mathrm{cm}$ or $23.5 \mathrm{n} \Omega \cdot \mathrm{m}$. The concentration of the alloy-forming element has to be expressed as an atom fraction, which is achieved by applying the following equation:
$$
x=\frac{M_{\mathrm{Au}} w}{(1-w) M_{\mathrm{Cu}}+w M_{\mathrm{Au}}}... | 40.69 | nΩ·m | No supplementary information needed. | single | medium | Materials: Metals | Metals | Optical | Composites | Shaping | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.12 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 2.14 | FORMULA | Write down the electronic configuration of a $\mathrm{Fe}(Z=26)$ atom, explaining the meaning of the different symbols used and the principles governing the filling order of electrons. | The electronic configuration of a $\mathrm{Fe}(Z=26)$ atom is determined by the principles of quantum mechanics, including the Pauli exclusion principle and the order in which different energy levels are filled. Starting with the lowest energy level ($n=1$), the configuration is built up as follows:
1. For $n=1$, the ... | 1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^6 3\mathrm{s}^2 3\mathrm{p}^6 4\mathrm{s}^2 3\mathrm{d}^6 | No supplementary information needed. | single | medium | Materials: Metals | Metals | Magnetic | Composites | Atomic Bonding | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 2.14 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.1 | NUM | What is the resistivity ($\rho$) of essentially pure Ge at 300 K? Assume that the mobilities of the electrons ($\mu_{\mathrm{n}}$) and the holes ($\mu_{\mathrm{p}}$) in Ge at 300 K are 3900 and $1900 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}$, respectively. The intrinsic carrier concentration ($n_{\mathrm{i}}$) is... | To calculate the resistivity ($\rho$) of essentially pure Ge at 300 K, we first compute the conductivity ($\sigma$) using the given mobilities of electrons ($\mu_{\mathrm{n}} = 3900 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}$) and holes ($\mu_{\mathrm{p}} = 1900 \mathrm{~cm}^{2} / \mathrm{V} \cdot \mathrm{s}$), and... | 43.1 | Ω⋅cm | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.1 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.2 | FORMULA | Develop a relationship between the wavelength of light ($\lambda$ in micrometers) emitted from an LED and the band gap ($E_g$ in electron volts). | The relationship between the band gap energy ($E_{\mathrm{g}}$ in eV) and the wavelength of light ($\lambda$ in micrometers) emitted from an LED can be derived from the equation $E_{\mathrm{g}}=\frac{h c}{\lambda}$, where $h$ is Planck's constant and $c$ is the speed of light. Simplifying this for practical application... | $\frac{1.24}{E_{\mathrm{g}}}$ | No supplementary information needed. | single | hard | Materials: Semiconductors | Semiconductors | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.2 | ||||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.3 | NUM | An LED is made using gallium nitride ($\mathrm{GaN}; E_{\mathrm{g}}=3.47 \mathrm{eV}$ at $T=0 \mathrm{~K}$). What is the wavelength emitted from this semiconductor LED? | Using Equation ($\lambda(\mathrm{in} \mu \mathrm{m})=\frac{1.24}{E_{\mathrm{g}}(\mathrm{in} \mathrm{eV})} \mu \mathrm{m}$), the wavelength of light emitted from the LED can be calculated as follows:
$$
\begin{aligned}
\lambda(\text { in } \mu \mathrm{m}) & =\frac{1.24}{E_{\mathrm{g}}(\text { in } \mathrm{eV})} \mu \ma... | 357 | nm | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.3 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.4 | NUM | Equation is given by:
$$\frac{m_{0} q^{4}}{2\left(4 \pi \varepsilon_{0}\right)^{2} \hbar^{2}}$$
where $m_{0}=9.11 \times 10^{-31} \mathrm{~kg}$ is the mass of an electron, $q=1.6 \times 10^{-19} \mathrm{C}$ is the charge of an electron, $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$ is the permittivi... | To show that the term expressed in Equation is equal to 13.6 eV, we substitute the given values into the equation:
$$\frac{m_{0} q^{4}}{2\left(4 \pi \varepsilon_{0}\right)^{2} \hbar^{2}} = \frac{\left(9.11 \times 10^{-31} \mathrm{~kg}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{4}}{2\left(4 \times 3.14 \times 8... | 13.6 | eV | * The expected value of the expression in electron volts (13.6 eV). | single | easy | Materials: Metals | Metals | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.4 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.5 | NUM | Assuming that Si acts as a donor in a GaAs sample, calculate the position of the donor energy level ($E_{\mathrm{d}}$) relative to the conduction band edge ($E_{\mathrm{c}}$). For GaAs, $\frac{m^{*}}{m_{0}}=0.067$ and $\varepsilon_{\mathrm{r}}=13.2$ for GaAs. | Using the given values in the equation: $$ E_{\mathrm{d}}=E_{\mathrm{c}}-13.6 \frac{0.067}{(13.2)^{2}} \mathrm{eV} $$ we find: $$ E_{\mathrm{d}}=E_{\mathrm{c}}-5.22 \times 10^{-3} \mathrm{eV} $$ Thus, the donor energy level for Si in GaAs is approximately $5.2 \mathrm{meV}$ below the conduction band edge. This calculat... | 5.22 | meV | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.5 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.6 | NUM | Calculate the thermal energy $\left(k_{\mathrm{B}} T\right)$ of carriers at 300 K. (Note: Boltzmann's constant $\left[k_{\mathrm{B}}\right]$ is $8.617 \times 10^{-5} \mathrm{eV} / \mathrm{K}$ or $1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$.) | Since we prefer to obtain the energy in electron volts, we use the value of kB in electron volts. The value of kBT at 300 K is ~0.026 eV . | 0.026 | eV | No supplementary information needed. | single | medium | Properties: Thermal | Thermal | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.6 | |||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.7 | NUM | Calculate the position of the donor energy level ($E_{\mathrm{d}}$) relative to the conduction band edge ($E_{\mathrm{c}}$) for phosphorus-doped silicon. The dielectric constant ($\varepsilon_{\mathrm{r}}$) of silicon is 11.8, and the ratio of the conductivity effective mass of electrons in silicon to the free electron... | To calculate the position of the donor energy level relative to the conduction band edge for phosphorus-doped silicon, we use the modified hydrogen-like model equation:
$$
E_{\mathrm{d}}=E_{\mathrm{c}}-13.6 \frac{\left(\frac{m_{\mathrm{o}}^{*}}{m_{0}}\right)}{\left(\varepsilon_{\mathrm{r}}\right)^{2}} \mathrm{eV}
$$
... | 0.025 | eV | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.7 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.9 | NUM | Calculate the resistivity of donor-doped GaAs with $N_{\mathrm{d}}=10^{13}$ atoms $/ \mathrm{cm}^{3}$. Assume that $T=300 \mathrm{~K}$ and that all the donors are ionized. For GaAs at $300 \mathrm{~K}, n_{\mathrm{i}}=2.0 \times 10^{6}$ electrons $/ \mathrm{cm}^{3}$. In this case, $N_{\mathrm{d}} \gg n_{\mathrm{i}}$; th... | The conductivity ($\sigma$) is calculated as follows:
$$
\begin{aligned}
\sigma= & \left(1.6 \times 10^{-19} \mathrm{C}\right)\left(10^{13}\right)(8000) \\
& \therefore \sigma=1.28 \times 10^{-3} \Omega^{-1} \cdot \mathrm{~cm}^{-1}
\end{aligned}
$$
The resistivity ($\rho$) is the inverse of conductivity, thus:
$$
\r... | 781 | Ω⋅cm | * Resistivity of intrinsic GaAs at 300 K ($3.51 \times 10^{8} \Omega \cdot \mathrm{~cm}$) | single | medium | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.9 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.11 | NUM | Calculate the band gap of (1) Si and (2) GaAs at 300 K using the Varshni parameters shown in Table.
\begin{tabular}{lccc}
Semiconductor & $\boldsymbol{E}_{\mathbf{g}}($ at $\mathbf{0} \mathbf{K})$ & $\boldsymbol{\alpha}\left(\mathbf{1 0}^{-4} \mathbf{e V} / \mathbf{K}\right)$ & $\boldsymbol{\beta}(\mathbf{K})$ \\
Si... | 1:
description: For Si, $E_{\mathrm{g}}$ at 0 K is $1.170 \mathrm{eV}, \alpha=4.73 \times 10^{-4} \mathrm{eV} / \mathrm{K}$, and $\beta=636 \mathrm{~K}$. Therefore, the band gap at 300 K is calculated as:
calculation: $$
\begin{aligned}
E_{\mathrm{g}} & =E_{\mathrm{g}}(\text { at } T=0 \mathrm{~K})-\frac{\alpha T^{... | (1.124, 1.422) | (eV, eV) | No supplementary information needed. | multiple | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.11 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 3.12 | NUM | What is the wavelength of light emitted from a GaN LED operating at 300 K? Assume that the band gap of GaN changes according to the Varshni parameters shown in Table.
Table Varshni Parameters for Semiconductors:
| Semiconductor | $\boldsymbol{E}_{\mathrm{g}}$ (at $\mathbf{0 ~ K}$ ) | $\boldsymbol{\alpha}\left(\mathb... | Using the Varshni parameters for GaN (see Table), we can show that the band gap of GaN at 300 K is 3.4 eV. For this band gap, the wavelength based on a GaN LED will be
$$
\begin{aligned}
\lambda(\text { in } \mu \mathrm{m}) & =\frac{1.24}{E_{\mathrm{g}}(\text { in } \mathrm{eV})} \mu \mathrm{m} \\
\therefore \lambda(\... | 365 | nm | No supplementary information needed. | single | medium | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 3.12 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 4.1 | NUM | In Cu, if we assume that each Cu atom donates one electron, the free electron concentration is $8.5 \times 10^{22} \mathrm{~cm}^{-3}$. (a) What is the value of $E_{\mathrm{F}, 0}$? (b) If the concentration of conduction electrons estimated by conductivity measurements is $1.5 \times 10^{23} \mathrm{~cm}^{-3}$, then fin... | 1. To find the Fermi energy level at $T=0 \mathrm{~K}$ for Cu, we use the formula:
$$
E_{\mathrm{F}, 0} = \left(\frac{\hbar^{2}}{2 m_{\mathrm{e}}}\right)\left(3 \pi^{2} n\right)^{2 / 3}
$$
where $\hbar = 1.05 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}$, $m_{\mathrm{e}} = 9.1 \times 10^{-30} \mathrm{~kg}$, and $n =... | (7, 1.76) | (eV, ) | No supplementary information needed. | multiple | hard | Materials: Metals | Metals | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 4.1 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 4.2 | NUM | Calculate the locations of the Fermi energy level for intrinsic Si located relative to the middle of the band gap. Assume that the effective masses for electrons and holes in Si are 1.08 and 0.56, respectively. | From Equation \[E_{\mathrm{F}, \mathrm{i}}-E_{\text{midgap}}=\frac{3}{4} k_{\mathrm{B}} T \ln \left(\frac{m_{\mathrm{p}}^{*}}{\mathrm{~m}_{\mathrm{n}}^{*}}\right)\], we get
\[E_{\mathrm{F}, \mathrm{i}}-E_{\text{midgap}}=\frac{3}{4}(0.026 \mathrm{eV}) \ln \left(\frac{0.56}{1.08}\right)=-12.8 \mathrm{meV}\]
Thus, the i... | -12.8 | meV | * The approximate value for half of the band gap of Si (550 meV or equivalent to a band gap of approximately 1.1 eV). | single | hard | Materials: Semiconductors | Semiconductors | Thermal | Composites | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 4.2 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 4.3 | NUM | A Si crystal is doped with $10^{16} / \mathrm{cm}^{3} \mathrm{Sb}$ atoms. At 300 K :
1. What is the concentration of electrons $(n)$ ?
2. What is the concentration of holes $(p)$ in this n-type Si?
3. Where is the Fermi energy level $\left(E_{\mathrm{F}}\right)$ for this material relative to the Fermi energy level for... | 1. The donor dopant $(\mathrm{Sb})$ concentration $\left(N_{\mathrm{d}}\right)$ is $10^{16}$ atoms $/ \mathrm{cm}^{3}$. Since $N_{\mathrm{d}} \gg n_{\mathrm{i}}$, we have $n \approx N_{\mathrm{d}}$. Assuming complete dopant ionization at 300 K, the electron concentration $(n) \approx N_{\mathrm{d}} = 10^{16}$ electrons... | (1 \times 10^{16}, 2.25 \times 10^{4}, 0.349) | (electrons/cm^3, holes/cm^3, eV) | * Intrinsic carrier concentration of Si at 300 K ($n_{\mathrm{i}}$) is $1.5 \times 10^{10} / \mathrm{cm}^{3}$. | multiple | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 4.3 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 4.4 | NUM | A Si crystal is doped with $10^{17} / \mathrm{cm}^{3}$ boron (B) atoms. At 300 K :
1. What is the concentration of holes $(p)$ ?
2. What is the concentration of electrons $(n)$ in this p-type Si?
3. Where is the Fermi energy level $\left(E_{\mathrm{F}}\right)$ for this material relative to the Fermi energy level for in... | 1. The acceptor dopant (B) concentration $\left(N_{\mathrm{a}}\right)$ is $10^{17}$ atoms $/ \mathrm{cm}^{3}$. Assuming complete dopant ionization at 300 K, the hole concentration $(p) \approx N_{\mathrm{a}} = 10^{17}$ holes $/ \mathrm{cm}^{3}$.
2. Using the relationship $n \times p = n_{i}^{2}$, we find:
$$
n \times... | (1 \times 10^{17}, 2.25 \times 10^{3}, -0.407) | (/cm^3, /cm^3, eV) | Based on the analysis, the following specific information is used in the Solution but is not explicitly provided in the Problem Description:
* Intrinsic carrier concentration ($n_i$) of Si at 300 K ($1.5 \times 10^{10} / \mathrm{cm}^3$)
* Thermal voltage ($kT/q$) at 300 K ($0.026 \mathrm{eV}$) | multiple | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 4.4 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 4.5 | NUM | A GaAs n-type sample is doped so that the concentration of electrons is $10^{17} \mathrm{cm}^{-3}$. What is the value of the $E_{\mathrm{F}} - E_{\mathrm{C}}$? Assume that $T=300 \mathrm{K}$ and that the semiconductor is degenerate. The density of states for GaAs at 300 K is $4.45 \times 10^{17} \mathrm{cm}^{-3}$ and $... | To calculate the Fermi energy level ($E_{\mathrm{F}}$) relative to the conduction band edge ($E_{\mathrm{C}}$) for the degenerate n-type GaAs sample, we use the Joyce-Dixon approximation:
$$
E_{\mathrm{F}}-E_{\mathrm{C}}=(0.026 \mathrm{eV})\left[\ln \frac{n}{N_{\mathrm{c}}}+\frac{1}{\sqrt{8}}\left[\frac{n}{N_{\mathrm{... | -0.037 | eV | No supplementary information needed. | single | hard | Materials: Semiconductors | Semiconductors | Electrical | Composites | Diffusion & Kinetics | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 4.5 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.1 | NUM | Consider a p-n junction in Si. Assume that the n- and p-sides have a dopant concentration of $N_{\mathrm{d}}=10^{16}$ and $N_{\mathrm{a}}=10^{17}$ atoms $/ \mathrm{cm}^{3}$, respectively. Estimate its contact potential $\left(V_{0}\right)$. Assume that $T=300 \mathrm{~K}, n_{\mathrm{i}}=1.5 \times 10^{10}$ electrons $/... | 1. For the n-type semiconductor, assuming complete donor ionization ($n \approx N_{\mathrm{d}}=10^{16}$ atoms $/ \mathrm{cm}^{3}$), the Fermi energy position relative to the intrinsic Fermi level is calculated as:
$$
E_{\mathrm{F}, \mathrm{n}}-E_{\mathrm{F}, \mathrm{i}}=(0.026 \mathrm{eV}) \ln \left(\frac{10^{16}}{1.5... | 0.756 | V | No supplementary information needed. | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.1 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.2 | NUM | Calculation of Contact Potential \(\left(V_{0}\right)\)
A step junction in Si is such that the n- and p-sides have dopant concentrations of \(N_{\mathrm{d}}=10^{16}\) and \(N_{\mathrm{a}}=10^{17}\) atoms \(/ \mathrm{cm}^{3}\), respectively. Calculate the contact potential \(\left(V_{0}\right)\), assuming that \(T=300... | From the given equation for the contact potential:
\[ q V_{0}=\frac{k_{\mathrm{B}} T}{q} \ln \left(\frac{N_{\mathrm{a}} N_{\mathrm{d}}}{n_{\mathrm{i}}^{2}}\right)=(0.026 \mathrm{eV}) \ln \left(\frac{10^{17} \text { atoms } / \mathrm{cm}^{3} \times 10^{16} \text { atoms } / \mathrm{cm}^{3}}{\left(1.5 \times 10^{10} \te... | 0.757 | V | No supplementary information needed. | single | hard | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.2 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.3 | NUM | A silicon abrupt (step) p–n junction is formed with N_D=1×10¹⁶ cm⁻³ and N_A=1×10¹⁷ cm⁻³. Assuming T=300 K (nᵢ=1.5×10¹⁰ cm⁻³) and zero applied bias, calculate the total depletion width w. | From Equations ($x_{\mathrm{n}}=\left[\left(\frac{2 \varepsilon V_0}{q}\right)\left(\frac{N_{\mathrm{a}}}{N_{\mathrm{d}}}\right)\left(\frac{1}{N_{\mathrm{a}}+N_{\mathrm{d}}}\right)\right]^{1 / 2}$) and ($x_{\mathrm{p}}=\left[\left(\frac{2 \varepsilon V_0}{q}\right)\left(\frac{N_{\mathrm{d}}}{N_{\mathrm{a}}}\right)\left... | 327.8 | nm | Based on the analysis of the provided Problem Description and Solution, the following specific necessary information used in the Solution is missing from the Problem Description:
* Permittivity of silicon (used as 11.8 × 8.85 × 10⁻¹⁴ F/cm).
* Built-in potential V₀ (used as 0.757 V and 0.755 V). | single | easy | Materials: Semiconductors | Semiconductors | Composites | Impact | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.3 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.4 | NUM | Assume $T \approx 300~\mathrm{K}$, intrinsic concentration $n_i \approx 1.5 \times 10^{10}~\mathrm{cm}^{-3}$, and relative permittivity $\varepsilon_r \approx 11.8$. A p-n junction in Si has donor doping level $N_{\mathrm{d}} = 10^{16}~\mathrm{cm}^{-3}$ and acceptor doping level $N_{\mathrm{a}} = 5 \times 10^{17}~\math... | 1. Calculation of the built-in potential $\left(V_{0}\right)$:
From Equation,
$$q V_{0}=k_{\mathrm{B}} T \ln \left(\frac{N_{\mathrm{a}} N_{\mathrm{d}}}{n_{\mathrm{i}}^{2}}\right)=(0.026 \mathrm{eV}) \ln \left(\frac{5 \times 10^{17} \text { atoms } / \mathrm{cm}^{3} \times 10^{16} \text { atoms } / \mathrm{cm}^{3}}{\lef... | (0.795, 4.91 \times 10^4) | (V, V/cm) | No supplementary information needed. | multiple | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.4 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.5 | NUM | The built-in potential $(V_{0})$ of a Si p-n junction diode is 0.8 V. Assume that the reverse-bias saturation current $(I_{0})$ is $10^{-13}$ amperes (A). A forward bias of 0.5 V is then applied. 1. What is the new height of the potential barrier seen on the electrostatic potential diagram? 2. What is the current flowi... | 1. When the applied voltage is 0.5 V, the new height of the barrier seen on the electrostatic potential will be $(V_{0}-V_{\mathrm{F}})=(0.8-0.5)=0.3 \mathrm{~V}$.
2. Since the forward bias applied $(V=0.5 \mathrm{~V})$ is much larger than $k T / q(0.0259 \mathrm{~V})$, that is, $q V \gg k T$, the exponential term is m... | (0.3, 2.42 \times 10^{-5}, 1.15 \times 10^{-3}) | (V, A, A) | * The value of $kT/q$ used in calculations, which is $0.0259 \mathrm{~V}$. | multiple | medium | Materials: Semiconductors | Semiconductors | Electrical | Composites | Plastic | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.5 | |||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.6 | NUM | A Si p-n junction functions as a diode and is connected to a resistor, $R_{\mathrm{L}}=1000 \Omega$. Assume that the knee voltage is 0.7 V. 1. What is the forward current if a 4.5 V forward bias is applied? 2. What is the forward current, assuming that the diode offers a dynamic resistance of $10 \Omega$ at the selecte... | 1. From Equation ($I_{\mathrm{F}}=\frac{\left(V_{\text {applied }}-V_{\text {knee }}\right)}{R_{\mathrm{L}}}$),
$$
I_{\mathrm{F}}=\frac{(4.5-0.7)}{1000 \Omega}=3.8 \mathrm{~mA}
$$
2. When the dynamic resistance of the diode $\left(r_{\mathrm{d}}^{\prime}\right)$ must be accounted for, we add that resistance to the $R... | (3.8, 3.762) | (mA, mA) | No supplementary information needed. | multiple | hard | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.6 | ||||
Electronic_Magnetic_and_Optical_Materials_Fulay_Example 5.7 | NUM | In a Si diode, $N_{\mathrm{a}}=10^{19} \mathrm{atoms} / \mathrm{cm}^{3}$ and $N_{\mathrm{d}}=5 \times 10^{15} \mathrm{atoms} / \mathrm{cm}^{3}$. If the critical breakdown field is $4 \times 10^{5} \mathrm{~V} / \mathrm{cm}$, what is the breakdown voltage for this diode at $T=300 \mathrm{~K}$? | The breakdown voltage for the diode is calculated using Equation ($V_{\mathrm{BD}}=\frac{\varepsilon E_{\text {critical }}^2}{2 q N_{\mathrm{d}}} V_{\mathrm{BR}}=\frac{\varepsilon E_{\text {critical }}^2}{2 q N_{\mathrm{d}}}$):
$$
V_{\mathrm{or}}=\frac{(11.8)\left(8.85 \times 10^{-14} \mathrm{~F} / \mathrm{cm}\right)\... | 103.5 | V | * Relative permittivity of Silicon ($\varepsilon_r$) | single | easy | Materials: Semiconductors | Semiconductors | Electrical | Composites | Electronic_Magnetic_and_Optical_Materials_Fulay | Example 5.7 |
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