imone/ARB huggingface dataset:
Question: A beam of neutral atoms passes through a Stern-Gerlach apparatus. Five equally-spaced lines are observed. What is the total angular momentum of the atom?
Answer: When unpolarized neutral atoms of total angular momentum $J$ pass through the Stern-Gerlach apparatus, the incident beam will split into $2 J+1$ lines. Thus $2 J+1=5$, giving $J=2$.
Question: Calculate the ratio of the mean densities of the earth and the sun from the following approximate data:

$\theta=$ angular diameter of the sun seen from the earth $=\frac{1}{2}^{\circ}$.

$l=$ length of $1^{\circ}$ of latitude on the earth's surface $=100 \mathrm{~km}$.

$t=$ one year $=3 \times 10^{7} \mathrm{~s}$.

$g=10 \mathrm{~ms}^{-2}$.
Answer: Let $r$ be the distance between the sun and the earth, $M_{e}$ and $M_{s}$ be the masses and $R_{e}$ and $R_{s}$ be the radii of the earth and the sun respectively, and $G$ be the gravitational constant. We then have

$$
\begin{aligned}
&\frac{G M_{e} M_{s}}{r^{2}}=M_{e} r \omega^{2}, \\
&\frac{2 R_{s}}{r}=\frac{1}{2} \frac{2 \pi}{360}=\frac{\pi}{360} \mathrm{rad},
\end{aligned}
$$

i.e.

$$
r=\frac{720 R_{s}}{\pi} .
$$

The above gives

$$
\frac{G M_{s}}{\left(720 R_{s} / \pi\right)^{3}}=\omega^{2}
$$

or

$$
\frac{G M_{s}}{R_{s}^{3}}=\left(\frac{720}{\pi}\right)^{3}\left(\frac{2 \pi}{3 \times 10^{7}}\right)^{2} .
$$

For a mass $m$ on the earth's surface,

$$
\frac{G m M_{e}}{R_{e}^{2}}=m g,
$$

giving

$$
\frac{G M_{e}}{R_{e}^{3}}=\frac{g}{R_{e}}=\frac{g}{\left(\frac{360 \times 100}{2 \pi}\right)}=\frac{g \pi}{18 \times 10^{3}}
$$

Hence

$$
\frac{\rho_{e}}{\rho_{s}}=\frac{g \pi}{18 \times 10^{3}}\left(\frac{720}{\pi}\right)^{-3}\left(\frac{2 \pi}{3 \times 10^{7}}\right)^{-2}=3.31
$$
Question: The latent heat of vaporization of water is about $2.44 \times 10^{6} \mathrm{~J} / \mathrm{kg}$ and the vapor density is $0.598 \mathrm{~kg} / \mathrm{m}^{3}$ at $100^{\circ} \mathrm{C}$. Find the rate of change of the boiling temperature with altitude near sea level in ${ }^{\circ} \mathrm{C}$ per $\mathrm{km}$. Assume the temperature of the air is $300 \mathrm{~K}$.

(Density of air at $0^{\circ} \mathrm{C}$ and 1 atm is $1.29 \mathrm{~kg} / \mathrm{m}^{3}$ ).
Answer: The Boltzmann distribution gives the pressure change with height:

$$
p(z)=p(0) \exp -\frac{m g z}{k T},
$$

where $p(0)$ is the pressure at sea level $z=0, m$ is the molecular weight of air, and $T_{0}=300 \mathrm{~K}$ is the temperature of the atmosphere. The ClausiusClapeyron equation can be written as

$$
\frac{d p}{d T}=\frac{L}{T\left(V_{2}-V_{1}\right)}=\frac{L}{T M\left(\frac{1}{\rho_{2}}-\frac{1}{\rho_{1}}\right)}=\frac{\alpha}{T} .
$$

with $\rho_{1}=1000 \mathrm{~kg} / \mathrm{m}^{3}, \rho_{2}=0.598 \mathrm{~kg} / \mathrm{m}^{3}$ and $L / M=2.44 \times 10^{6} \mathrm{~J} / \mathrm{kg}$, we have

$$
\alpha=\frac{L \rho_{1} \rho_{2}}{M\left(\rho_{1}-\rho_{2}\right)}=1.40 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} .
$$

So the rate of change of the boiling point with height is

$$
\frac{d T}{d z}=\frac{d T}{d p} \cdot \frac{d p}{d z}=\frac{T}{\alpha} \cdot\left(\frac{-m g}{k T_{0}}\right) p(z) .
$$

Using the equation of state for ideal gas $p=\rho k T_{0} / m$, we have near the sea level

$$
\frac{d T}{d z}=-\rho g T(0) / \alpha,
$$

where $\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}$ is the density of air, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ and $T(0)=100^{\circ} \mathrm{C}$.

Thus $\frac{d T}{d z}=-0.87^{\circ} \mathrm{C} / \mathrm{km}$.
Question: A 55 year old man can focus objects clearly from $100 \mathrm{~cm}$ to $300 \mathrm{~cm}$. Representing the eye as a simple lens $2 \mathrm{~cm}$ from the retina, what is the focal length of the lens at the far point (focussed at $300 \mathrm{~cm}$)?
Answer: $\frac{1}{f_{\text {far }}}=\frac{1}{u_{\text {far }}}+\frac{1}{v}, \quad u_{\text {far }}=300 \mathrm{~cm}, v=2 \mathrm{~cm}$.

Solving the equation yields $f_{\text {far }}=1.987 \mathrm{~cm}$.
Question: A defective satellite of mass $950 \mathrm{~kg}$ is being towed by a spaceship in empty space. The two vessels are connected by a uniform $50 \mathrm{~m}$ rope whose mass per unit length is $1 \mathrm{~kg} / \mathrm{m}$. The spaceship is accelerating in a straight line with acceleration $5 \mathrm{~m} / \mathrm{sec}^{2}$.

What is the force exerted by the spaceship on the rope?
Answer: $$
\begin{aligned}
F &=\left(m_{\text {rope }}+m_{\text {gatellite }}\right) \cdot a \\
&=(950+50) \times 5=5 \times 10^{3} \mathrm{~N} .
\end{aligned}
$$
Question: The index of refraction of air at $300 \mathrm{~K}$ and 1 atmosphere pressure is $1.0003$ in the middle of the visible spectrum. Assuming an isothermal atmosphere at $300 \mathrm{~K}$, calculate by what factor the earth's atmosphere would have to be more dense to cause light to bend around the earth with the earth's curvature at sea level. (In cloudless skies we could then watch sunset all night, in principle, but with an image of the sun drastically compressed vertically.) You may assume that the index of refraction $n$ has the property that $n-1$ is proportional to the density. (Hint: Think of Fermat's Principle.) The 1/e height of this isothermal atmosphere is 8700 metres.
Answer: We are given that

$$
n(r)-1=\rho e^{-\frac{r-f}{s} r b},
$$

where $R=6400 \times 10^{3} \mathrm{~m}$ is the earth's radius and $\rho$ is the density coefficient of air. Then

$$
\begin{gathered}
n(r)=1+\rho e^{-\frac{5 \pi}{500}}, \\
\frac{d n(r)}{d r}=n^{\prime}(r)=-\frac{1}{8700} \rho e^{-\frac{5-8}{500}} .
\end{gathered}
$$

It is also given that air is so dense that it makes light bend around the earth with the earth's curvature at sea level, as shown in Fig. 1.11.

Fig. $1.11$

The optical path length from $A$ to $B$ is

$$
l=n(r) r \theta .
$$

According to Fermat's Principle, the optical path length from A to B should be an extremum. Therefore,

$$
\frac{d l}{d r}=\left[n^{\prime}(r) r+n(r)\right] \theta=0,
$$

i.e.,

$$
n^{\prime}(r)=\frac{-n(r)}{r} .
$$

Substituting (3) into (2) yields

$$
\frac{1}{8700} \rho e^{-\frac{r-\pi}{800}}=\frac{n(r)}{r} .
$$

At sea level, $r=R=6400 \times 10^{3} \mathrm{~m}$. This with (1) and (4) yields

$$
\frac{\rho \times 6400 \times 10^{3}}{8700}=1+\rho
$$

giving

$$
\rho=0.00136 .
$$

At sea level, i.e., at $300 \mathrm{~K}$ and 1 atmosphere pressure, $n_{0}-1=\rho_{0}=0.0003$. Therefore

$$
\frac{\rho}{\rho_{0}}=4.53 .
$$

Thus only if the air were $4.53$ times as dense as the real air would light bend around the earth with the earth's curvature at sea level.
Question: A 100-ohm resistor is held at a constant temperature of $300 \mathrm{~K}$. A current of 10 amperes is passed through the resistor for $300 \mathrm{sec}$.

What is the change in the internal energy of the universe?
Answer: As the temperature of the resistor is constant, its state does not change. The entropy is a function of state. Hence the change in the entropy of the resistor is zero: $\Delta S_{1}=0$.The heat that flows from the resistor to the external world (a heat source of constant temperature) is

$$
I^{2} R t=3 \times 10^{6} \mathrm{~J} .
$$

The increase of entropy of the heat source is $\Delta S_{2}=3 \times 10^{6} / 300=10^{4} \mathrm{~J} / \mathrm{K}$. Thus the total change of entropy is $\Delta S=\Delta S_{1}+\Delta S_{2}=10^{4} \mathrm{~J} / \mathrm{K}$.The increase of the internal energy of the universe is

$$
\Delta U=3 \times 10^{6} \mathrm{~J} \text {. }
$$
Question: An electron is confined in the ground state of a one-dimensional harmonic oscillator such that $\sqrt{\left\langle(x-\langle x\rangle)^{2}\right\rangle}=10^{-10} \mathrm{~m}$. Find the energy (in $\mathrm{eV})$ required to excite it to its first excited state.

[Hint: The virial theorem can help.]
Answer: The virial theorem for a one-dimensional harmonic oscillator states that $\langle T\rangle=\langle V\rangle$. Thus $E_{0}=\langle H\rangle=\langle T\rangle+\langle V\rangle=2\langle V\rangle=m_{e} \omega^{2}\left\langle x^{2}\right\rangle$, or, for the ground state,

$$
\frac{\hbar \omega}{2}=m_{e} \omega^{2}\left\langle x^{2}\right\rangle,
$$

giving

$$
\omega=\frac{\hbar}{2 m_{e}\left\langle x^{2}\right\rangle} .
$$

As $\langle x\rangle=0$ for a harmonic oscillator, we have

$$
\sqrt{\left\langle(x-\langle x\rangle)^{2}\right\rangle}=\sqrt{\left\langle x^{2}\right\rangle-\langle x\rangle^{2}}=\sqrt{\left\langle x^{2}\right\rangle}=10^{-10} \mathrm{~m} .
$$

The energy required to excite the electron to its first excited state is therefore

$$
\begin{aligned}
\Delta E=\hbar \omega &=\frac{\hbar^{2}}{2 m_{e}\left\langle x^{2}\right\rangle}=\frac{\hbar^{2} c^{2}}{2 m_{e} c^{2}\left\langle x^{2}\right\rangle} \\
&=\frac{\left(6.58 \times 10^{-16}\right)^{2} \times\left(3 \times 10^{8}\right)^{2}}{2 \times 0.51 \times 10^{-20}}=3.8 \mathrm{eV} .
\end{aligned}
$$
Question: $6.0 \times 10^{22}$ atoms of helium gas occupy $2.0$ litres at atmospheric pressure. What is the temperature of the gas?


Answer: Using the equation of state for an ideal gas, we get

$$
T=p V / n k=241 \mathrm{~K} .
$$
Question: What is the cohesive energy with respect to separated ions for crystalline $\mathrm{NaCl}$ ? Give the approximate value and a derivation using a very simple model. Lattice constant $a=5.6 \AA$.
Answer: For a crystal consisting of $N$ ions, each of charge $\pm e$, the cohesive energy is

$$
U(r)=-\frac{N}{2} \sum_{j}^{\prime}\left[\pm \frac{e^{2}}{r_{l j}}-\frac{b}{r_{l j}^{n}}\right]
$$

where $b$ is the Madelung constant and the prime indicates that $j=l$ is excluded from the summation. With $r_{l j}=\alpha_{j} R$ the above becomes

$$
U(R)=-\frac{N}{2}\left(\frac{\alpha e^{2}}{R}-\frac{\beta}{R^{n}}\right),
$$

where

$$
\alpha=\sum_{j}^{\prime}(\pm) \alpha_{j}^{-1}, \quad B=b \sum_{j}^{\prime} \alpha_{j}^{-n}
$$

At equilibrium $U(R)$ is a minimum and $R=R_{0}$ is given by

$$
\left(\frac{\partial U}{\partial R}\right)_{R=R_{0}}=0
$$

which yields

$$
R_{0}=\left(\frac{\alpha e^{2}}{n B}\right)^{\frac{1}{n-1}} .
$$

and thus

$$
U\left(R_{0}\right)=-\frac{N \alpha e^{2}}{2 R_{0}}\left(1-\frac{1}{n}\right) .
$$

The two terms in $U(R)$ are a Coulomb potential and a repulsive potential. By comparing the calculated Coulomb potential and the observed total binding energy, $n$ can be estimated to be about 10 . So to a $10 \%$ accuracy we have

$$
U\left(R_{0}\right) \approx-\frac{N \alpha e^{2}}{2 R_{0}} .
$$

Applying Ewald's method to the known $\mathrm{NaCl}$ structure, we can calculate $\alpha$ and obtain $\alpha=1.7476$. Then with $R_{0}=\frac{a}{2}=2.8 \AA$, we find

$$
U\left(R_{0}\right)=178 \mathrm{kcal} / \mathrm{mol},
$$

as for crystalline $\mathrm{NaCl}, n=8$.
Question: Sunlight is normally incident on the surface of water with index of refraction $n=1.33$.

If the incident flux is $1 \mathrm{~kW} / \mathrm{m}^{2}$, what is the pressure that sunlight exerts on the surface of the water?
Answer: Fresnel's formulae give for normally incidence

$$
\begin{aligned}
&R=\left(\frac{n_{2}-n_{1}}{n_{2}+n_{1}}\right)^{2}=\left(\frac{n-1}{n+1}\right)^{2}=0.02 \\
&T=\frac{4 n_{1} n_{2}}{\left(n_{2}+n_{1}\right)^{2}}=\frac{4 n}{(n+1)^{2}}=0.98 \\
&R+T=1
\end{aligned}
$$

From the point of view of the photon theory, light pressure is the result of a transfer of the momentum of the incoming light to the target. If $W$ is the incident flux density and $R$ the reflection coefficient, the change of momentum per unit area is

$$
\frac{W R}{c}-\left(-\frac{W}{c}\right)=\frac{(1+R) W}{c} .
$$

By definition this is the light pressure $p$

$$
p=\frac{(1+0.02) \times 1000}{3 \times 10^{8}}=3.34 \times 10^{-6} \mathrm{~N} / \mathrm{m}^{2} .
$$
Question: A given type of fuel cell produces electrical energy $W$ by the interaction of $\mathrm{O}_{2}$ fed into one electrode and $\mathrm{H}_{2}$ to the other. These gases are fed in at 1 atmosphere pressure and $298 \mathrm{~K}$, and react isothermally and isobarically to form water. Assuming that the reaction occurs reversibly and that the internal resistance of the cell is negligible, calculate the e.m.f. of the cell. Given: one Faraday $=96,500$ coulombs $/ \mathrm{g}$ mole.

Enthalpies in joules/g mole at 1 atmospheric and $298 \mathrm{~K}$ for oxygen, hydrogen, and water are respectively $17,200,8,100$ and $-269,300$.

Entropies in joules/mole. $\mathrm{K}$ at 1 atmosphere and $298 \mathrm{~K}$ for oxygen, hydrogen, and water are respectively 201,128 and $66.7$.
Answer: The chemical equation is

$$
\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}=\mathrm{H}_{2} \mathrm{O} \text {. }
$$

In the reversible process at constant temperature and pressure, the decrease of Gibbs function of the system is equal to the difference between the total external work and the work the system does because of the change of volume. Thus

$$
-\Delta g=\varepsilon \Delta q,
$$

or

$$
-\sum_{\mathrm{i}}\left(\Delta h_{\mathrm{i}}-T \Delta S_{\mathrm{i}}\right)=\varepsilon \Delta q \text {. }
$$

If 1 mole of water forms, there must have been electric charges of $2 \mathrm{~F}$ flowing in the circuit, i.e., $\Delta g=2 \mathrm{~F}$. Thus the e.m.f. is

$$
\varepsilon=\frac{1}{2 \mathrm{~F}}\left[T S_{\mathrm{W}}-h_{\mathrm{W}}-T\left(S_{\mathrm{H}}+\frac{1}{2} S_{0}\right)+\left(h_{\mathrm{H}}+\frac{1}{2} h_{0}\right)\right]
$$

As given $, S_{0}=201 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, S_{\mathrm{H}}=128 \mathrm{~J} / \mathrm{mol} \mathrm{K}$, $S_{\mathrm{W}}=66.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, \dot{h}_{0}=-17200 \mathrm{~J} / \mathrm{mol}$ $h_{\mathrm{H}}=8100 \mathrm{~J} / \mathrm{mol}, h_{\mathrm{W}}=-269300 \mathrm{~J} / \mathrm{mol}$, and $T=298 \mathrm{~K}$, We have $\varepsilon=1.23 \mathrm{~V}$
Question: Calculate the ratio of the intensities of the reflected wave and the incident wave for a light wave incident normally on the surface of a deep body of water with index of refraction $n=1.33$.
Answer: Fresnel's equations give for light incident normally

$$
R=\frac{I_{\mathrm{r}}}{I_{\mathrm{i}}}=\left(\frac{n_{2}-n_{1}}{n_{2}+n_{1}}\right)^{2}=0.02 .
$$
Question: At room temperature, $k_{\mathrm{B}} T / e=26 \mathrm{mV}$. A sample of cadmium sulfide displays a mobile carrier density of $10^{16} \mathrm{~cm}^{-3}$ and a mobility coefficient $\mu=10^{2} \mathrm{~cm}^{2} /$ volt sec.

If the charge carriers have an effective mass equal to $0.1$ times the mass of a free electron, what is the average time between successive scatterings?
Answer: The electrical conductivity is given by $\sigma=n e \mu$. With $n=10^{22} \mathrm{~m}^{-3}$, $e=1.6 \times 10^{-19} \mathrm{C}, \mu=10^{-2} \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$, we have for the material $\sigma=$ $16 \Omega^{-1} \mathrm{~m}^{-1}$.The law of equipartition of energy

$$
\frac{1}{2} m \bar{v}_{x}^{2}=\frac{1}{2} m \bar{v}_{y}^{2}=\frac{1}{2} m \bar{v}_{z}^{2}=\frac{1}{2} k_{\mathrm{B}} T
$$

gives

$$
\frac{1}{2} m \bar{v}^{2}=\frac{3}{2} k_{\mathrm{B}} T,
$$

or

$$
\bar{v}^{2}=\sqrt{\frac{3 k_{\mathrm{B}} T}{m}} .
$$

The rms distance $l$ between successive trappings is given by

$$
l^{2}=\bar{v}^{2} t^{2} .
$$

Hence

$$
l=\sqrt{\frac{3 k_{\mathrm{B}} T}{m}} t=\sqrt{3\left(\frac{k_{\mathrm{B}} T}{e}\right) \frac{e}{m}} t .
$$

With $\frac{k_{\mathrm{B}} T}{e}=26 \times 10^{-3} \mathrm{~V}, \frac{e}{m}=1.76 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}, t=10^{-5} \mathrm{~s}$, we have $l=1.17 \mathrm{~m}$.The free electron model of metals gives

$$
\sigma=\frac{n e^{2}\langle\tau\rangle}{m^{*}},
$$

where $m^{*}$ is the effective mass of an electron. Then the average time between successive scatterings is

$$
\langle\tau\rangle=\frac{0.1 \sigma}{n e}\left(\frac{m}{e}\right)=5.7 \times 10^{-15} \mathrm{~s} .
$$
Question: For a rock salt $(\mathrm{NaCl})$ crystal placed in front of the tube, calculate the Bragg angle for a first order reflection maximum at $\lambda=0.5 \AA$. $\left(\rho_{\mathrm{NaCl}}=\right.$ $\left.2.165 \mathrm{~g} / \mathrm{cm}^{3}\right)$
Answer: When a high voltage is applied to the picture tube, electrons emitted from the negative electrode will be accelerated by the electric field to strike the screen target. If the energy of the electrons exceeds a certain value, they can knock off inner-shell electrons in the target atoms and make holes in the inner shells. Then as the outer-shell electrons fall in to fill up these holes, X-rays are emitted.
The maximum energy of the X-ray photons produced, $h \nu_{\max }$, is equal to the energy $\mathrm{eV}$ of the incident electrons. Hence the minimum wavelength of the X-rays is

$$
\lambda_{\min }=\frac{h c}{e V}=\frac{12000}{V}=\frac{12000}{25 \times 10^{3}}=0.48 \AA .
$$Bragg's law

$$
2 d \sin \theta=n \lambda
$$

gives the angle $\theta$ for the first order diffraction $(n=1)$ maximum:

$$
\sin \theta=\frac{\lambda}{2 d},
$$

where $d$ is the distance between two neighboring ions in the $\mathrm{NaCl}$ crystal.

As $\mathrm{NaCl}$ crystal has a simple cubic structure with $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions arranged alternately, there are $N_{0} \mathrm{Na}^{+}$ions and $N_{0} \mathrm{Cl}^{-}$ions in a mole of $\mathrm{NaCl}$, where $N_{0}$ is Avogadro's number. As $\mathrm{NaCl}$ has a molar weight $M=58.45 \mathrm{~g} / \mathrm{mol}$ and a density $\rho=2.165 \mathrm{~g} / \mathrm{cm}^{3}$, its crystal has

$$
d=\left(\frac{\frac{M}{\rho}}{2 N_{0}}\right)^{1 / 3}=\left(\frac{\frac{58.45}{2.165 \times 10^{6}}}{2 \times 6.02 \times 10^{23}}\right)^{1 / 3}=2.82 \AA .
$$

This gives

$$
\sin \theta=\frac{0.5}{2 \times 2.82}=0.088
$$

and hence

$$
\theta=5^{\circ} \text {. }
$$
Question: A charge $q=2 \mu \mathrm{C}$ is placed at $a=10 \mathrm{~cm}$ from an infinite grounded conducting plane sheet. Find the force on the charge $q$.
Answer: The method of images requires that an image charge $-q$ is placed symmetrically with respect to the plane sheet. This means that the total induced charge on the surface of the conductor is $-q$.The force acting on $+q$ is

$$
F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{(2 a)^{2}}=9 \times 10^{9} \times \frac{\left(2 \times 10^{-6}\right)^{2}}{0.2^{2}}=0.9 \mathrm{~N}
$$

where we have used $\varepsilon_{0}=\frac{1}{4 \pi \times 9 \times 10^{9}} C^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$.
Question: Consider an opaque screen with 5 equally spaced narrow slits (spacing $d$) and with monochromatic plane-wave light (wavelength $\lambda$) incident normally. Assume $d / \lambda=10$. Approximately what is the ratio in intensity of the least intense to the most intense peak?
Answer: For multiple-slit interference,

$$
I \propto\left(\frac{\sin \frac{N \delta}{2}}{\sin \frac{\delta}{2}}\right)^{2}=\left(\frac{\sin \frac{5 \pi d \sin \theta}{\lambda}}{\sin \frac{\pi d \ln \theta}{\lambda}}\right)^{2}=\left(\frac{\sin (50 \pi \sin \theta)}{\sin (10 \pi \sin \theta)}\right)^{2} .
$$

For $\theta=0$ to $\frac{1}{5}$ rad we may take the approximation $\sin \theta \approx \theta$. Thus

$$
I \propto\left(\frac{\sin (50 \pi \theta)}{\sin (10 \pi \theta)}\right)^{2} .
$$

Therefore, intensity maxima occur when $10 \pi \theta=m \pi$, i.e., $\theta=m / 10$, where $m=0, \pm 1, \pm 2, \ldots$. Intensity minima occur when $50 \pi \theta=n \pi$, i.e., $\theta=n / 50$ where $n$ is an integer such that $n \neq 0, \pm 5, \pm 10, \ldots$

Fig. 2.54 Figure $2.54$ shows the intensity distribution. The least intense peak is midway between two adjacent most intense peaks, for which

$$
\theta=\frac{1}{2} \times \frac{1}{10}=0.05 \mathrm{rad} .
$$

The ratio in intensity of the least to the most intense peak is

$$
\begin{aligned}
\frac{I(\theta=0.05)}{I(\theta=0)} &=\left(\frac{\sin (50 \pi \times 0.05)}{\sin (10 \pi \times 0.05)}\right)^{2} /\left(\frac{\sin (50 \pi \times 0)}{\sin (10 \pi \times 0)}\right)^{2} \\
&=\left[\frac{\sin (0.5 \pi)}{\sin (0.5 \pi)}\right]^{2} /\left[\lim _{\theta \rightarrow 0} \frac{\sin (50 \pi \varepsilon)}{\sin (10 \pi \varepsilon)}\right]^{2}=\frac{1}{25} .
\end{aligned}
$$

The sketch shows the angular distance of the first subsidiary peak from $\theta=0$ to be

$$
\frac{3}{2} \times \frac{1}{50}=0.03 \mathrm{rad} .
$$
Question: What is the smallest possible time necessary to freeze $2 \mathrm{~kg}$ of water at $0^{\circ} \mathrm{C}$ if a 50 watt motor is available and the outside air (hot reservoir) is at $27^{\circ} \mathrm{C}$ ?
Answer: When $2 \mathrm{~kg}$ of water at $0^{\circ} \mathrm{C}$ becomes ice, the heat released is

$$
Q_{2}=1.44 \times 2 \times 10^{3} / 18=1.6 \times 10^{2} \mathrm{kcal} .
$$

The highest efficiency of the motor is

$$
\varepsilon=\frac{T_{2}}{T_{1}-T_{2}}=\frac{Q_{2}}{W_{\min }} .
$$

Thus,

$$
W_{\min }=Q_{2} \frac{T_{1}-T_{2}}{T_{2}} .
$$

If we use the motor of $P=50 \mathrm{~W}$, the smallest necessary time is

$$
\tau=\frac{W_{\min }}{P}=\frac{Q_{2}}{P} \cdot \frac{T_{1}-T_{2}}{T_{2}} .
$$

With $T_{1}=300 \mathrm{~K}, T_{2}=273 \mathrm{~K}$, we find

$$
\tau=1.3 \times 10^{3} \mathrm{~s} .
$$
Question: A gas of molecular hydrogen $\mathrm{H}_{2}$, is originally in equilibrium at a temperature of $1,000 \mathrm{~K}$. It is cooled to $20 \mathrm{~K}$ so quickly that the nuclear spin states of the molecules do not change, although the translational and rotational degrees of freedom do readjust through collisions. What is the approximate internal energy per molecule in terms of temperature units $\mathrm{K}$ ?

Note that the rotational part of the energy for a diatomic molecule is $A l(l+1)$ where $l$ is the rotational quantum number and $A \sim 90 \mathrm{~K}$ for $\mathrm{H}_{2}$. Vibrational motion can be neglected.
Answer: Originally the temperature is high and the para- and orthohydrogen molecules are in equilibrium in a ratio of about $1: 3$. When the system is quickly cooled, for a rather long period the nuclear spin states remain the same. The ratio of parahydrogen to orthohydrogen is still $1: 3$. Now the para- and orthohydrogen are no longer in equilibrium but, through collisions, each component is in equilibrium by itself. At the low temperature of $20 \mathrm{~K}$, $\exp (-\beta A) \sim \exp (-90 / 20) \ll 1$, so that each is in its ground state. Thus $\bar{E}_{r, \mathrm{p}}=0, \bar{E}_{\mathrm{r}, \mathrm{o}}=A(1+1) \cdot 1=2 A=180 \mathrm{~K}$, giving

$$
\bar{E}_{\mathrm{r}}=\frac{1}{4} \bar{E}_{\mathrm{r}, \mathrm{p}}+\frac{3}{4} \bar{E}_{\mathrm{r}, \mathrm{O}}=135 \mathrm{~K} .
$$

From equipartition of energy, we have

$$
\bar{E}_{\mathrm{t}}=\frac{3}{2} k T=30 \mathrm{~K} .
$$

The average energy of a molecule is

$$
\bar{E}=\bar{E}_{\mathrm{t}}+\bar{E}_{\mathrm{r}}=165 \mathrm{~K} .
$$
Question: Čerenkov radiation is emitted by a high energy charged particle which moves through a medium with a velocity greater than the velocity of electromagnetic wave propagation in the medium.

Hydrogen gas at one atmosphere and at $20^{\circ} \mathrm{C}$ has an index of refraction $n=1+1.35 \times 10^{-4}$. What is the minimum kinetic energy in $\mathrm{MeV}$ which an electron (of rest mass $0.5 \mathrm{MeV} / \mathrm{c}^{2}$) would need in order to emit Čerenkov radiation in traversing a medium of hydrogen gas at $20^{\circ} \mathrm{C}$ and one atmosphere?
Answer: As shown in Fig. 5.13, the radiation emitted by the charge at $Q^{\prime}$ at time $t^{\prime}$ arrives at $P$ at time $t$ when the charge is at Q. As the radiation propagates at the speed $c / n$ and the particle has speed $v$ where $v>c / n$, we have

$$
\mathrm{Q}^{\prime} \mathrm{P}=\frac{c}{n}\left(t-t^{\prime}\right), \quad \mathrm{Q}^{\prime} \mathrm{Q}=v\left(t-t^{\prime}\right),
$$

or

$$
\frac{\mathrm{Q}^{\prime} \mathrm{P}}{\mathrm{Q}^{\prime} \mathrm{Q}}=\cos \theta=\frac{c}{v n}=\frac{1}{\beta n},
$$

where $\beta=\frac{v}{c}$. At all the points intermediate between $Q^{\prime}$ and $Q$ the radiation emitted will arrive at the line QP at time $t$. Hence QP forms the wavefront of all radiation emitted prior to $t$.

Fig. $5.13$

As $|\cos \theta| \leq 1$, we require $\beta \geq \frac{1}{n}$ for emission of Čerenkov radiation. Hence we require

$$
\gamma=\frac{1}{\sqrt{1-\beta^{2}}} \geq\left(1-\frac{1}{n^{2}}\right)^{-\frac{1}{2}}=\frac{n}{\sqrt{n^{2}-1}} .
$$

Thus the particle must have a kinetic energy greater than

$$
\begin{aligned}
T &=(\gamma-1) m_{0} c^{2} \\
&=\left[\frac{n}{\sqrt{(n+1)(n-1)}}-1\right] m_{0} c^{2} \\
& \approx\left(\frac{1}{\sqrt{2 \times 1.35 \times 10^{-4}}}-1\right) \times 0.5 \\
& \approx 29.93 \mathrm{MeV} .
\end{aligned}
$$