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By planar diffusion of antimony (Sb) into p-type germanium (Ge), a p-n junction is obtained at a depth of $3 \times 10^{-3} \mathrm{~cm}$ below the surface. What is the donor concentration in the bulk germanium if diffusion is carried out for three hours at $790^{\circ} \mathrm{C}$? Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places, and express it in units of $1/\mathrm{cm}^3$. The surface concentration of antimony is held constant at a value of $8 \times 10^{18}$ $\mathrm{cm}^{-3} ; D_{790^{\circ} \mathrm{C}}=4.8 \times 10^{-11} \mathrm{~cm}^{2} / \mathrm{s}$. | \includegraphics[scale=0.5]{set_37_img_00.jpg}
\nonessentialimage
\[
\begin{aligned}
&\frac{c}{c_{s}}=\operatorname{erfc} \frac{x}{2 \sqrt{D t}}=\operatorname{erfc} \frac{3 \times 10^{-3}}{2 \sqrt{D t}}=\operatorname{erfc}(2.083) \\
&\frac{c}{c_{s}}=1-\operatorname{erf}(2.083), \therefore 1-\frac{c}{c_{s}}=0.9964 \\
&\frac{c}{c_{s}}=3.6 \times 10^{-3}, \therefore c=2.88 \times 10^{16} \mathrm{~cm}^{-3}
\end{aligned}
\]
The donor concentration in germanium is $\boxed{2.88e16} / \mathrm{cm}^{3}$. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 187 | [
"2.88 \\times 10^{16}\n",
"2.88 \\times 10^{16}"
] |
Preamble: Calculate the vacancy fraction in copper (Cu) in $\mathrm{~cm}^{-3}$ at the following temperatures. Measurements have determined the values of the enthalpy of vacancy formation, $\Delta \mathrm{H}_{\mathrm{V}}$, to be $1.03 \mathrm{eV}$ and the entropic prefactor, A, to be 1.1. Please format your answers as $n \times 10^x$ where $n$ is to 2 decimal places.
$20^{\circ} \mathrm{C}$. | number of sites / unit volume (also known as site density) is given by:
\[
\begin{aligned}
\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{V}_{\text {molar }}} & \therefore \text { site density }=6.02 \times 10^{23} / 7.11 \mathrm{~cm}^{3}=8.47 \times 10^{22} \\
& \rightarrow \text { vacancy density }=\mathrm{f}_{\mathrm{v}} \times \text { site density }
\end{aligned}
\]
$f_{V}=A e^{-\frac{\Delta H_{V}}{k_{B} T}}=1.1 \times e^{-\frac{1.03 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-22} \times(20+273)}}=2.19 \times 10^{-18}$
vacancy density at $20^{\circ} \mathrm{C}= \boxed{1.85e5} \mathrm{~cm}^{-3}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 206 | [
"1.85 \\times 10^{5}\n"
] |
Preamble: For the element copper (Cu) determine:
Subproblem 0: the distance of second nearest neighbors (in meters). Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places.
Solution: The answer can be found by looking at a unit cell of $\mathrm{Cu}$ (FCC).
\includegraphics[scale=0.5]{set_23_img_00.jpg}
\nonessentialimage
Nearest neighbor distance is observed along $<110>$; second-nearest along $<100>$. The second-nearest neighbor distance is found to be "a".
Cu: atomic volume $=7.1 \times 10^{-6} \mathrm{~m}^{3} /$ mole $=\frac{\mathrm{N}_{\mathrm{A}}}{4} \mathrm{a}^{3}$ ( $\mathrm{Cu}: \mathrm{FCC} ; 4$ atoms/unit cell) $a=\sqrt[3]{\frac{7.1 \times 10^{-6} \times 4}{6.02 \times 10^{23}}}= \boxed{3.61e-10} \mathrm{~m}$
Final answer: The final answer is 3.61e-10. I hope it is correct.
Subproblem 1: the interplanar spacing of $\{110\}$ planes (in meters). Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | $d_{h k l}=\frac{a}{\sqrt{h^{2}+k^{2}+1^{2}}}$
\[
d_{110}=\frac{3.61 \times 10^{-10}}{\sqrt{2}}= \boxed{2.55e-10} \mathrm{~m}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 198 | [
"3.61e-10"
] |
Preamble: A first-order chemical reaction is found to have an activation energy $\left(E_{A}\right)$ of 250 $\mathrm{kJ} /$ mole and a pre-exponential (A) of $1.7 \times 10^{14} \mathrm{~s}^{-1}$.
Determine the rate constant at $\mathrm{T}=750^{\circ} \mathrm{C}$. Round your answer to 1 decimal place, in units of $\mathrm{s}^{-1}$. | $\mathrm{k}=\mathrm{Ae} \mathrm{e}^{-\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{RT}}}=1.7 \times 10^{14} \times \mathrm{e}^{-\frac{2.5 \times 10^{5}}{8.31 \times 10^{23}}}= \boxed{28.8} \mathrm{~s}^{-1}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 214 | [
"28.8\n"
] |
A metal is found to have BCC structure, a lattice constant of $3.31 \AA$, and a density of $16.6 \mathrm{~g} / \mathrm{cm}^{3}$. Determine the atomic weight of this element in g/mole, and round your answer to 1 decimal place. | $B C C$ structure, so $\mathrm{n}=2$
\[
\begin{aligned}
&a=3.31 \AA=3.31 \times 10^{-10} \mathrm{~m} \\
&\rho=16.6 \mathrm{~g} / \mathrm{cm}^{3} \\
&\frac{\text { atomic weight }}{\rho} \times 10^{-6}=\frac{N_{A}}{n} \times a^{3}
\end{aligned}
\]
\[
\begin{aligned}
&\text { atomic weight }=\frac{\left(6.023 \times 10^{23} \text { atoms } / \text { mole }\right)\left(3.31 \times 10^{-10} \mathrm{~m}\right)^{3}}{(2 \text { atoms } / \text { unit cell })\left(10^{-6} \mathrm{~m}^{3} / \mathrm{cm}^{3}\right)} \times 16.6 \mathrm{~g} / \mathrm{cm}^{3} \\
&= \boxed{181.3} \mathrm{~g} / \text { mole }
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 196 | [
"181.3\n"
] |
Determine the atomic (metallic) radius of Mo in meters. Do not give the value listed in the periodic table; calculate it from the fact that Mo's atomic weight is $=95.94 \mathrm{~g} /$ mole and $\rho=10.2 \mathrm{~g} / \mathrm{cm}^{3}$. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | Mo: atomic weight $=95.94 \mathrm{~g} /$ mole
\[
\rho=10.2 \mathrm{~g} / \mathrm{cm}^{3}
\]
BCC, so $n=2$ atoms/unit cell
\[
\begin{aligned}
&\mathrm{a}^{3}=\frac{(95.94 \mathrm{~g} / \mathrm{mole})(2 \text { atoms/unit cell })}{\left(10.2 \mathrm{~g} / \mathrm{cm}^{3}\right)\left(6.023 \times 10^{23} \text { atoms } / \mathrm{mole}\right)} \times 10^{-6} \frac{\mathrm{m}^{3}}{\mathrm{~cm}^{3}} \\
&=3.12 \times 10^{-29} \mathrm{~m}^{3} \\
&a=3.22 \times 10^{-10} \mathrm{~m}
\end{aligned}
\]
For BCC, $a \sqrt{3}=4 r$, so $r= \boxed{1.39e-10} \mathrm{~m}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 191 | [
"1.39 \\times 10^{-10}\n"
] |
Subproblem 0: What is the working temperature for silica glass in Celsius?
Solution: \boxed{1950}.
Final answer: The final answer is 1950. I hope it is correct.
Subproblem 1: What is the softening temperature for silica glass in Celsius?
Solution: \boxed{1700}.
Final answer: The final answer is 1700. I hope it is correct.
Subproblem 2: What is the working temperature for Pyrex in Celsius?
Solution: \boxed{1200}.
Final answer: The final answer is 1200. I hope it is correct.
Subproblem 3: What is the softening temperature for Pyrex in Celsius?
Solution: \boxed{800}.
Final answer: The final answer is 800. I hope it is correct.
Subproblem 4: What is the working temperature for soda-lime glass in Celsius?
Solution: \boxed{900}.
Final answer: The final answer is 900. I hope it is correct.
Subproblem 5: What is the softening temperature for soda-lime glass in Celsius? | \boxed{700}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 217 | [
"700\n"
] |
What is the maximum wavelength $(\lambda)$ (in meters) of radiation capable of second order diffraction in platinum (Pt)? Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | The longest wavelength capable of $1^{\text {st }}$ order diffraction in Pt can be identified on the basis of the Bragg equation: $\lambda=2 \mathrm{~d} \sin \theta . \lambda_{\max }$ will diffract on planes with maximum interplanar spacing (in compliance with the selection rules): $\{111\}$ at the maximum value $\theta\left(90^{\circ}\right)$. We determine the lattice constant a for $\mathrm{Pt}$, and from it obtain $\mathrm{d}_{\{111\}}$. Pt is FCC with a value of atomic volume or $V_{\text {molar }}=9.1 \mathrm{~cm}^{3} /$ mole.
\[
\mathrm{V}_{\text {molar }}=\frac{N_{\mathrm{A}}}{4} \mathrm{a}^{3} ; \mathrm{a}=\sqrt[3]{\frac{9.1 \times 10^{-6} \times 4}{\mathrm{~N}_{\mathrm{A}}}}=3.92 \times 10^{-10} \mathrm{~m}
\]
If we now look at $2^{\text {nd }}$ order diffraction, we find $2 \lambda=2 \mathrm{~d}_{\{111\}} \sin 90^{\circ}$
\[
\therefore \lambda_{\max }=\mathrm{d}_{\{111\}}=\frac{\mathrm{a}}{\sqrt{3}}=\frac{3.92 \times 10^{-10}}{\sqrt{3}}= \boxed{2.26e-10} \mathrm{~m}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 218 | [
"2.26 \\times 10^{-10}"
] |
What is the activation energy of a process which is observed to increase by a factor of three when the temperature is increased from room temperature $\left(20^{\circ} \mathrm{C}\right)$ to $40^{\circ} \mathrm{C}$ ? Round your answer to 1 decimal place, and express it in $\mathrm{~kJ} / \mathrm{mole}$. | \[
\mathrm{k}_{1}=A \mathrm{e}^{\frac{-E_{A}}{R T_{1}}} ; k_{2}=3 k_{1}=A e^{\frac{-E_{A}}{R T_{2}}} \rightarrow \frac{1}{3}=e^{-\frac{E_{A}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)}
\]
\[
\begin{aligned}
&\ln 3=\frac{E_{A}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) \\
&E_{A}=\frac{R \times \ln 3}{\frac{1}{293}-\frac{1}{313}}=4.19 \times 10^{4}= \boxed{41.9} \mathrm{~kJ} / \mathrm{mole}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 219 | [
"41.9\n"
] |
Preamble: Iron $\left(\rho=7.86 \mathrm{~g} / \mathrm{cm}^{3}\right.$ ) crystallizes in a BCC unit cell at room temperature.
Calculate the radius in cm of an iron atom in this crystal. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | In $\mathrm{BCC}$ there are 2 atoms per unit cell, so $\frac{2}{\mathrm{a}^{3}}=\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{V}_{\text {molar }}}$, where $\mathrm{V}_{\text {molar }}=\mathrm{A} / \rho ; \mathrm{A}$ is the atomic mass of iron.
\[
\begin{aligned}
&\frac{2}{a^{3}}=\frac{N_{A} \times p}{A} \\
&\therefore a=\left(\frac{2 A}{N_{A} \times \rho}\right)^{\frac{1}{3}}=\frac{4}{\sqrt{3}} r \\
&\therefore r= \boxed{1.24e-8} \mathrm{~cm}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 197 | [
"1.24 \\times 10^{-8}\n"
] |
Preamble: A formation energy of $2.0 \mathrm{eV}$ is required to create a vacancy in a particular metal. At $800^{\circ} \mathrm{C}$ there is one vacancy for every 10,000 atoms.
At what temperature (in Celsius) will there be one vacancy for every 1,000 atoms? Format your answer as an integer. | We need to know the temperature dependence of the vacancy density:
\[
\frac{1}{10^{4}}=A e^{-\frac{\Delta H_{v}}{k T_{1}}} \quad \text { and } \frac{1}{10^{3}}=A e^{-\frac{\Delta H_{v}}{k T_{x}}}
\]
From the ratio: $\frac{\frac{1}{10^{4}}}{\frac{1}{10^{3}}}=\frac{10^{3}}{10^{4}}=\frac{\mathrm{Ae}^{-\Delta \mathrm{H}_{v} / \mathrm{k} T_{1}}}{\mathrm{Ae}^{-\Delta \mathrm{H}_{v} / \mathrm{kT} \mathrm{x}}}$ we get $-\ln 10=-\frac{\Delta \mathrm{H}_{\mathrm{v}}}{\mathrm{k}}\left(\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{\mathrm{x}}}\right)$
\[
\begin{aligned}
&\therefore \quad\left(\frac{1}{T_{1}}-\frac{1}{T_{x}}\right)=\frac{k \ln 10}{\Delta H_{v}} \\
&\frac{1}{T_{x}}=\frac{1}{T_{1}}-\frac{k \ln 10}{\Delta H_{v}}=\frac{1}{1073}-\frac{1.38 \times 10^{-23} \times \ln 10}{2 \times 1.6 \times 10^{-19}}=8.33 \times 10^{-4} \\
&T_{x}=1200 \mathrm{~K}= \boxed{928}^{\circ} \mathrm{C}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 209 | [
"928"
] |
A cubic metal $(r=0.77 \AA$ ) exhibits plastic deformation by slip along $<111>$ directions. Determine its planar packing density (atoms $/ \mathrm{m}^{2}$) for its densest family of planes. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | Slip along $<111>$ directions suggests a BCC system, corresponding to $\{110\},<111>$ slip. Therefore:
\[
\begin{aligned}
&a \sqrt{3}=4 r \\
&a=\frac{4 r}{\sqrt{3}}=1.78 \times 10^{-10} \mathrm{~m}
\end{aligned}
\]
Densest planes are $\{110\}$, so we find:
\[
\frac{2 \text { atoms }}{a^{2} \sqrt{2}}=\boxed{4.46e19} \text { atoms } / \mathrm{m}^{2}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 215 | [
"4.46 \\times 10^{19}\n"
] |
If no electron-hole pairs were produced in germanium (Ge) until the temperature reached the value corresponding to the energy gap, at what temperature (Celsius) would Ge become conductive? Please format your answer as $n \times 10^x$ where n is to 1 decimal place. $\left(\mathrm{E}_{\mathrm{th}}=3 / 2 \mathrm{kT}\right)$ | \[
\begin{aligned}
&E_{t h}=\frac{3 K T}{2} ; E_{g}=0.72 \times 1.6 \times 10^{-19} \mathrm{~J} \\
&T=\frac{0.72 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}}=5565 \mathrm{~K}=5.3 \times 10^{3}{ }^{\circ} \mathrm{C}
\end{aligned}
\]
The temperature would have to be $\boxed{5.3e3}{ }^{\circ} \mathrm{C}$, about $4400^{\circ} \mathrm{C}$ above the melting point of Ge. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 213 | [
"5.3 \\times 10^{3}\n"
] |
For $\mathrm{NaF}$ the repulsive (Born) exponent, $\mathrm{n}$, is 8.7. Making use of data given in your Periodic Table, calculate the crystal energy ( $\left.\Delta \mathrm{E}_{\text {cryst }}\right)$ in kJ/mole, to 1 decimal place. | \[
\Delta E=\frac{e^{2} N_{A} M}{4 \pi \varepsilon_{0} r_{0}}\left(1-\frac{1}{n}\right)
\]
The assumption must be made that the distance of separation of Na- $F$ is given by the sum of the ionic radii (that in a crystal they touch each other - a not unreasonable assumption). Thus, $r_{0}=0.95 \times 10^{-10}+1.36 \times 10^{-10} \mathrm{~m}=2.31 \AA$ and you must also assume $M$ is the same as for $\mathrm{NaCl}=1.747$ : $\mathrm{E}_{\text {cryst }}=-\frac{\left(1.6 \times 10^{-19}\right)^{2} 6.02 \times 10^{23} \times 1.747}{4 \pi 8.85 \times 10^{-12} \times 2.31 \times 10^{-10}}\left(1-\frac{1}{8.7}\right)$
\\
$\mathrm{E}_{\text {cryst }}=\boxed{927.5} /$ kJ/mole | Introduction to Solid State Chemistry (3.091 Fall 2010) | 210 | [
"927.5\n"
] |
Determine the inter-ionic equilibrium distance in meters between the sodium and chlorine ions in a sodium chloride molecule knowing that the bond energy is $3.84 \mathrm{eV}$ and that the repulsive exponent is 8. Please format your answer as $n \times 10^x$ where $n$ is to 1 decimal place. | $\mathrm{E}_{\mathrm{equ}}=-3.84 \mathrm{eV}=-3.84 \times 1.6 \times 10^{-19} \mathrm{~J}=-\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0} r_{0}}\left(1-\frac{1}{\mathrm{n}}\right)$
\\
$r_{0}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{4 \pi 8.85 \times 10^{-12} \times 6.14 \times 10^{-19}}\left(1-\frac{1}{8}\right)=
\boxed{3.3e-10} \mathrm{~m}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 208 | [
"3.3 \\times 10^{-10}\n"
] |
Preamble: Calculate the molecular weight in g/mole (to 2 decimal places) of each of the substances listed below.
Subproblem 0: $\mathrm{NH}_{4} \mathrm{OH}$
Solution: $\mathrm{NH}_{4} \mathrm{OH}$ :
$5 \times 1.01=5.05(\mathrm{H})$
$1 \times 14.01=14.01(\mathrm{~N})$
$1 \times 16.00=16.00(\mathrm{O})$
$\mathrm{NH}_{4} \mathrm{OH}= \boxed{35.06}$ g/mole
Final answer: The final answer is 35.06. I hope it is correct.
Subproblem 1: $\mathrm{NaHCO}_{3}$ | $\mathrm{NaHCO}_{3}: 3 \times 16.00=48.00(\mathrm{O})$
$1 \times 22.99=22.99(\mathrm{Na})$
$1 \times 1.01=1.01$ (H)
$1 \times 12.01=12.01$ (C)
$\mathrm{NaHCO}_{3}= \boxed{84.01}$ g/mole | Introduction to Solid State Chemistry (3.091 Fall 2010) | 211 | [
"35.06\n"
] |
Subproblem 0: What is the working temperature for silica glass in Celsius?
Solution: \boxed{1950}.
Final answer: The final answer is 1950. I hope it is correct.
Subproblem 1: What is the softening temperature for silica glass in Celsius?
Solution: \boxed{1700}.
Final answer: The final answer is 1700. I hope it is correct.
Subproblem 2: What is the working temperature for Pyrex in Celsius?
Solution: \boxed{1200}.
Final answer: The final answer is 1200. I hope it is correct.
Subproblem 3: What is the softening temperature for Pyrex in Celsius?
Solution: \boxed{800}.
Final answer: The final answer is 800. I hope it is correct.
Subproblem 4: What is the working temperature for soda-lime glass in Celsius? | \boxed{900}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 199 | [
"900\n"
] |
Determine the differences in relative electronegativity $(\Delta x$ in $e V)$ for the systems ${H}-{F}$ and ${C}-{F}$ given the following data:
$\begin{array}{cl}\text { Bond Energy } & {kJ} / \text { mole } \\ {H}_{2} & 436 \\ {~F}_{2} & 172 \\ {C}-{C} & 335 \\ {H}-{F} & 565 \\ {C}-{H} & 410\end{array}$
\\
Please format your answer to 2 decimal places. | According to Pauling, the square of the difference in electro negativity for two elements $\left(X_{A}-X_{B}\right)^{2}$ is given by the following relationship: $\left(X_{A}-X_{B}\right)^{2}=[$ Bond Energy $(A-B)-\sqrt{\text { Bond Energy AA. Bond Energy } B B}] \times \frac{1}{96.3}$
If bond energies are given in ${kJ}$.\\
$\left(X_{H}-X_{F}\right)^{2}=[565-\sqrt{436 \times 172}] \frac{1}{96.3}=3.02$
\[
\begin{aligned}
& \left({X}_{{H}}-{X}_{{F}}\right)=\sqrt{3.02}=1.7 \\
& \left(X_{C}-X_{H}\right)^{2}=[410-\sqrt{335 \times 436}] \frac{1}{96.3}=0.29 \\
& \left(X_{C}-X_{H}\right)=\sqrt{0.29}= \boxed{0.54}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 221 | [
"0.54",
"0.54\n"
] |
Preamble: Moldavia is a small country that currently trades freely in the world barley market. Demand and supply for barley in Moldavia is governed by the following schedules:
Demand: $Q^{D}=4-P$
Supply: $Q^{S}=P$
The world price of barley is $\$ 1 /$ bushel.
Subproblem 0: Calculate the free trade equilibrium price of barley in Moldavia, in dollars per bushel.
Solution: In free trade, Moldavia will import barley because the world price of $\$ 1 /$ bushel is lower than the autarkic price of $\$ 2$ /bushel. Free trade equilibrium price will be \boxed{1} dollar per bushel.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: Calculate the free trade equilibrium quantity of barley in Moldavia (in bushels). | In free trade, Moldavia will import barley because the world price of $\$ 1 /$ bushel is lower than the autarkic price of $\$ 2$ /bushel. Free trade equilibrium quantity will be \boxed{3} bushels, of which 1 is produced at home and 2 are imported. | Principles of Microeconomics (14.01 Fall 2011) | 246 | [
"1\n"
] |
Preamble: Two lasers generate radiation of (1) $9.5 \mu {m}$ and (2) $0.1 \mu {m}$ respectively.
Subproblem 0: Determine the photon energy (in eV, to two decimal places) of the laser generating radiation of $9.5 \mu {m}$.
Solution: \[
\begin{aligned}
{E} &={h} v=\frac{{hc}}{\lambda} {J} \times \frac{1 {eV}}{1.6 \times 10^{-19} {~J}} \\
{E}_{1} &=\frac{{hc}}{9.5 \times 10^{-6}} \times \frac{1}{1.6 \times 10^{-19}} {eV}= \boxed{0.13} {eV}
\end{aligned}
\]
Final answer: The final answer is 0.13. I hope it is correct.
Subproblem 1: Determine the photon energy (in eV, to one decimal place) of the laser generating radiation of $0.1 \mu {m}$. | \[
\begin{aligned}
{E} &={h} v=\frac{{hc}}{\lambda} {J} \times \frac{1 {eV}}{1.6 \times 10^{-19} {~J}} \\
{E}_{2} &=\frac{{hc}}{0.1 \times 10^{-6}} \times \frac{1}{1.6 \times 10^{-19}} {eV}= \boxed{12.4} {eV}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 227 | [
"0.13"
] |
Preamble: The decay rate of ${ }^{14} \mathrm{C}$ in living tissue is $15.3$ disintegrations per minute per gram of carbon. Experimentally, the decay rate can be measured to $\pm 0.1$ disintegrations per minute per gram of carbon. The half-life of ${ }^{14} \mathrm{C}$ is 5730 years.
What is the maximum age of a sample that can be dated, in years? | Radioactive decay is a $1^{\text {st }}$ order reaction which can be modeled as:
\[
-\frac{d c}{d t}=k c \text { or } c=c_{0} e^{-k t}
\]
With a little algebra we can get an expression for the relationship between time, $\mathrm{t}$, and the instant value of the decay rate.
At any time, t, we can write $\quad-\frac{\mathrm{dc}}{\mathrm{dt}}=\mathrm{kc}=\mathrm{kc}_{0} \mathrm{e}^{-\mathrm{kt}}$
and at time zero,
\[
-\frac{d c}{d t}=k c_{0}
\]
Divide eq. 1 by eq. 2 to get
where to reduce clutter let $r=\frac{d c}{d t}$
Take the logarithm of both sides of eq. 3 and substitute $k=\frac{\ln 2}{t_{1 / 2}}$.
With rearrangement, this gives $\quad t=-\frac{t_{1 / 2}}{\ln 2} \times \ln \frac{r_{t}}{r_{0}}$
So, for the oldest specimen we would measure the minimum instant decay rate of $0.1 \pm 0.1$ disintegrations per minute per gram. Set this equal to $r_{t}$ in eq. 4 and solve for $t$ to get $\boxed{41585} \pm 5730$ years. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 233 | [
"41585"
] |
Preamble: The number of electron-hole pairs in intrinsic germanium (Ge) is given by:
\[
n_{i}=9.7 \times 10^{15} \mathrm{~T}^{3 / 2} \mathrm{e}^{-\mathrm{E}_{g} / 2 \mathrm{KT}}\left[\mathrm{cm}^{3}\right] \quad\left(\mathrm{E}_{\mathrm{g}}=0.72 \mathrm{eV}\right)
\]
What is the density of pairs at $\mathrm{T}=20^{\circ} \mathrm{C}$, in inverse $\mathrm{cm}^3$? Please format your answer as $n \times 10^x$ where n is to 2 decimal places. | Recall: $\mathrm{T}$ in thermally activated processes is the absolute temperature: $\mathrm{T}^{\circ} \mathrm{K}=$ $\left(273.16+\mathrm{t}^{\circ} \mathrm{C}\right)$; Boltzmann's constant $=\mathrm{k}=1.38 \times 10^{-23} \mathrm{~J} /{ }^{\circ} \mathrm{K}$
$\mathrm{T}=293.16 \mathrm{~K}:$
\[
\begin{aligned}
&n_{i}=9.7 \times 10^{15} \times 293.16^{\frac{3}{2}} \times e^{-\frac{0.72 \times 16 \times 10^{-19}}{2 \times 1.38 \times 10^{-23} \times 293.16}} \\
&=9.7 \times 10^{15} \times 5019 \times 6.6 \times 10^{-7} \\
&n_{i}= \boxed{3.21e13} / \mathrm{cm}^{3}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 222 | [
"3.21 \\times 10^{13}",
"3.21 \\times 10^{13}\n"
] |
Preamble: For light with a wavelength $(\lambda)$ of $408 \mathrm{~nm}$ determine:
Subproblem 0: the frequency in $s^{-1}$. Please format your answer as $n \times 10^x$, where $n$ is to 3 decimal places.
Solution: To solve this problem we must know the following relationships:
\[
\begin{aligned}
v \lambda &=c
\end{aligned}
\]
$v$ (frequency) $=\frac{c}{\lambda}=\frac{3 \times 10^{8} m / s}{408 \times 10^{-9} m}= \boxed{7.353e14} s^{-1}$
Final answer: The final answer is 7.353e14. I hope it is correct.
Subproblem 1: the wave number in $m^{-1}$. Please format your answer as $n \times 10^x$, where $n$ is to 2 decimal places. | To solve this problem we must know the following relationships:
\[
\begin{aligned}
1 / \lambda=\bar{v}
\end{aligned}
\]
$\bar{v}$ (wavenumber) $=\frac{1}{\lambda}=\frac{1}{408 \times 10^{-9} m}=\boxed{2.45e6} m^{-1}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 223 | [
"7.353 \\times 10^{14}"
] |
How much oxygen (in kg, to 3 decimal places) is required to completely convert 1 mole of $\mathrm{C}_{2} \mathrm{H}_{6}$ into $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ ? | To get the requested answer, let us formulate a ``stoichiometric'' equation (molar quantities) for the reaction: $\mathrm{C}_{2} \mathrm{H}_{6}+70 \rightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}_{\text {. Each } \mathrm{C}_{2} \mathrm{H}_{6}}$ (ethane) molecule requires 7 oxygen atoms for complete combustion. In molar quantities: 1 mole of $\mathrm{C}_{2} \mathrm{H}_{6}=2 \times 12.01+6 \times 1.008=30.07 \mathrm{~g}$ requires
$7 \times 15.9984 \mathrm{~g}=1.12 \times 10^{2}$ oxygen $=\boxed{0.112} kg$ oxygen
We recognize the oxygen forms molecules, $\mathrm{O}_{2}$, and therefore a more appropriate formulation would be: $\mathrm{C}_{2} \mathrm{H}_{6}+7 / 2 \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}$. The result would be the same. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 220 | [
"0.112\n"
] |
Determine the highest linear density of atoms (atoms/m) encountered in vanadium (V). Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | \[
\begin{aligned}
&\mathrm{V}: \quad \text { atomic weight }=50.94 \mathrm{~g} / \text { mole } \\
&\rho=5.8 \mathrm{~g} / \mathrm{cm}^{3}
\end{aligned}
\]
$B C C$, so $n=2$
The highest density would be found in the [111] direction. To find "a":
\[
\begin{aligned}
&\frac{\text { atomic weight }}{\rho}=a^{3} \frac{N_{A}}{n} \rightarrow a^{3}=\frac{50.94 \times 2}{5.8 \times 6.023 \times 10^{23}} \\
&a=3.08 \times 10^{-8} \mathrm{~cm}=3.08 \times 10^{-10} \mathrm{~m}
\end{aligned}
\]
The length in the [111] direction is $\mathrm{a} \sqrt{3}$, so there are:
\[
\begin{aligned}
&2 \text { atoms } / \mathrm{a} \sqrt{3}=2 \text { atoms/ }\left(3.08 \times 10^{-10} \mathrm{~m} \times \sqrt{3}\right) \\
&= \boxed{3.75e9} \text { atoms } / \mathrm{m}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 238 | [
"3.75 \\times 10^9",
"3.75 \\times 10^{9}"
] |
A slab of plate glass containing dissolved helium (He) is placed in a vacuum furnace at a temperature of $400^{\circ} \mathrm{C}$ to remove the helium from the glass. Before vacuum treatment, the concentration of helium is constant throughout the glass. After 10 minutes in vacuum at $400^{\circ} \mathrm{C}$, at what depth (in $\mu \mathrm{m}$) from the surface of the glass has the concentration of helium decreased to $1 / 3$ of its initial value? The diffusion coefficient of helium in the plate glass at the processing temperature has a value of $3.091 \times 10^{-6} \mathrm{~cm}^{2} / \mathrm{s}$. | \includegraphics[scale=0.5]{set_37_img_01.jpg}
\nonessentialimage
\[
\begin{aligned}
&c=A+B \text { erf } \frac{x}{2 \sqrt{D t}} ; c(0, t)=0=A ; c(\infty, t)=c_{0}=B \\
&\therefore c(x, t)=c_{0} \operatorname{erf} \frac{x}{2 \sqrt{D t}}
\end{aligned}
\]
What is $\mathrm{x}$ when $\mathrm{c}=\mathrm{c}_{0} / 3$ ?
\[
\begin{gathered}
\frac{c_{0}}{3}=c_{0} \operatorname{erf} \frac{x}{2 \sqrt{D t}} \rightarrow 0.33=\operatorname{erf} \frac{x}{2 \sqrt{D t}} ; \operatorname{erf}(0.30)=0.3286 \approx 0.33 \\
\therefore \frac{x}{2 \sqrt{D t}}=0.30 \rightarrow x=2 \times 0.30 \times \sqrt{3.091 \times 10^{-6} \times 10 \times 60}=2.58 \times 10^{-2} \mathrm{~cm}=\boxed{258} \mu \mathrm{m}
\end{gathered}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 225 | [
"258"
] |
Preamble: $\mathrm{Bi}_{2} \mathrm{~S}_{3}$ dissolves in water according to the following reaction:
\[
\mathrm{Bi}_{2} \mathrm{~S}_{3}(\mathrm{~s}) \Leftrightarrow 2 \mathrm{Bi}^{3+}(\mathrm{aq})+3 \mathrm{~s}^{2-}(\mathrm{aq})
\]
for which the solubility product, $\mathrm{K}_{\mathrm{sp}}$, has the value of $1.6 \times 10^{-72}$ at room temperature.
At room temperature how many moles of $\mathrm{Bi}_{2} \mathrm{~S}_{3}$ will dissolve in $3.091 \times 10^{6}$ liters of water? Please format your answer as $n \times 10^x$ where $n$ is to 1 decimal place. | $\mathrm{Bi}_{2} \mathrm{~S}_{3}=2 \mathrm{Bi}^{3+}(\mathrm{aq})+3 \mathrm{~S}^{2-}(\mathrm{aq})$
\[
\therefore\left[\mathrm{Bi}^{3+}\right]=2 \mathrm{C}_{\mathrm{s}} \text { and }\left[\mathrm{s}^{2}\right]=3 \mathrm{C}_{\mathrm{s}}
\]
\[
\begin{aligned}
& \therefore \mathrm{K}_{\mathrm{sp}}=\left(2 \mathrm{C}_{\mathrm{s}}\right)^{2}\left(3 \mathrm{C}_{\mathrm{s}}\right)^{3}=4 \mathrm{C}_{\mathrm{s}}^{2} \cdot 27 \mathrm{C}_{\mathrm{s}}^{3}=108 \mathrm{C}_{\mathrm{s}}^{5} \\
& \therefore C_{\mathrm{s}}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5}=1.715 \times 10^{-15} \mathrm{~mol} / \mathrm{L} \\
& \therefore \text { in } 3.091 \times 10^{6} \mathrm{~L} \Rightarrow \boxed{5.3e-9} \mathrm{~mol} \mathrm{Bi}_{2} \mathrm{~S}_{3}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 228 | [
"5.3 \\times 10^{-9}"
] |
Determine the amount (in grams) of boron (B) that, substitutionally incorporated into $1 \mathrm{~kg}$ of germanium (Ge), will establish a charge carrier density of $3.091 \mathrm{x}$ $10^{17} / \mathrm{cm}^{3}$. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | The periodic table gives the molar volume of Ge as $13.57 \mathrm{~cm}^{3}$ and 1 mole of Ge weighs $72.61 \mathrm{~g}$, so set up the ratio $\frac{72.61}{13.6}=\frac{1000 \mathrm{~g}}{\mathrm{x}}$ and solve for $\mathrm{x}$ to get $187.30$ $\mathrm{cm}^{3}$ for the total volume. The addition of boron gives 1 charge carrier/B atom.
$\rightarrow \mathrm{B}$ concentration in Si must be $3.091 \times 10^{17} \mathrm{~B} / \mathrm{cm}^{3}$
$\mathrm{N}_{\mathrm{A}}$ of $B$ atoms weighs $10.81 \mathrm{~g}$
$\therefore 3.091 \times 10^{17} \mathrm{~B}$ atoms weigh $\frac{3.091 \times 10^{17}}{6.02 \times 10^{23}} \times 10.81=5.55 \times 10^{-6} \mathrm{~g}$
$\therefore$ for every $1 \mathrm{~cm}^{3}$ of Ge, add $5.55 \times 10^{-6} \mathrm{~g} \mathrm{~B}$
$\rightarrow$ for $187.30 \mathrm{~cm}^{3}$ of Ge, add $187.30 \times 5.55 \times 10^{-6}= \boxed{1.04e-3} \mathrm{~g} \mathrm{~B}$ | Introduction to Solid State Chemistry (3.091 Fall 2010) | 236 | [
"1.04 \\times 10^{-3}",
"1.04 \\times 10^{-3}\n"
] |
Calculate the volume in mL of $0.25 \mathrm{M} \mathrm{NaI}$ that would be needed to precipitate all the $\mathrm{g}^{2+}$ ion from $45 \mathrm{~mL}$ of a $0.10 \mathrm{M} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ solution according to the following reaction:
\[
2 \mathrm{NaI}(\mathrm{aq})+\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{HgI}_{2}(\mathrm{~s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})
\] | \[
\begin{aligned}
&2 \mathrm{NaI}(\mathrm{aq})+\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{HgI}_{2}(\mathrm{~s})+\mathrm{NaNO}_{3}(\mathrm{aq}) \\
&\frac{0.10 \mathrm{~mol} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}}{1 \mathrm{~L}} \times 0.045 \mathrm{~L}=4.5 \times 10^{-3} \mathrm{~mol} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2} \\
&4.5 \times 10^{-3} \mathrm{~mol} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2} \times \frac{2 \mathrm{~mol} \mathrm{NaI}}{1 \mathrm{~mol} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}}=9.00 \times 10^{-3} \mathrm{~mol} \mathrm{NaI} \\
&\frac{9.00 \times 10^{-3} \mathrm{~mol} \mathrm{NaI}}{0.25 \frac{\mathrm{mol} \mathrm{NaI}}{\mathrm{L}}}=3.6 \times 10^{-2} \mathrm{~L} \times \frac{1000 \mathrm{ml}}{1 \mathrm{~L}}=\boxed{36} \mathrm{~mL} \mathrm{NaI}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 224 | [
"36\n"
] |
Subproblem 0: What is the working temperature for silica glass in Celsius?
Solution: \boxed{1950}.
Final answer: The final answer is 1950. I hope it is correct.
Subproblem 1: What is the softening temperature for silica glass in Celsius? | \boxed{1700}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 226 | [
"1950\n"
] |
Whiskey, suspected to be of the "moonshine" variety, is analyzed for its age by determining its amount of naturally occurring tritium (T) which is a radioactive hydrogen isotope $\left({ }^{3} \mathrm{H}\right)$ with a half-life of $12.5$ years. In this "shine" the activity is found to be $6 \%$ of that encountered in fresh bourbon. What is the age (in years) of the whiskey in question? | \[
\begin{aligned}
&\frac{c_{o}}{c}=e^{k t} ; c=0.06 c_{0} \\
&\ln \frac{c_{0}}{0.06 c_{0}}=k t_{x} \\
&\ln 0.06=-k_{x} \\
&t_{x}=-\frac{\ln 0.06}{\frac{\ln 2}{t_{1 / 2}}}=\frac{\ln 0.06}{\frac{0.693}{12.5}}= \boxed{50.7} \text { years }
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 229 | [
"50.7\n"
] |
Subproblem 0: What is the working temperature for silica glass in Celsius?
Solution: \boxed{1950}.
Final answer: The final answer is 1950. I hope it is correct.
Subproblem 1: What is the softening temperature for silica glass in Celsius?
Solution: \boxed{1700}.
Final answer: The final answer is 1700. I hope it is correct.
Subproblem 2: What is the working temperature for Pyrex in Celsius?
Solution: \boxed{1200}.
Final answer: The final answer is 1200. I hope it is correct.
Subproblem 3: What is the softening temperature for Pyrex in Celsius? | \boxed{800}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 230 | [
"800\n"
] |
What is the working temperature for silica glass in Celsius? | \boxed{1950}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 202 | [
"1950^\\circ C\n"
] |
Subproblem 0: Is an energy level of $-1.362 \times 10^{-19} {~J}$ an allowed electron energy state in atomic hydrogen?
Solution: $E_{e l} =-\frac{1}{n^{2}} {~K}$ \\
$-1.362 \times 10^{-19} {~J}=-\frac{1}{{n}^{2}} \times 2.18 \times 10^{-18} {~J}$\\
${n} &=\sqrt{\frac{2.18 \times 10^{-18}}{1.362 \times 10^{-19}}}=4.00$\\
The answer is \boxed{Yes}.
Final answer: The final answer is Yes. I hope it is correct.
Subproblem 1: If your answer is yes, determine its principal quantum number $(n)$. If your answer is no, determine ${n}$ for the "nearest allowed state". | n = \boxed{4}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 237 | [
"4",
"4\n"
] |
Determine the energy gap (in eV) between the electronic states $n=7$ and $n=8$ in hydrogen. Please format your answer as $n \times 10^x$ where $n$ is to 1 decimal place. | Here we need to know the "basis" of the Rydberg equation [ $E_{e l}=-\left(1 / n^{2}\right) K$ ] and $1 {eV}=1.6 \times 10^{-19} {~J}$ :
\[
\begin{aligned}
&\Delta {E}_{{el}}={K}\left(\frac{1}{{n}_{{i}}^{2}}-\frac{1}{{n}_{{f}}^{2}}\right)=2.18 \times 10^{-18}\left(\frac{1}{49}-\frac{1}{64}\right)=1.043 \times 10^{-20} {~J} \\
&\Delta {E}_{{el}}=1.043 \times 10^{-20} {~J} \times \frac{1 {eV}}{\left(1.6 \times 10^{-19} {~J}\right)}= \boxed{6.5e-2} {eV}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 232 | [
"6.5 \\times 10^{-2}"
] |
To increase its corrosion resistance, chromium $(\mathrm{Cr})$ is diffused into steel at $980^{\circ} \mathrm{C}$. If during diffusion the surface concentration of chromium remains constant at $100 \%$, how long will it take (in days) to achieve a $\mathrm{Cr}$ concentration of $1.8 \%$ at a depth of $0.002 \mathrm{~cm}$ below the steel surface? Round your answer to 1 decimal place. $\left(D_{o}=0.54 \mathrm{~cm}^{2} / \mathrm{s} ; E_{A}=286 \mathrm{~kJ} / \mathrm{mol}\right.$ ) | A solution to Fick's second law for the given boundary conditions is:
$\frac{c}{c_{s}}=1-\operatorname{erf} \frac{x}{2 \sqrt{D t}}$, from which we get erf $\frac{x}{2 \sqrt{D t}}=1-0.018=0.982$
From the error function tables, $0.982$ is the erf of $1.67$. This means that
\[
\frac{0.002}{2 \sqrt{D t}}=\frac{0.001}{\sqrt{D t}}=1.67
\]
\[
\begin{aligned}
& \mathrm{D}=\mathrm{D}_{0} \mathrm{e}^{\left(\frac{-286 \times 10^{5}}{8.314 \times 1253}\right)}=6.45 \times 10^{-13} \mathrm{~cm}^{2} / \mathrm{s} \\
& \therefore \mathrm{t}=\frac{0.001^{2}}{1.67^{2} \times 6.45 \times 10^{-13}}=5.56 \times 10^{5} \mathrm{sec}=\boxed{6.4} \text { days }
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 204 | [
"6.4",
"6.4\n"
] |
Subproblem 0: Determine the inter-ionic equilibrium distance in meters between the sodium and chlorine ions in a sodium chloride molecule knowing that the bond energy is $3.84 \mathrm{eV}$ and that the repulsive exponent is 8. Please format your answer as $n \times 10^x$ where $n$ is to 1 decimal place.
Solution: $\mathrm{E}_{\mathrm{equ}}=-3.84 \mathrm{eV}=-3.84 \times 1.6 \times 10^{-19} \mathrm{~J}=-\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0} r_{0}}\left(1-\frac{1}{\mathrm{n}}\right)$
\\
$r_{0}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{4 \pi 8.85 \times 10^{-12} \times 6.14 \times 10^{-19}}\left(1-\frac{1}{8}\right)=
\boxed{3.3e-10} \mathrm{~m}$
Final answer: The final answer is 3.3e-10. I hope it is correct.
Subproblem 1: At the equilibrium distance, how much (in percent) is the contribution to the attractive bond energy by electron shell repulsion? | Shell "repulsion" obviously constitutes a "negative" contribution to the bond energy. Looking at the energy equation we find:
\[
\begin{array}{ll}
\text { the attractive term as: } & -E \times(1)=-E \\
\text { the repulsion term as: } & -E \times(-1 / n)=E / n=E / 8
\end{array}
\]
The contribution to the bond energy by the repulsion term $=1 / 8 \times 100 = \boxed{12.5}\%$ Since the bond energy is negative, the $12.5 \%$ constitute a reduction in bond strength. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 242 | [
"3.3 \\times 10^{-10}\n",
"3.3e-10\n"
] |
An electron beam strikes a crystal of cadmium sulfide (CdS). Electrons scattered by the crystal move at a velocity of $4.4 \times 10^{5} \mathrm{~m} / \mathrm{s}$. Calculate the energy of the incident beam. Express your result in eV, and as an integer. CdS is a semiconductor with a band gap, $E_{g}$, of $2.45$ eV. | \includegraphics[scale=0.5]{set_18_img_01.jpg}
\nonessentialimage
\[
\begin{aligned}
&E_{\text {incident } e^{-}}=E_{\text {emitted } \mathrm{v}}+E_{\text {scattered } e^{-}}=E_{g}+\frac{\mathrm{mv}^{2}}{2} \\
&=2.45 \mathrm{eV}+\frac{1}{2} \times \frac{9.11 \times 10^{-31} \mathrm{~kg} \times\left(4.4 \times 10^{5} \mathrm{~m} / \mathrm{s}\right)^{2}}{1.6 \times 10^{-19} \mathrm{eV} / \mathrm{J}} \\
&=2.45 \mathrm{eV}+0.55 \mathrm{eV}=\boxed{3} \mathrm{eV}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 241 | [
"3\n"
] |
Preamble: Consider the market for apple juice. In this market, the supply curve is given by $Q_{S}=$ $10 P_{J}-5 P_{A}$ and the demand curve is given by $Q_{D}=100-15 P_{J}+10 P_{T}$, where $J$ denotes apple juice, $A$ denotes apples, and $T$ denotes tea.
Subproblem 0: Assume that $P_{A}$ is fixed at $\$ 1$ and $P_{T}=5$. Calculate the equilibrium price in the apple juice market.
Solution: We have the system of equations $Q=10 P_{J}-5 \cdot 1$ and $Q=100-15 P_{J}+10 \cdot 5$. Solving for $P_{J}$ we get that $P_{J}=\boxed{6.2}$.
Final answer: The final answer is 6.2. I hope it is correct.
Subproblem 1: Assume that $P_{A}$ is fixed at $\$ 1$ and $P_{T}=5$. Calculate the equilibrium quantity in the apple juice market. | We have the system of equations $Q=10 P_{J}-5 \cdot 1$ and $Q=100-15 P_{J}+10 \cdot 5$. Solving for $Q$ we get that $Q=\boxed{57}$. | Principles of Microeconomics (14.01 Fall 2011) | 247 | [
"6.2\n"
] |
Estimate the ionic radius of ${Cs}^{+}$ in Angstroms to 2 decimal places. The lattice energy of $\mathrm{CsCl}$ is $633 \mathrm{~kJ} / \mathrm{mol}$. For $\mathrm{CsCl}$ the Madelung constant, $\mathrm{M}$, is $1.763$, and the Born exponent, $\mathrm{n}$, is 10.7. The ionic radius of $\mathrm{Cl}^{-}$is known to be $1.81 \AA$. | \[
\mathrm{E}_{\text {lattice }}=\frac{\mathrm{Mq}_{1} \mathrm{q}_{2}}{4 \pi \varepsilon_{0} r_{\mathrm{o}}}\left(1-\frac{1}{\mathrm{n}}\right) \text { and } \mathrm{r}_{\mathrm{o}}=\mathrm{r}_{\mathrm{Cs}^{+}}+\mathrm{r}_{\mathrm{Cl}}
\]
Solve first for $r_{0}$
\[
\begin{aligned}
r_{0} &=\frac{M q_{1} q_{2} N_{A v}}{4 \pi \varepsilon_{0} E_{\text {lattice }}}\left(1-\frac{1}{n}\right)=\frac{1.763\left(1.6 \times 10^{-19}\right)^{2} 6.02 \times 10^{23}}{4 \pi 8.85 \times 10^{-12} 6.33 \times 10^{5}}\left(1-\frac{1}{10.7}\right) \\
&=3.50 \times 10^{-10} \mathrm{~m}=3.50 \AA=r_{\mathrm{Cs}^{+}}+r_{\mathrm{Cr}} \\
\therefore & r_{\mathrm{Cs}^{+}}=3.50-1.81=\boxed{1.69} \AA
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 234 | [
"1.69\n"
] |
Subproblem 0: What is the working temperature for silica glass in Celsius?
Solution: \boxed{1950}.
Final answer: The final answer is 1950. I hope it is correct.
Subproblem 1: What is the softening temperature for silica glass in Celsius?
Solution: \boxed{1700}.
Final answer: The final answer is 1700. I hope it is correct.
Subproblem 2: What is the working temperature for Pyrex in Celsius? | \boxed{1200}. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 205 | [
"1200\n"
] |
Preamble: A first-order chemical reaction is found to have an activation energy $\left(E_{A}\right)$ of 250 $\mathrm{kJ} /$ mole and a pre-exponential (A) of $1.7 \times 10^{14} \mathrm{~s}^{-1}$.
Subproblem 0: Determine the rate constant at $\mathrm{T}=750^{\circ} \mathrm{C}$. Round your answer to 1 decimal place, in units of $\mathrm{s}^{-1}$.
Solution: $\mathrm{k}=\mathrm{Ae} \mathrm{e}^{-\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{RT}}}=1.7 \times 10^{14} \times \mathrm{e}^{-\frac{2.5 \times 10^{5}}{8.31 \times 10^{23}}}= \boxed{28.8} \mathrm{~s}^{-1}$
Final answer: The final answer is 28.8. I hope it is correct.
Subproblem 1: What percent of the reaction will be completed at $600^{\circ} \mathrm{C}$ in a period of 10 minutes? | Requires knowledge of $k_{600}$ :
\[
\begin{aligned}
&\mathrm{k}_{600}=1.7 \times 10^{14} \times \mathrm{e}^{-\frac{2.5 \times 10^{5}}{8.31 \times 873}}=0.184 \\
&\frac{\mathrm{c}}{\mathrm{c}_{0}}=\mathrm{e}^{-\mathrm{kt}}=\mathrm{e}^{-0.184 \times 600}=1.3 \times 10^{-48} \approx 0
\end{aligned}
\]
$c=0$ means the reaction is essentially $ \boxed{100} \%$ complete. | Introduction to Solid State Chemistry (3.091 Fall 2010) | 231 | [
"100\n"
] |
Preamble: There are two algebraic conditions describing a firm that is at a capital level that minimizes its costs in the long-term.
Write the condition which involves the SRAC, or short-run average cost? | \boxed{SRAC=LRAC}, short-run average cost equals long-run average cost. | Principles of Microeconomics (14.01 Fall 2011) | 250 | [
"SRAC=LRAC\n"
] |
Given the ionic radii, $\mathrm{Cs}^{+}=1.67 \AA, \mathrm{Cl}^{-}=1.81 \AA$, and the Madelung constant $\mathrm{M}(\mathrm{CsCl})=1.763$, determine to the best of your ability the molar Crystal energy ( $\Delta \mathrm{E}_{\text {cryst }}$ ) for $\mathrm{CsCl}$. Please format your answer as $n \times 10^x$ where n is to 2 decimal places; answer in $\mathrm{J} / \text{mole}$. | Given the radii $\mathrm{Cs}^{+}=1.67 \AA$ and $\mathrm{Cl}^{-}=1.81 \AA$, we can assume that $\mathrm{r}_{0}$ is the sum of the two. However, we need to know the exponential constant of the repulsive term which is not provided. Considering only the attractive force:
\[
\begin{array}{ll}
\Delta \mathrm{E}_{\text {cryst }}=\frac{-\mathrm{e}^{2} \mathrm{~N}_{\mathrm{A}} \mathrm{MQ}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} r_{0}} & \text { where: } \mathrm{Q}_{1}=\mathrm{Q}_{2}=1 \\
& \mathrm{M}=1.763 \\
& \mathrm{~N}_{\mathrm{A}}=6.02 \times 10^{23} \text { particle/mole }
\end{array}
\]
\[
\begin{aligned}
& \Delta \mathrm{E}_{\text {cryst }}=\frac{-\left(1.6 \times 10^{-19} \mathrm{coul}\right)^{2} \times 6.02 \times 10^{23} \times 1.763 \times 1 \times 1}{4 \pi 8.85 \times 10^{-12} \times(1.81+1.67) \times 10^{-10} \mathrm{~m}} \\
& = \boxed{7.02e5} \mathrm{~J} / \text { mole }
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 235 | [
"7.02 \\times 10^5"
] |
Preamble: Consider the market for apple juice. In this market, the supply curve is given by $Q_{S}=$ $10 P_{J}-5 P_{A}$ and the demand curve is given by $Q_{D}=100-15 P_{J}+10 P_{T}$, where $J$ denotes apple juice, $A$ denotes apples, and $T$ denotes tea.
Assume that $P_{A}$ is fixed at $\$ 1$ and $P_{T}=5$. Calculate the equilibrium price in the apple juice market. | We have the system of equations $Q=10 P_{J}-5 \cdot 1$ and $Q=100-15 P_{J}+10 \cdot 5$. Solving for $P_{J}$ we get that $P_{J}=\boxed{6.2}$. | Principles of Microeconomics (14.01 Fall 2011) | 256 | [
"6.2\n"
] |
Preamble: Sebastian owns a coffee factory in Argentina. His production function is:
\[
F(K, L)=(K-1)^{\frac{1}{4}} L^{\frac{1}{4}}
\]
Consider the cost of capital to be $r$ and the wage to be $w$. Both inputs are variable, and Sebastian faces no fixed costs.
What is the marginal rate of technical substitution of labor for capital? | \[
M R T S=\frac{M P_{L}}{M P_{K}}=\boxed{\frac{K-1}{L}}
\] | Principles of Microeconomics (14.01 Fall 2011) | 249 | [
"\\frac{K-1}{L}\n"
] |
Preamble: In Cambridge, shoppers can buy apples from two sources: a local orchard, and a store that ships apples from out of state. The orchard can produce up to 50 apples per day at a constant marginal cost of 25 cents per apple. The store can supply any remaining apples demanded, at a constant marginal cost of 75 cents per unit. When apples cost 75 cents per apple, the residents of Cambridge buy 150 apples in a day.
Assume that the city of Cambridge sets the price of apples within its borders. What price should it set, in cents? | The city should set the price of apples to be $\boxed{75}$ cents since that is the marginal cost when residents eat at least 50 apples a day, which they do when the price is 75 cents or less. | Principles of Microeconomics (14.01 Fall 2011) | 257 | [
"75\n"
] |
Preamble: You manage a factory that produces cans of peanut butter. The current market price is $\$ 10 /$ can, and you know the following about your costs (MC stands for marginal cost, and ATC stands for average total cost):
\[
\begin{array}{l}
MC(5)=10 \\
ATC(5)=6 \\
MC(4)=4 \\
ATC(4)=4
\end{array}
\]
A case of food poisoning breaks out due to your peanut butter, and you lose a lawsuit against your company. As punishment, Judge Judy decides to take away all of your profits, and considers the following two options to be equivalent:
i. Pay a lump sum in the amount of your profits.
ii. Impose a tax of $\$\left[P-A T C\left(q^{*}\right)\right]$ per can since that is your current profit per can, where $q^{*}$ is the profit maximizing output before the lawsuit.
How much is the tax, in dollars per can? | You maximize profits where $P=M C$, and since $P=10=M C(5)$ you would set $q^{*}=5$.
\[
\pi / q=(P-A T C)=(10-6)=4
\]
The tax would be $\$ \boxed{4} /$ can. | Principles of Microeconomics (14.01 Fall 2011) | 258 | [
"4\n"
] |
Preamble: A consumer's preferences are representable by the following utility function:
\[
u(x, y)=x^{\frac{1}{2}}+y
\]
Subproblem 0: Obtain the marginal rate of substitution of the consumer at an arbitrary point $(X,Y)$, where $X>0$ and $Y>0$.
Solution: \[ M R S=-\frac{\frac{1}{2} x^{-\frac{1}{2}}}{1}=\boxed{-\frac{1}{2} X^{-\frac{1}{2}}} \]
Final answer: The final answer is -\frac{1}{2} X^{-\frac{1}{2}}. I hope it is correct.
Subproblem 1: Suppose the price of the second good $(y)$ is 1 , and the price of the first good $(x)$ is denoted by $p>0$. If the consumer's income is $m>\frac{1}{4p}$, in the optimal consumption bundle of the consumer (in terms of $m$ and $p$ ), what is the quantity of the first good $(x)$? | The consumer solves $\max x^{\frac{1}{2}}+y$ so that $p x+y=m$. We look for stationary values of the Lagrangian $L=x^{\frac{1}{2}}+y+\lambda(m-p x-y)$. The first-order conditions for stationarity are
\[
\begin{aligned}
&\frac{\partial L}{\partial x}=\frac{1}{2} x^{-\frac{1}{2}}-\lambda p=0 \\
&\frac{\partial L}{\partial y}=1-\lambda=0 \\
&\frac{\partial L}{\partial \lambda}=m-p x-y=0
\end{aligned}
\]
Combining the first two equations above gives $\frac{1}{2 x^{\frac{1}{2}}}=p$, or $x^{*}=\frac{1}{4 p^{2}}$. Substituting $x^{*}$ into the budget constraint gives $y=m-p x^{*}=m-\frac{1}{4 p}$.
Case 1) $m \geq \frac{1}{4 p} \longrightarrow x^{*}=\frac{1}{4 p^{2}}$ and $y=m-\frac{1}{4 p} \geq 0$.
Case 2) $m \leq \frac{1}{4 p} \longrightarrow x^{*}=\frac{m}{p}$ and $y=0$.
Since we know $m>\frac{1}{4p}$, we use case 1, in which case our optimal consumption bundle $(x*,y*)$ is $(\frac{1}{4p^2},m-\frac{1}{4p})$. So the answer is $\boxed{\frac{1}{4p^2}}$. | Principles of Microeconomics (14.01 Fall 2011) | 255 | [
"\\frac{1}{4p^2}\n"
] |
What algebraic condition describes a firm that is at an output level that maximizes its profits, given its capital in the short-term? Use standard acronyms in your condition. | The required condition is \boxed{MR=SRMC}, or marginal revenue is equal to short-run marginal cost. | Principles of Microeconomics (14.01 Fall 2011) | 245 | [
"MR=SRMC\n"
] |
Strontium fluoride, $\mathrm{SrF}_{2}$, has a $\mathrm{K}_{\mathrm{sp}}$ value in water of $2.45 \times 10^{-9}$ at room temperature.
Calculate the solubility of $\mathrm{SrF}_{2}$ in water. Express your answer in units of molarity. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places. | \[
\begin{aligned}
&\mathrm{SrF}_{2}=\mathrm{Sr}^{2+}+2 \mathrm{~F}^{-} \quad \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Sr}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}, \quad \text { but }[\mathrm{F}]=2\left[\mathrm{Sr}^{2+}\right]=2 \mathrm{c}_{\mathrm{s}} \\
&\therefore \mathrm{K}_{\mathrm{sp}}=\mathrm{c}_{\mathrm{s}}\left(2 \mathrm{c}_{\mathrm{s}}\right)^{2}=4 \mathrm{c}_{\mathrm{s}}^{3} \quad \therefore \quad \mathrm{c}_{\mathrm{s}}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{1 / 3}= \boxed{8.49e-4} \mathrm{M}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 239 | [
"8.49 \\times 10^{-4}\n"
] |
Preamble: There are two algebraic conditions describing a firm that is at a capital level that minimizes its costs in the long-term.
Subproblem 0: Write the condition which involves the SRAC, or short-run average cost?
Solution: \boxed{SRAC=LRAC}, short-run average cost equals long-run average cost.
Final answer: The final answer is SRAC=LRAC. I hope it is correct.
Subproblem 1: Write the condition which involves SRMC, or short-run marginal cost? | \boxed{SRMC=LRMC}, or short-run marginal cost equals long-run levels. | Principles of Microeconomics (14.01 Fall 2011) | 251 | [
"SRAC=LRAC\n"
] |
Preamble: This problem deals with the H\"uckel MO theory of $\pi$-conjugated systems.
To answer each question, you will need to construct the Hückel MOs for each of the molecules pictured, divide them into sets of occupied and unoccupied orbitals, and determine the relevant properties, such as ground state energy, bond order, etc.
NOTE: For all parts we take $\alpha=\alpha_{\mathrm{C}}=-11.2 \mathrm{eV}$ and $\beta=\beta_{\mathrm{CC}}=-0.7 \mathrm{eV}$.
Determine the ionization potential of benzene (remember, ionization potential $\left[\mathrm{IP}=\mathrm{E}\left(\mathrm{B}^{+}\right)-\mathrm{E}(\mathrm{B})\right]$), in $\mathrm{eV}$, rounded to one decimal place. The benzene molecule is shown below:
\chemfig{C*6((-H)-C(-H)=C(-H)-C(-H)=C(-H)-C(-H)=)} | Let's build the Hückel MO Hamiltonian from the 6 carbon atoms. The differences between benzene and hexatriene are only connectivity:
\[
H_{\text {benzene }}=\left(\begin{array}{cccccc}
\alpha & \beta & 0 & 0 & 0 & \beta \\
\beta & \alpha & \beta & 0 & 0 & 0 \\
0 & \beta & \alpha & \beta & 0 & 0 \\
0 & 0 & \beta & \alpha & \beta & 0 \\
0 & 0 & 0 & \beta & \alpha & \beta \\
\beta & 0 & 0 & 0 & \beta & \alpha
\end{array}\right)
\]
We now substitute $\alpha$ and $\beta$ with the values above and find the eigenvalues of the Hamiltonian numerically. The eigenvalues of $\mathrm{H}_{\text {benzene }}$ (in $\mathrm{eV}$ ) are
\[
E^{\mu}=\{-12.6,-11.9,-11.9,-10.5,-10.5,-9.8\}
\].
The ionization potential in this model is simply the energy of the HOMO of the ground state of each molecule (this is the orbital from which the electron is ejected). Since there are $6 \pi$-electrons, we can fill the three lowest MOs and the HOMO will be the third lowest. Therefore, the IP of benzene is $\boxed{11.9} \mathrm{eV}$ | Physical Chemistry (5.61 Fall 2017) | 266 | [
"11.9\n",
"11.9"
] |
You wish to dope a single crystal of silicon (Si) with boron (B). The specification reads $5 \times 10^{16}$ boron atoms/ $\mathrm{cm}^{3}$ at a depth of $25 \mu \mathrm{m}$ from the surface of the silicon. What must be the effective concentration of boron in units of atoms/ $\mathrm{cm}^{3}$ if you are to meet this specification within a time of 90 minutes? Round your answer to 4 decimal places. Assume that initially the concentration of boron in the silicon crystal is zero. The diffusion coefficient of boron in silicon has a value of $7.23 \times 10^{-9} \mathrm{~cm}^{2} / \mathrm{s}$ at the processing temperature. | \[
\begin{aligned}
&c(x, t)=A+B \text { erf } \frac{x}{2 \sqrt{D t}} ; c(0, t)=c_{s}=A ; c(x, 0)=c_{i}=0 \\
&c(\infty, t)=c_{i}=0=A+B \rightarrow A=-B \\
&\therefore c(x, t)=c_{s}-c_{s} \operatorname{erf} \frac{x}{2 \sqrt{D t}}=c_{s} \operatorname{erfc} \frac{x}{2 \sqrt{D t}} \rightarrow 5 \times 10^{16}=c_{s} \text { erfc } \frac{25 \times 10^{-4}}{2 \sqrt{7.23 \times 10^{-9} \times 90 \times 60}} \\
&\therefore c_{s}=\frac{5 \times 10^{16}}{\operatorname{erfc} \frac{25 \times 10^{-4}}{2 \sqrt{7.23 \times 10^{-9} \times 5400}}}=6.43 \times 10^{16} \mathrm{~cm}^{-3} \\
&\operatorname{erfc}(0.20)=1-\operatorname{erf}(0.20)=1-0.2227=\boxed{0.7773}
\end{aligned}
\] | Introduction to Solid State Chemistry (3.091 Fall 2010) | 240 | [
"6.43 \\times 10^{16}",
"6.43 \\times 10^{16} \\text{ cm}^{-3}"
] |
Preamble: A consumer's preferences are representable by the following utility function:
\[
u(x, y)=x^{\frac{1}{2}}+y
\]
Obtain the marginal rate of substitution of the consumer at an arbitrary point $(X,Y)$, where $X>0$ and $Y>0$. | \[ M R S=-\frac{\frac{1}{2} x^{-\frac{1}{2}}}{1}=\boxed{-\frac{1}{2} X^{-\frac{1}{2}}} \] | Principles of Microeconomics (14.01 Fall 2011) | 243 | [
"-\\frac{1}{2}X^{-\\frac{1}{2}}\n"
] |
Preamble: Xiaoyu spends all her income on statistical software $(S)$ and clothes (C). Her preferences can be represented by the utility function: $U(S, C)=4 \ln (S)+6 \ln (C)$.
Compute the marginal rate of substitution of software for clothes. | We have that $M R S=\frac{\frac{4}{S}}{\frac{6}{C}}=\boxed{\frac{2}{3} \frac{C}{S}}$. | Principles of Microeconomics (14.01 Fall 2011) | 244 | [
"\\frac{2}{3} \\frac{C}{S}\n"
] |
Preamble: You have been asked to analyze the market for steel. From public sources, you are able to find that last year's price for steel was $\$ 20$ per ton. At this price, 100 million tons were sold on the world market. From trade association data you are able to obtain estimates for the own price elasticities of demand and supply on the world markets as $-0.25$ for demand and $0.5$ for supply. Assume that steel has linear demand and supply curves throughout, and that the market is competitive.
Solve for the equations of demand in this market. Use $P$ to represent the price of steel in dollars per ton, and $X_{d}$ to represent the demand in units of millions of tons. | Assume that this is a competitive market and assume that demand and supply are linear. Thus, $X_{d}=a-b P$ and $X_{s}=c+d P$. We know from the equation for own-price elasticity of demand that
\[
E_{Q_{X} P_{X}}=\frac{d X_{d}}{d P_{X}} \frac{P_{X}}{X_{d}}=-b \frac{P_{X}}{X_{d}}=-b \frac{20}{100}=-0.25
\]
Solving for $b$, then, we have $b=1.25$. Substituting back into the equation for demand, $X_{d}=$ $a-1.25 P$ or $100=a-1.25(20)$. Solving for $a$ we have $a=125$. Hence, the equation for last year's demand is $\boxed{X_{d}=125-1.25 P}$. | Principles of Microeconomics (14.01 Fall 2011) | 260 | [
"X_{d}=125-1.25 P"
] |
Given that the work function of chromium is $4.40 \mathrm{eV}$, calculate the kinetic energy of electrons in Joules emitted from a clean chromium surface that is irradiated with ultraviolet radiation of wavelength $200 \mathrm{~nm}$. | The chromium surface is irradiated with $200 \mathrm{~nm}$ UV light. These photons have energy
\[
\begin{aligned}
E &=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{34} \mathrm{~J} \cdot \mathrm{s}\right)\left(3 \times 10^{8} \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{200 \times 10^{-9} \mathrm{~m}} \\
&=9.94 \times 10^{-19} \mathrm{~J} \\
&=6.20 \mathrm{eV}
\end{aligned}
\]
The photo-ejected electron has kinetic energy
\[
K E=E_{\text {photon }}-\phi_{o}=6.20 \mathrm{eV}-4.40 \mathrm{eV}=1.80 \mathrm{eV}=\boxed{2.88e-19} \mathrm{~J}
\] | Physical Chemistry (5.61 Fall 2017) | 264 | [
"2.88 \\times 10^{-19}\n"
] |
Preamble: Suppose, in the short run, the output of widgets is supplied by 100 identical competitive firms, each having a cost function:
\[
c_{s}(y)=\frac{1}{3} y^{3}+2
\]
The demand for widgets is given by:
\[
y^{d}(p)=6400 / p^{\frac{1}{2}}
\]
Subproblem 0: Obtain the short run industry supply function for widgets.
Solution: Since $P=M C=y^{2}$, the supply function of each firm is given by $y_{i}^{s}=p^{\frac{1}{2}}$.
The industry supply function is $y^{s}(p)=100 y_{i}^{s}(p)=\boxed{100 p^{\frac{1}{2}}}$.
Final answer: The final answer is 100 p^{\frac{1}{2}}. I hope it is correct.
Subproblem 1: Obtain the short run equilibrium price of widgets. | $y^{s}=y^{d} \longrightarrow 100 p^{\frac{1}{2}}=\frac{6400}{p^{\frac{1}{2}}} \longrightarrow p=\boxed{64}$. | Principles of Microeconomics (14.01 Fall 2011) | 254 | [
"100p^{\\frac{1}{2}}\n",
"100 p^{\\frac{1}{2}}\n"
] |
Preamble: Suppose, in the short run, the output of widgets is supplied by 100 identical competitive firms, each having a cost function:
\[
c_{s}(y)=\frac{1}{3} y^{3}+2
\]
The demand for widgets is given by:
\[
y^{d}(p)=6400 / p^{\frac{1}{2}}
\]
Obtain the short run industry supply function for widgets. | Since $P=M C=y^{2}$, the supply function of each firm is given by $y_{i}^{s}=p^{\frac{1}{2}}$.
The industry supply function is $y^{s}(p)=100 y_{i}^{s}(p)=\boxed{100 p^{\frac{1}{2}}}$. | Principles of Microeconomics (14.01 Fall 2011) | 252 | [
"100p^{\\frac{1}{2}}\n"
] |
Preamble: Suppose, in the short run, the output of widgets is supplied by 100 identical competitive firms, each having a cost function:
\[
c_{s}(y)=\frac{1}{3} y^{3}+2
\]
The demand for widgets is given by:
\[
y^{d}(p)=6400 / p^{\frac{1}{2}}
\]
Subproblem 0: Obtain the short run industry supply function for widgets.
Solution: Since $P=M C=y^{2}$, the supply function of each firm is given by $y_{i}^{s}=p^{\frac{1}{2}}$.
The industry supply function is $y^{s}(p)=100 y_{i}^{s}(p)=\boxed{100 p^{\frac{1}{2}}}$.
Final answer: The final answer is 100 p^{\frac{1}{2}}. I hope it is correct.
Subproblem 1: Obtain the short run equilibrium price of widgets.
Solution: $y^{s}=y^{d} \longrightarrow 100 p^{\frac{1}{2}}=\frac{6400}{p^{\frac{1}{2}}} \longrightarrow p=\boxed{64}$.
Final answer: The final answer is 64. I hope it is correct.
Subproblem 2: Obtain the the output of widgets supplied by each firm. | $y^{s}=y^{d} \longrightarrow 100 p^{\frac{1}{2}}=\frac{6400}{p^{\frac{1}{2}}} \longrightarrow p=64$. Hence $y^{*}=100 \cdot 8=800$ and $y_{i}=\boxed{8}.$ | Principles of Microeconomics (14.01 Fall 2011) | 248 | [
"8\n"
] |
Preamble: Moldavia is a small country that currently trades freely in the world barley market. Demand and supply for barley in Moldavia is governed by the following schedules:
Demand: $Q^{D}=4-P$
Supply: $Q^{S}=P$
The world price of barley is $\$ 1 /$ bushel.
Calculate the free trade equilibrium price of barley in Moldavia, in dollars per bushel. | In free trade, Moldavia will import barley because the world price of $\$ 1 /$ bushel is lower than the autarkic price of $\$ 2$ /bushel. Free trade equilibrium price will be \boxed{1} dollar per bushel. | Principles of Microeconomics (14.01 Fall 2011) | 253 | [
"1\n"
] |
Preamble: Suppose there are exactly two consumers (Albie and Bubbie) who demand strawberries. Suppose that Albie's demand for strawberries is given by
\[
q_{a}(p)=p^{\alpha} f_{a}\left(I_{a}\right)
\]
and Bubbie's demand is given by
\[
q_{b}(p)=p^{\beta} f_{b}\left(I_{b}\right)
\]
where $I_{a}$ and $I_{b}$ are Albie and Bubbie's incomes, and $f_{a}(\cdot)$ and $f_{b}(\cdot)$ are two unknown functions.
Find Albie's (own-price) elasticity of demand, $\epsilon_{q_{a}, p}$. Use the sign convention that $\epsilon_{y, x}=\frac{\partial y}{\partial x} \frac{x}{y}$. | \[
\epsilon_{q_{a}, p}=\frac{\partial q_{a}}{\partial p} \frac{p}{q_{a}(p)}=\left[\alpha p^{\alpha-1} f_{a}\left(I_{a} s\right)\right] \frac{p}{p^{\alpha} f_{a}\left(I_{a}\right)}=\boxed{\alpha}
\] | Principles of Microeconomics (14.01 Fall 2011) | 259 | [
"\\alpha\n"
] |
A baseball has diameter $=7.4 \mathrm{~cm}$. and a mass of $145 \mathrm{~g}$. Suppose the baseball is moving at $v=1 \mathrm{~nm} /$ second. What is its de Broglie wavelength
\[
\lambda=\frac{h}{p}=\frac{h}{m \nu}
\]
? Give answer in meters. | \[
\begin{aligned}
D_{\text {ball }} &=0.074 m \\
m_{\text {ball }} &=0.145 \mathrm{~kg} \\
v_{\text {ball }} &=1 \mathrm{~nm} / \mathrm{s}=1 \times 10^{-9} \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
Using de Broglie:
\[
\lambda_{\text {ball }}=\frac{h}{p}=\frac{h}{m \nu}=\frac{6.626 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} / \mathrm{s}}{0.145 \mathrm{~kg} \cdot 1 \times 10^{-9} \mathrm{~m} / \mathrm{s}}=\boxed{4.6e-24} \mathrm{~m}=\lambda_{\text {ball }}
\] | Physical Chemistry (5.61 Fall 2017) | 267 | [
"4.6 \\times 10^{-24}\n"
] |
Harmonic Oscillator Subjected to Perturbation by an Electric Field: An electron is connected by a harmonic spring to a fixed point at $x=0$. It is subject to a field-free potential energy
\[
V(x)=\frac{1}{2} k x^{2} .
\]
The energy levels and eigenstates are those of a harmonic oscillator where
\[
\begin{aligned}
\omega &=\left[k / m_{e}\right]^{1 / 2} \\
E_{v} &=\hbar \omega(v+1 / 2) \\
\psi_{v}(x) &=(v !)^{-1 / 2}\left(\hat{\boldsymbol{a}}^{\dagger}\right)^{v} \psi_{v=0}(x) .
\end{aligned}
\]
Now a constant electric field, $E_{0}$, is applied and $V(x)$ becomes
\[
V(x)=\frac{1}{2} k x^{2}+E_{0} e x \quad(e>0 \text { by definition }) .
\]
Write an expression for the energy levels $E_{v}$ as a function of the strength of the electric field. | The total potential, including the interaction with the electric field is
\[
V(x)=\frac{m \omega^{2}}{2} x^{2}+E_{0} e x .
\]
We find its minimum to be
\[
\begin{aligned}
\frac{d V}{d x}=m \omega^{2} x &+E_{0} e=0 \\
\Rightarrow x_{\min } &=\frac{E_{0} e}{m \omega^{2}}, \\
V\left(x_{\min }\right) &=\frac{m \omega^{2}}{2} \frac{E_{0}^{2} e^{2}}{m^{2} \omega^{2}}-\frac{E_{0}^{2} e^{2}}{m \omega^{2}} \\
&=\frac{E_{0}^{2} e^{2}}{2 m \omega^{2}} .
\end{aligned}
\]
Defining the displacement from the minimum $x^{\prime}=x-x_{\min }$, we arrive at
\[
\begin{aligned}
V\left(x^{\prime}\right) &=\frac{m \omega^{2}}{2}\left(x^{\prime}-\frac{E_{0} e}{m \omega^{2}}\right)^{2}+E_{0} e\left(x^{\prime}-\frac{E_{0} e}{m \omega^{2}}\right) \\
&=\frac{m \omega^{2}}{2} x^{\prime 2}-\frac{E_{0}^{2} e^{2}}{2 m \omega^{2}}
\end{aligned}
\]
Thus, we see that the system is still harmonic! All we have done is to shift the minimum position and minimum energy, but the potential is still quadratic. The harmonic frequency $\omega$ remains unchanged.
Since the potential now is a harmonic oscillator with frequency $\omega$ and a constant offset, we can easily write down the energy levels:
\[
E_{v}=\boxed{\hbar \omega(v+1 / 2)-\frac{E_{0}^{2} e^{2}}{2 m \omega^{2}}}
\] | Physical Chemistry (5.61 Fall 2017) | 261 | [
"\\hbar \\omega(v+1 / 2)-\\frac{E_{0}^{2} e^{2}}{2 m \\omega^{2}}\n",
"\\hbar \\omega(v+1 / 2)-\\frac{E_{0}^{2} e^{2}}{2 m \\omega^{2}}"
] |
Preamble: The following concern the independent particle model. You may find the following set of Coulomb and exchange integrals useful (energies in $\mathrm{eV}$):
$\mathrm{J}_{1 s 1 s}=17.0 Z$
$\mathrm{~J}_{1 s 2 s}=4.8 Z$
$\mathrm{~K}_{1 s 2 s}=0.9 Z$
$\mathrm{~J}_{2 s 2 s}=3.5 Z$
$\mathrm{J}_{1 s 2 p}=6.6 Z$
$\mathrm{~K}_{1 s 2 p}=0.5 Z$
$\mathrm{~J}_{2 s 2 p}=4.4 Z$
$\mathrm{~K}_{2 s 2 p}=0.8 Z$
$\mathrm{J}_{2 p_{i}, 2 p_{i}}=3.9 Z$
$\mathrm{~J}_{2 p_{i}, 2 p_{k}}=3.5 Z$
$\mathrm{~K}_{2 p_{i}, 2 p_{k}}=0.2 Z i \neq k$
Using the independent particle model, what is the energy difference between the $1 s^{2} 2 p_{x}^{2}$ configuration and the $1 s^{2} 2 s^{2}$ configuration? Give your answer in eV, in terms of $Z$, and round to a single decimal place. | We are asked to calculate the energy difference between a $1 s^{2} 2 p_{x}^{2}$ and a $1 s^{2} 2 s^{2}$ configuration. Let's compute the energy for each using the independent particle model
\[
\begin{aligned}
E\left[1 s^{2} 2 p_{x}^{2}\right]=& \sum_{i} E_{i}+\sum_{i, j}^{i>j} \widetilde{J}_{i j}-\widetilde{K}_{i j} \\
=& 2 E_{1 s}+2 E_{2 p} \\
&+\widetilde{J}_{1 s \alpha, 1 s \beta}+\widetilde{J}_{1 s \alpha, 2 p_{x} \alpha}+\widetilde{J}_{1 s \alpha, 2 p_{x} \beta}+\widetilde{J}_{1 s \beta, 2 p_{x} \alpha}+\widetilde{J}_{1 s \beta, 2 p_{x} \beta}+\widetilde{J}_{2 p_{x} \alpha, 2 p_{x} \beta} \\
&-\widetilde{K}_{1 s \alpha, 1 s \beta}-\widetilde{K}_{1 s \alpha, 2 p_{x} \alpha}-\widetilde{K}_{1 s \alpha, 2 p_{x} \beta}-\widetilde{K}_{1 s \beta, 2 p_{x} \alpha}-\widetilde{K}_{1 s \beta, 2 p_{x} \beta}-\widetilde{K}_{2 p_{x} \alpha, 2 p_{x} \beta} \\
=& 2 E_{1 s}+2 E_{2 p}+J_{1 s, 1 s}+4 J_{1 s, 2 p}+J_{2 p_{i}, 2 p_{i}}-2 K_{1 s, 2 p} \\
E\left[1 s^{2} 2 s^{2}\right]=& 2 E_{1 s}+2 E_{2 s}+J_{1 s, 1 s}+4 J_{1 s, 2 s}+J_{2 s, 2 s}-2 K_{1 s, 2 s} \\
\Rightarrow \Delta E=& 4\left(J_{1 s, 2 p}-J_{1 s, 2 s}\right)+\left(J_{2 p_{i}, 2 p_{i}}-J_{2 s, 2 s}\right)-2\left(K_{1 s, 2 p}-K_{1 s, 2 s}\right) \\
=& Z[4(6.6-4.8)-(3.9-3.5)-2(0.5-0.9)] \\
=&+\boxed{7.6 Z} \mathrm{eV}
\end{aligned}
\] | Physical Chemistry (5.61 Fall 2017) | 262 | [
"7.6Z"
] |
Preamble: Consider the Particle in an Infinite Box ``superposition state'' wavefunction,
\[
\psi_{1,2}=(1 / 3)^{1 / 2} \psi_{1}+(2 / 3)^{1 / 2} \psi_{2}
\]
where $E_{1}$ is the eigen-energy of $\psi_{1}$ and $E_{2}$ is the eigen-energy of $\psi_{2}$.
Subproblem 0: Suppose you do one experiment to measure the energy of $\psi_{1,2}$. List the possible result(s) of your measurement.
Solution: Since the only eigenergies are $E_{1}$ and $E_{2}$, the possible outcomes of the measurement are $\boxed{E_{1},E_{2}}$.
Final answer: The final answer is E_{1},E_{2}. I hope it is correct.
Subproblem 1: Suppose you do many identical measurements to measure the energies of identical systems in state $\psi_{1,2}$. What average energy will you observe? | \[
\langle E\rangle =\boxed{\frac{1}{3} E_{1}+\frac{2}{3} E_{2}}
\]
This value of $\langle E\rangle$ is between $E_{1}$ and $E_{2}$ and is the weighted average energy. | Physical Chemistry (5.61 Fall 2017) | 268 | [
"E_{1},E_{2}\n"
] |
Compute the momentum of one $500 \mathrm{~nm}$ photon using $p_{\text {photon }}=E_{\text {photon }} / c$ where $c$ is the speed of light, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$, and $\nu=c / \lambda$. Express your answer in kilogram meters per second, rounding your answer to three decimal places. | \[
\begin{aligned}
p_{\text {proton }} &=E_{\text {proton }} / c \\
p &=\text { Momentum } \\
E &=\text { Energy }=h \nu \\
c &=\text { Speed of light, } 3 \times 10^{8} \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
\[
\begin{aligned}
& p_{\mathrm{PH}}=\frac{h \nu}{c} \quad \nu=c / \lambda \\
& p_{\mathrm{PH}}=h / \lambda(\lambda \text { in meters }), 500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m} \\
& p_{\mathrm{PH}}=h / 500 \times 10^{-9}=6.626 \times 10^{-34} / 500 \times 10^{-9}=\boxed{1.325e-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
\] | Physical Chemistry (5.61 Fall 2017) | 265 | [
"1.325 \\times 10^{-27}\n",
"1.325 \\times 10^{-27}"
] |
Preamble: Evaluate the following integrals for $\psi_{J M}$ eigenfunctions of $\mathbf{J}^{2}$ and $\mathbf{J}_{z}$.
$\int \psi_{22}^{*}\left(\widehat{\mathbf{J}}^{+}\right)^{4} \psi_{2,-2} d \tau$ | \[
\begin{gathered}
\int \psi_{22}^{*}\left(\hat{J}_{+}\right)^{4} \psi_{2,-2} d \tau=\int \psi_{22}^{*} \sqrt{2(2+1)-(-2)(-2+1)}\left(\hat{J}_{+}\right)^{3} \psi_{2,-1} d \tau \\
=\int \psi_{22}^{*} \sqrt{2(2+1)-(-2)(-2+1)} \sqrt{2(2+1)-(-1)(-1+1)}\left(\hat{J}_{+}\right)^{2} \psi_{2,0} d \tau \\
=\int \psi_{22}^{*} \sqrt{2(2+1)-(-2)(-2+1)} \sqrt{2(2+1)-(-1)(-1+1)} \\
\times \sqrt{2(2+1)-(0)(0+1)}\left(\hat{J}_{+}\right) \psi_{2,1} d \tau \\
=\int \psi_{22}^{*} \sqrt{2(2+1)-(-2)(-2+1)} \sqrt{2(2+1)-(-1)(-1+1)} \\
\times \sqrt{2(2+1)-(0)(0+1)} \sqrt{2(2+1)-(1)(1+1)} \psi_{22} d \tau \\
=\sqrt{4} \times \sqrt{6} \times \sqrt{6} \times \sqrt{4} \int \psi_{22}^{*} \psi_{22} d \tau \\
=\boxed{24}
\end{gathered}
\] | Physical Chemistry (5.61 Fall 2017) | 270 | [
"24"
] |
Preamble: Consider the 3-level $\mathbf{H}$ matrix
\[
\mathbf{H}=\hbar \omega\left(\begin{array}{ccc}
10 & 1 & 0 \\
1 & 0 & 2 \\
0 & 2 & -10
\end{array}\right)
\]
Label the eigen-energies and eigen-functions according to the dominant basis state character. The $\widetilde{10}$ state is the one dominated by the zero-order state with $E^{(0)}=10, \tilde{0}$ by $E^{(0)}=0$, and $-\widetilde{10}$ by $E^{(0)}=-10$ (we will work in units where $\hbar \omega = 1$, and can be safely ignored).
Use non-degenerate perturbation theory to derive the energy $E_{\widetilde{10}}$. Carry out your calculations to second order in the perturbing Hamiltonian, and round to one decimal place. | $E_{\widetilde{10}} = 10 + \frac{1^2}{10 - 0} = \boxed{10.1}.$ | Physical Chemistry (5.61 Fall 2017) | 271 | [
"10.1\n"
] |
Preamble: Consider the Particle in an Infinite Box ``superposition state'' wavefunction,
\[
\psi_{1,2}=(1 / 3)^{1 / 2} \psi_{1}+(2 / 3)^{1 / 2} \psi_{2}
\]
where $E_{1}$ is the eigen-energy of $\psi_{1}$ and $E_{2}$ is the eigen-energy of $\psi_{2}$.
Suppose you do one experiment to measure the energy of $\psi_{1,2}$. List the possible result(s) of your measurement. | Since the only eigenergies are $E_{1}$ and $E_{2}$, the possible outcomes of the measurement are $\boxed{E_{1},E_{2}}$. | Physical Chemistry (5.61 Fall 2017) | 269 | [
"E_1, E_2"
] |
Preamble: A pulsed Nd:YAG laser is found in many physical chemistry laboratories.
For a $2.00 \mathrm{~mJ}$ pulse of laser light, how many photons are there at $1.06 \mu \mathrm{m}$ (the Nd:YAG fundamental) in the pulse? PAnswer to three significant figures. | For $1.06 \mu \mathrm{m}$ Light
Energy of one photon $=E_{p}=h \nu ; \nu=c / \lambda ; E_{p}=h c / \lambda$
\[
\begin{aligned}
\lambda &=1.06 \mu \mathrm{m}=1.06 \times 10^{-6} \mathrm{~m} \\
c &=3 \times 10^{8} \mathrm{~m} / \mathrm{s} \\
h &=\text { Planck's constant }=6.626 \times 10^{-34} \mathrm{~kg} \mathrm{} \mathrm{m}^{2} / \mathrm{s}
\end{aligned}
\]
$E_{p}=1.88 \times 10^{-19} \mathrm{~J}$
$1.88 \times 10^{-19} \mathrm{~J} /$ photon, we want photons/pulse.
\[
\frac{1}{1.88 \times 10^{19} \mathrm{~J} / \text { photon }} \times \frac{2.00 \times 10^{-3}}{\text { pulse }}=\boxed{1.07e16} \mathrm{photons} / \mathrm{pulse}
\] | Physical Chemistry (5.61 Fall 2017) | 263 | [
"1.07 \\times 10^{16}\n"
] |
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