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.

Solution: $y^{s}=y^{d} \longrightarrow 100 p^{\frac{1}{2}}=\frac{6400}{p^{\frac{1}{2}}} \longrightarrow p=64$. Hence $y^{*}=100 \cdot 8=800$ and $y_{i}=\boxed{8}.$

Answer: 8


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.

Subproblem 0: What is the marginal rate of technical substitution of labor for capital?

Solution:
\[
M R T S=\frac{M P_{L}}{M P_{K}}=\boxed{\frac{K-1}{L}}
\]

Answer: \frac{K-1}{L}


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.

Answer: SRAC=LRAC


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?

Solution: \boxed{SRMC=LRMC}, or short-run marginal cost equals long-run levels.

Answer: SRMC=LRMC


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 $y^{s}(p)=100 y_{i}^{s}(p)=\boxed{100 p^{\frac{1}{2}}}$.

Answer: 100 p^{\frac{1}{2}}


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.

Answer: 1


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: $y^{s}=y^{d} \longrightarrow 100 p^{\frac{1}{2}}=\frac{6400}{p^{\frac{1}{2}}} \longrightarrow p=\boxed{64}$.

Answer: 64


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 $(x)$?

Solution:
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 $\boxed{\frac{1}{4p^2}}$.

Answer: \frac{1}{4p^2}


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 $P_{J}=\boxed{6.2}$.

Answer: 6.2


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.

Subproblem 0: Assume that the city of Cambridge sets the price of apples within its borders. What price should it set, in cents?

Solution: 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.

Answer: 75


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):
\[
\begin{array}{l}
MC(5)=10 \\
ATC(5)=6 \\
MC(4)=4 \\
ATC(4)=4
\end{array}
\]

Subproblem 0: 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?

Solution:
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.

Answer: 4


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.

Subproblem 0: Find Albie's (own-price) elasticity of demand, $\epsilon_{q_{a}, p}$. Use the sign convention that $\epsilon_{y, x}=\frac{\partial y}{\partial x} \frac{x}{y}$.

Solution:
\[
\epsilon_{q_{a}, p}=\frac{\partial q_{a}}{\partial p} \frac{p}{q_{a}(p)}=\left[\alpha p^{\alpha-1} f_{a}\left(I_{a} s\right)\right] \frac{p}{p^{\alpha} f_{a}\left(I_{a}\right)}=\boxed{\alpha}
\]

Answer: \alpha


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.

Subproblem 0: 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.

Solution:
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 $\boxed{X_{d}=125-1.25 P}$.

\section{Physical Chemistry (5.61 Fall 2017}

Answer: X_{d}=125-1.25 P


Subproblem 0: 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.

Solution:
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:
\[
E_{v}=\boxed{\hbar \omega(v+1 / 2)-\frac{E_{0}^{2} e^{2}}{2 m \omega^{2}}}
\]

Answer: \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 $\mathrm{eV}$):
$\mathrm{J}_{1 s 1 s}=17.0 Z$
$\mathrm{~J}_{1 s 2 s}=4.8 Z$
$\mathrm{~K}_{1 s 2 s}=0.9 Z$
$\mathrm{~J}_{2 s 2 s}=3.5 Z$
$\mathrm{J}_{1 s 2 p}=6.6 Z$
$\mathrm{~K}_{1 s 2 p}=0.5 Z$
$\mathrm{~J}_{2 s 2 p}=4.4 Z$
$\mathrm{~K}_{2 s 2 p}=0.8 Z$
$\mathrm{J}_{2 p_{i}, 2 p_{i}}=3.9 Z$
$\mathrm{~J}_{2 p_{i}, 2 p_{k}}=3.5 Z$
$\mathrm{~K}_{2 p_{i}, 2 p_{k}}=0.2 Z i \neq k$

Subproblem 0: 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.

Solution:
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
\[
\begin{aligned}
E\left[1 s^{2} 2 p_{x}^{2}\right]=& \sum_{i} E_{i}+\sum_{i, j}^{i>j} \widetilde{J}_{i j}-\widetilde{K}_{i j} \\
=& 2 E_{1 s}+2 E_{2 p} \\
&+\widetilde{J}_{1 s \alpha, 1 s \beta}+\widetilde{J}_{1 s \alpha, 2 p_{x} \alpha}+\widetilde{J}_{1 s \alpha, 2 p_{x} \beta}+\widetilde{J}_{1 s \beta, 2 p_{x} \alpha}+\widetilde{J}_{1 s \beta, 2 p_{x} \beta}+\widetilde{J}_{2 p_{x} \alpha, 2 p_{x} \beta} \\
&-\widetilde{K}_{1 s \alpha, 1 s \beta}-\widetilde{K}_{1 s \alpha, 2 p_{x} \alpha}-\widetilde{K}_{1 s \alpha, 2 p_{x} \beta}-\widetilde{K}_{1 s \beta, 2 p_{x} \alpha}-\widetilde{K}_{1 s \beta, 2 p_{x} \beta}-\widetilde{K}_{2 p_{x} \alpha, 2 p_{x} \beta} \\
=& 2 E_{1 s}+2 E_{2 p}+J_{1 s, 1 s}+4 J_{1 s, 2 p}+J_{2 p_{i}, 2 p_{i}}-2 K_{1 s, 2 p} \\
E\left[1 s^{2} 2 s^{2}\right]=& 2 E_{1 s}+2 E_{2 s}+J_{1 s, 1 s}+4 J_{1 s, 2 s}+J_{2 s, 2 s}-2 K_{1 s, 2 s} \\
\Rightarrow \Delta E=& 4\left(J_{1 s, 2 p}-J_{1 s, 2 s}\right)+\left(J_{2 p_{i}, 2 p_{i}}-J_{2 s, 2 s}\right)-2\left(K_{1 s, 2 p}-K_{1 s, 2 s}\right) \\
=& Z[4(6.6-4.8)-(3.9-3.5)-2(0.5-0.9)] \\
=&+\boxed{7.6 Z} \mathrm{eV}
\end{aligned}
\]

Answer: 7.6 Z


Preamble: A pulsed Nd:YAG laser is found in many physical chemistry laboratories.

Subproblem 0: 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.

Solution:
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.
\[
\frac{1}{1.88 \times 10^{19} \mathrm{~J} / \text { photon }} \times \frac{2.00 \times 10^{-3}}{\text { pulse }}=\boxed{1.07e16} \mathrm{photons} / \mathrm{pulse}
\]

Answer: 1.07e16


Subproblem 0: 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 $200 \mathrm{~nm}$.

Solution:
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
\[
K E=E_{\text {photon }}-\phi_{o}=6.20 \mathrm{eV}-4.40 \mathrm{eV}=1.80 \mathrm{eV}=\boxed{2.88e-19} \mathrm{~J}
\]

Answer: 2.88e-19


Subproblem 0: 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.

Solution:
\[
\begin{aligned}
p_{\text {proton }} &=E_{\text {proton }} / c \\
p &=\text { Momentum } \\
E &=\text { Energy }=h \nu \\
c &=\text { Speed of light, } 3 \times 10^{8} \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
\[
\begin{aligned}
& p_{\mathrm{PH}}=\frac{h \nu}{c} \quad \nu=c / \lambda \\
& p_{\mathrm{PH}}=h / \lambda(\lambda \text { in meters }), 500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m} \\
& p_{\mathrm{PH}}=h / 500 \times 10^{-9}=6.626 \times 10^{-34} / 500 \times 10^{-9}=\boxed{1.325e-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
\]

Answer: 1.325e-27


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 $\beta=\beta_{\mathrm{CC}}=-0.7 \mathrm{eV}$.

Subproblem 0: 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)=)}

Solution:
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 $\boxed{11.9} \mathrm{eV}$

Answer: 11.9


Subproblem 0: 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.

Solution:
\[
\begin{aligned}
D_{\text {ball }} &=0.074 m \\
m_{\text {ball }} &=0.145 \mathrm{~kg} \\
v_{\text {ball }} &=1 \mathrm{~nm} / \mathrm{s}=1 \times 10^{-9} \mathrm{~m} / \mathrm{s}
\end{aligned}
\]
Using de Broglie:
\[
\lambda_{\text {ball }}=\frac{h}{p}=\frac{h}{m \nu}=\frac{6.626 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} / \mathrm{s}}{0.145 \mathrm{~kg} \cdot 1 \times 10^{-9} \mathrm{~m} / \mathrm{s}}=\boxed{4.6e-24} \mathrm{~m}=\lambda_{\text {ball }}
\]

Answer: 4.6e-24


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?

Solution:
\[
\langle E\rangle =\boxed{\frac{1}{3} E_{1}+\frac{2}{3} E_{2}}
\]
This value of $\langle E\rangle$ is between $E_{1}$ and $E_{2}$ and is the weighted average energy.

Answer: \frac{1}{3} E_{1}+\frac{2}{3} E_{2}


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 $\boxed{E_{1},E_{2}}$.

Answer: E_{1},E_{2}


Preamble: Evaluate the following integrals for $\psi_{J M}$ eigenfunctions of $\mathbf{J}^{2}$ and $\mathbf{J}_{z}$.

Subproblem 0: $\int \psi_{22}^{*}\left(\widehat{\mathbf{J}}^{+}\right)^{4} \psi_{2,-2} d \tau$

Solution:
\[
\begin{gathered}
\int \psi_{22}^{*}\left(\hat{J}_{+}\right)^{4} \psi_{2,-2} d \tau=\int \psi_{22}^{*} \sqrt{2(2+1)-(-2)(-2+1)}\left(\hat{J}_{+}\right)^{3} \psi_{2,-1} d \tau \\
=\int \psi_{22}^{*} \sqrt{2(2+1)-(-2)(-2+1)} \sqrt{2(2+1)-(-1)(-1+1)}\left(\hat{J}_{+}\right)^{2} \psi_{2,0} d \tau \\
=\int \psi_{22}^{*} \sqrt{2(2+1)-(-2)(-2+1)} \sqrt{2(2+1)-(-1)(-1+1)} \\
\times \sqrt{2(2+1)-(0)(0+1)}\left(\hat{J}_{+}\right) \psi_{2,1} d \tau \\
=\int \psi_{22}^{*} \sqrt{2(2+1)-(-2)(-2+1)} \sqrt{2(2+1)-(-1)(-1+1)} \\
\times \sqrt{2(2+1)-(0)(0+1)} \sqrt{2(2+1)-(1)(1+1)} \psi_{22} d \tau \\
=\sqrt{4} \times \sqrt{6} \times \sqrt{6} \times \sqrt{4} \int \psi_{22}^{*} \psi_{22} d \tau \\
=\boxed{24}
\end{gathered}
\]

Answer: 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).

Subproblem 0: 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.

Solution:
$E_{\widetilde{10}} = 10 + \frac{1^2}{10 - 0} = \boxed{10.1}.$

\end{document}

Answer: 10.1