imone/ARB huggingface dataset:
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.

The carriers are continuously trapped into immobile sites and then being thermally reionized into mobile states. If the average free lifetime in a mobile state is $10^{-5}$ second, what is the rms distance a carrier diffuses between successive trappings?
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 $l=1.17 \mathrm{~m}$.
Question: A closely wound search coil has an area of $4 \mathrm{~cm}^{2}, 160$ turns and a resistance of $50 \Omega$. It is connected to a ballistic galvanometer whose resistance is $30 \Omega$. When the coil rotates quickly from a position parallel to a uniform magnetic field to one perpendicular, the galvanometer indicates a charge of $4 \times 10^{-5} \mathrm{C}$. What is the flux density of the magnetic field?
Answer: Suppose the coil rotates from a position parallel to the uniform magnetic field to one perpendicular in time $\Delta t$. Since $\Delta t$ is very short, we have

$$
\varepsilon=\frac{\Delta \phi}{\Delta t}=i(R+r)
$$

As $q=i \Delta t$, the increase of the magnetic flux is

$$
\Delta \phi=q(R+r)=B A N,
$$

since the coil is now perpendicular to the field. Hence the magnetic flux density is

$$
\begin{aligned}
B &=\frac{(R+r) q}{A N}=\frac{(50+30) \times\left(4 \times 10^{-5}\right)}{4 \times 10^{-4} \times 160} \\
&=0.05 \mathrm{~T}=50 \mathrm{Gs} .
\end{aligned}
$$
Question: A solenoid has an air core of length $0.5 \mathrm{~m}$, cross section $1 \mathrm{~cm}^{2}$, and 1000 turns. A secondary winding wrapped around the center of the solenoid has 100 turns. A constant current of $1 \mathrm{~A}$ flows in the secondary winding and the solenoid is connected to a load of $10^{3}$ ohms. The constant current is suddenly stopped. How much charge flows through the resistance? (Neglect end effects.)
Answer: Let the current in the winding of the solenoid be $i$. The magnetic induction inside the solenoid is then $B=\mu_{0} n i$ with direction along the axis, $n$ being the number of turns per unit length of the winding.

The total magnetic flux linkage is

$$
\psi=N \phi=N B S=N^{2} \mu_{0} S i / l .
$$

Hence the self-inductance is

$$
\begin{aligned}
L &=\frac{\psi}{i}=N^{2} \mu_{0} S / l \\
& \approx \frac{1000^{2} \times 4 \pi \times 10^{-7} \times 10^{-4}}{1 / 2}=2.513 \times 10^{-4} \mathrm{H} .
\end{aligned}
$$

The total magnetic flux linkage in the secondary winding produced by the currrent $i$ is $\psi^{\prime}=N^{\prime} \phi$, giving the mutual inductance as

$$
M=\frac{\psi^{\prime}}{i}=\frac{N N^{\prime} \mu_{o} S}{l}=2.513 \times 10^{-5} \mathrm{H} .
$$

Because of the magnetic flux linkage $\psi^{\prime}=M I, I$ being the current in the secondary, an emf will be induced in the solenoid when the constant current $I$ in the secondary is suddenly stopped. Kirchhoff's law gives for the induced current $i$ in the solenoid

$$
-\frac{d \psi^{\prime}}{d t}=R i+L \frac{d i}{d t},
$$

or

$$
-d \psi^{\prime}=R i d t+L d i=R d q+L d i
$$

Integrating over $t$ from $t=0$ to $t=\infty$ gives $-\Delta \psi^{\prime}=R q$, since $i(0)=$ $i(\infty)=0$. Thus the total charge passing through the resistance is

$$
q=\frac{-\Delta \psi^{\prime}}{R}=\frac{M I}{R}=\frac{2.513 \times 10^{-5} \times 1}{10^{3}}=2.76 \times 10^{-7} \mathrm{C}
$$
Question: The electric field of an electromagnetic wave in vacuum is given by

$$
\begin{gathered}
E_{x}=0, \\
E_{y}=30 \cos \left(2 \pi \times 10^{8} t-\frac{2 \pi}{3} x\right), \\
E_{z}=0,
\end{gathered}
$$

where $E$ is in volts/meter, $t$ in seconds, and $x$ in meters. Determine the wavelength $\lambda$.
Answer: $$
k=\frac{2 \pi}{3} \mathrm{~m}^{-1}, \quad \omega=2 \pi \times 10^{8} \mathrm{~s}^{-1} .
$$
$f=\frac{\omega}{2 \pi}=10^{8} \mathrm{~Hz}$.$\lambda=\frac{2 \pi}{k}=3 \mathrm{~m}$.
Question: A common lecture demonstration is as follows: hold or clamp a onemeter long thin aluminium bar at the center, strike one end longitudinally (i.e. parallel to the axis of the bar) with a hammer, and the result is a sound wave of frequency $2500 \mathrm{~Hz}$.

From this experiment, calculate the speed of sound in aluminium.
Answer: The point where the bar is struck is an antinode and the point where it is held a node. With the bar held at the center and its one end struck, the wavelength $\lambda$ is related to its length $L$ by $\lambda=2 L$. Hence the speed of sound propagation in the aluminium bar is

$$
v_{\mathrm{Al}}=\nu \lambda=2 \nu L=2 \times 2500 \times 1=5000 \mathrm{~m} / \mathrm{s} \text {. }
$$

The speed of sound in a solid is

$$
v=\sqrt{\frac{Y}{\rho}},
$$

where $Y$ is the Young's modulus of its material and $\rho$ its density. The speed of sound in a fluid is

$$
v=\sqrt{\frac{M}{\rho}}
$$

where $M$ is its bulk modulus and $\rho$ its density. For adiabatic compression of a gas, $M=\gamma p$, where $p$ is its pressure and $\gamma$ the ratio of its principal specific heats; $\gamma=1.4$ for air, a diatomic gas. Hence

$$
\frac{v_{\text {air }}}{v_{\mathrm{Al}}}=\sqrt{\frac{1.4 p \rho_{\mathrm{Al}}}{Y \rho_{\mathrm{air}}}} .
$$

With

$$
\begin{aligned}
p &=1.013 \times 10^{6} \mathrm{dyn} / \mathrm{cm}^{2} \quad(\text { standard atmosphere) }\\
Y &=7.05 \times 10^{11} \mathrm{dyn} / \mathrm{cm}^{2} \\
\rho_{\mathrm{Al}} &=2.7 \mathrm{~g} / \mathrm{cm}^{3} \\
\rho_{\text {air }} &=1.165 \times 10^{-3} \mathrm{~g} / \mathrm{cm}^{3} \quad\left(\text { at } 30^{\circ} \mathrm{C}\right) \\
v_{\text {air }} &=6.83 \times 10^{-2} \times 5000=341 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$$v_{\mathrm{Al}}=5000 \mathrm{~m} / \mathrm{s}$.
Question: In a measurement of $e / m$ for electron using a Thomson type apparatus, i.e., crossed electric and magnetic fields in a cathode-ray tube, it is noted that if the accelerating potential difference is sufficiently large, the ratio $\mathrm{e} / \mathrm{m}$ becomes one-half as large as the accepted value. Take $e / m_{0}=1.8 \times 10^{11}$ $\mathrm{C} / \mathrm{kg}$.

Find the accelerating potential difference $V$ which causes $e / m$ to be one-half its accepted value. Take $c=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$.
Answer: A Thomson type apparatus is shown schematically in Fig. 2.79, where $V_{1}$ is the accelerating voltage and $V_{2}$ is the deflecting voltage.

Fig. $2.79$ With the addition of a magnetic field $B$ as shown, the electromagnetic field has the action of a velocity-filter. With given values of $V_{1}$ and $V_{2}$, we adjust the magnitude of $B$ so that the electrons strike the center $O$ of the screen. At this time the velocity of the electron is $v=E / B$ (since $e E=e v B$ ). Afterward the magnetic field $B$ is turned off and the displacement $y_{2}$ of the electrons on the screen is measured. The ratio $e / m$ is calculated as follows:

$$
\begin{aligned}
y_{1} &=\frac{1}{2} \cdot \frac{e E}{m}\left(\frac{L}{v}\right)^{2}, \\
y_{2}=\frac{D+\frac{L}{2}}{L / 2} y_{1} &=\frac{e E}{m v^{2}}\left(\frac{L^{2}}{2}+L D\right)=\frac{e}{m} \cdot \frac{d B^{2}}{V_{2}}\left(\frac{L^{2}}{2}+L D\right),
\end{aligned}
$$

giving

$$
e / m=\frac{V_{2} y_{2}}{d B^{2}\left(\frac{L^{2}}{2}+L D\right)} .
$$

When the accelerating voltage is very large, relativistic effects must be considered. From energy conversation

$$
e V_{1}+m_{0} c^{2}=m c^{2},
$$

we find

$$
V_{1}=\left(\frac{m}{e}-\frac{m_{0}}{e}\right) c^{2} .
$$

As $\frac{e}{m}=\frac{1}{2} \frac{e}{m_{0}}$, the accelerating voltage is

$$
V_{1}=\frac{m_{0} c^{2}}{e}=\frac{9 \times 10^{16}}{1.8 \times 10^{11}}=5 \times 10^{5} \mathrm{~V}
$$
Question: In high energy proton-proton collisions, one or both protons may "diffractively dissociate" into a system of a proton and several charged pions. The reactions are

(1) $p+p \rightarrow p+(p+n \pi)$,

(2) $p+p \rightarrow(p+n \pi)+(p+m \pi)$.

Here $n$ and $m$ count the number of produced pions.

In the laboratory frame, an incident proton of total energy $E$ (the projectile) strikes a proton at rest (the target). Find the incident proton energy $E_{0}$ that is the minimum energy for reaction (2) to take place when both protons dissociate into a proton and 4 pions.

$$
m_{\pi}=0.140 \mathrm{GeV}, \quad m_{p}=0.938 \mathrm{GeV} .
$$
Answer: The energy for the reaction

$$
p+p \rightarrow p+(p+4 \pi)
$$

is minimum when all the final particles are at rest in an inertial frame, particularly the center of mass frame $\Sigma^{\prime}$. Then in the laboratory frame $\Sigma$,

$$
E^{2}-p^{2}=\left(E_{0}+m_{p}\right)^{2}-\left(E_{0}^{2}-m_{p}^{2}\right)=2 m_{p} E_{0}+2 m_{p}^{2},
$$

and in $\Sigma^{\prime}$,

$$
E^{\prime 2}-p^{\prime 2}=\left(2 m_{p}+4 m_{\pi}\right)^{2}
$$

so that

$$
2 m_{p} E_{0}=2 m_{p}^{2}+16 m_{p} m_{\pi}+16 m_{\pi}^{2}
$$

giving

$$
E_{0}=\frac{m_{p}^{2}+8 m_{p} m_{\pi}+8 m_{\pi}^{2}}{m_{p}}=2.225 \mathrm{GeV}
$$

as the minimum energy the incident proton must have to cause the reaction.Since both the initial particles are protons and the final state particles are the same as before, the minimum energy remains the same, $2.225 \mathrm{GeV}$.For the reaction

$$
p+p \rightarrow(p+4 \pi)+(p+4 \pi)
$$

we have

$$
\left(E_{0}+m_{p}\right)^{2}-\left(E_{0}^{2}-m_{p}^{2}\right)=\left(2 m_{p}+8 m_{\pi}\right)^{2},
$$

giving the minimum incident energy as

$$
E_{0}=\frac{m_{p}^{2}+16 m_{p} m_{\pi}+32 m_{\pi}^{2}}{m_{p}}=3.847 \mathrm{GeV} .
$$
Question: Consider a spinless particle represented by the wave function

$$
\psi=K(x+y+2 z) e^{-\alpha r}
$$

where $r=\sqrt{x^{2}+y^{2}+z^{2}}$, and $K$ and $\alpha$ are real constants.

What is the total angular momentum of the particle?

You may find the following expressions for the first few spherical harmonics useful:

$$
\begin{aligned}
&Y_{0}^{0}=\sqrt{\frac{1}{4 \pi}}, \quad Y_{1}^{\pm 1}=\mp \sqrt{\frac{3}{8 \pi}} \sin \theta e^{\pm i \phi}, \\
&Y_{1}^{0}=\sqrt{\frac{3}{4 \pi}} \cos \theta, \quad Y_{2}^{\pm 1}=\mp \sqrt{\frac{15}{8 \pi}} \sin \theta \cos \theta e^{\pm i \phi} .
\end{aligned}
$$
Answer: The wave function may be rewritten in spherical coordinates as

$$
\psi=K r(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta) e^{-\alpha r}
$$

its angular part being

$$
\psi(\theta, \phi)=K^{\prime}(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta),
$$

where $K^{\prime}$ is the normalization constant such that

$$
K^{\prime 2} \int_{\theta}^{\pi} d \theta \int_{0}^{2 \pi} \sin \theta(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta)^{2} d \phi=1 \text {. }
$$

Since

$$
\cos \phi=\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right), \quad \sin \phi=\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right)
$$

we have

$$
\begin{aligned}
\psi(\theta, \phi) &=K^{\prime}\left[\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right) \sin \theta+\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right) \sin \theta+\cos 2 \theta\right], \\
&=K^{\prime}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$

The normalization condition and the orthonormality of $Y_{l}^{m}$ then give

$$
K^{\prime 2}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}+\frac{1}{2} \cdot \frac{8 \pi}{3}+4 \cdot \frac{4 \pi}{3}\right]=1
$$

or

$$
K^{\prime}=\sqrt{\frac{1}{8 \pi}},
$$

and thus

$$
\begin{aligned}
\psi(\theta, \phi)=& \sqrt{\frac{1}{8 \pi}}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}\right.\\
&\left.+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$

The total angular momentum of the particle is

$$
\sqrt{\left\langle\mathbf{L}^{2}\right\rangle}=\sqrt{l(l+1)} \hbar=\sqrt{2} \hbar .
$$

as the wave function corresponds to $l=1$.
Question: In high energy proton-proton collisions, one or both protons may "diffractively dissociate" into a system of a proton and several charged pions. The reactions are

(1) $p+p \rightarrow p+(p+n \pi)$,

(2) $p+p \rightarrow(p+n \pi)+(p+m \pi)$.

Here $n$ and $m$ count the number of produced pions.

In the laboratory frame, an incident proton of total energy $E$ (the projectile) strikes a proton at rest (the target). Find the incident proton energy $E_{0}$ that is the minimum energy for reaction (1) to take place when the target dissociates into a proton and 4 pions.

$$
m_{\pi}=0.140 \mathrm{GeV}, \quad m_{p}=0.938 \mathrm{GeV} .
$$
Answer: The quantity $E^{2}-p^{2}$ for a system, where we have taken $c=1$ for convenience, is invariant under Lorentz transformation. If the system undergoes a nuclear reaction that conserves energy and momentum, the quantity will also remain the same after the reaction. In particular for a particle of rest mass $m$,

$$
E^{2}-p^{2}=m^{2} .
$$

The energy for the reaction

$$
p+p \rightarrow p+(p+4 \pi)
$$

is minimum when all the final particles are at rest in an inertial frame, particularly the center of mass frame $\Sigma^{\prime}$. Then in the laboratory frame $\Sigma$,

$$
E^{2}-p^{2}=\left(E_{0}+m_{p}\right)^{2}-\left(E_{0}^{2}-m_{p}^{2}\right)=2 m_{p} E_{0}+2 m_{p}^{2},
$$

and in $\Sigma^{\prime}$,

$$
E^{\prime 2}-p^{\prime 2}=\left(2 m_{p}+4 m_{\pi}\right)^{2}
$$

so that

$$
2 m_{p} E_{0}=2 m_{p}^{2}+16 m_{p} m_{\pi}+16 m_{\pi}^{2}
$$

giving

$$
E_{0}=\frac{m_{p}^{2}+8 m_{p} m_{\pi}+8 m_{\pi}^{2}}{m_{p}}=2.225 \mathrm{GeV}
$$

as the minimum energy the incident proton must have to cause the reaction.
Question: A Fresnel sone plate is made by dividing a photographic image into 5 separate sones.

The first sone consists of an opaque circular disc of radius $r_{1}$. The second is a concentric transparent ring from $r_{1}$ to $r_{2}$ followed by an opaque ring from $r_{2}$ to $r_{3}$, a second transparent ring from $r_{3}$ to $r_{4}$ and a final sone from $r_{4}$ to infinity that is black. The radii $r_{1}$ to $r_{4}$ are in the ratio $r_{1}: r_{2}: r_{3}: r_{4}=1: \sqrt{2}: \sqrt{3}: \sqrt{4}$.

The sone plate is placed in the $x-y$ plane and illuminated by plane monochromatic light waves of wavelength $5,000 \AA$. The most intense spot of light behind the plate is seen on the axis of the sone plate 1 meter behind it.

What is the radius $r_{1}$?
Answer: The focal lengths of the sones are given by

$$
f=\frac{r_{j}^{2}}{j \lambda} .
$$

For $j=1, f=1 \mathrm{~m}, \lambda=5000 \AA$, we have $r_{1}=0.707 \mathrm{~mm}$.
Question: What is the drift velocity of electrons in a $1 \mathrm{~mm} \mathrm{Cu}$ wire carrying $10 \mathrm{~A}$?
Answer: It is $10^{-2} \mathrm{~cm} / \mathrm{sec}$.
Question: The ground state of the realistic helium atom is of course nondegenerate. However, consider a hypothetical helium atom in which the two electrons are replaced by two identical, spin-one particles of negative charge. Neglect spin-dependent forces. For this hypothetical atom, what is the degeneracy of the ground state? 
Answer: The two new particles are Bosons; thus the wave function must be symmetrical. In the ground state, the two particles must stay in 1s orbit. Then the space wave function is symmetrical, and consequently the spin wave function is symmetrical too. As $s_{1}=1$ and $s_{2}=1$, the total $S$ has three possible values:

$S=2$, the spin wave function is symmetric and its degeneracy is $2 S+1=5$.

$S=1$, the spin wave function is antisymmetric and its degeneracy is $2 S+1=3$.

$S=0$, the spin wave function is symmetric and its degeneracy is $2 S+1=1$

If the spin-dependent forces are neglected, the degeneracy of the ground state is $5+3+1=9$.
Question: Consider the pion photoproduction reaction

$$
\gamma+p \rightarrow p+\pi^{\circ}
$$

where the rest energy is $938 \mathrm{MeV}$ for the proton and $135 \mathrm{MeV}$ for the neutral pion.

If the initial proton is at rest in the laboratory, find the laboratory threshold gamma-ray energy for this reaction to "go".
Answer: The quantity $E^{2}-P^{2} c^{2}$ is invariant under Lorentz transformation and for an isolated system is the same before and after a reaction. The threshold $\gamma$-ray energy is that for which the final state particles are all at rest in the center of mass frame. Thus

$$
\left(E_{\gamma}+m_{p} c^{2}\right)^{2}-\left(\frac{E_{\gamma}}{c}\right)^{2} c^{2}=\left(m_{p}+m_{\pi}\right)^{2} c^{4}
$$

where $E_{\gamma}$ is the energy of the photon and $\frac{E_{\gamma}}{c}$ its momentum, giving

$$
E_{\gamma}=\frac{\left(m_{\pi}^{2}+2 m_{\pi} m_{p}\right) c^{4}}{2 m_{p} c^{2}}=144.7 \mathrm{MeV}
$$

as the threshold $\gamma$-ray energy.
Question: A monochromatic transverse wave with frequency $\nu$ propagates in a direction which makes an angle of $60^{\circ}$ with the $x$-axis in the reference frame $K$ of its source. The source moves in the $x$-direction at speed $v=\frac{4}{5} c$ towards an observer at rest in the $K^{\prime}$ frame (where his $x^{\prime}$-axis is parallel to the $x$-axis). The observer measures the frequency of the wave.

What is the angle of observation in the $K^{\prime}$ frame?
Answer: The frame $K$ of the light source moves with velocity $\beta$ c relative to $K^{\prime}$, the observer's frame. The (inverse) transformation of the components of the wave 4 -vector is given by

$$
k_{x}^{\prime} \mathrm{c}=\gamma\left(k_{x} \mathrm{c}+\beta \omega\right), \quad k_{y}^{\prime} \mathrm{c}=k_{y} \mathrm{c}, \quad k_{z}^{\prime} \mathrm{c}=k_{z} \mathrm{c}, \quad \omega^{\prime}=\gamma\left(\omega+\beta k_{x} \mathrm{c}\right)
$$

where $\gamma=\left(1-\beta^{2}\right)^{-\frac{1}{2}}$. The angular frequency of the wave in $K$ is $\omega=2 \pi \nu$. If the angle between the light and the $x$-axis is $\theta$, then

$$
k_{x}=k \cos \theta, \quad k_{y}=k \sin \theta, \quad k_{z}=0, \quad \omega=k \mathrm{c} .
$$

Thus

$$
\omega^{\prime}=\gamma(\omega+\beta \omega \cos \theta)=\gamma(1+\beta \cos \theta) \omega,
$$

or

$$
\nu^{\prime}=\frac{(1+\beta \cos \theta) \nu}{\sqrt{1-\beta^{2}}} .
$$

The above can also be written as

$$
k^{\prime}=\gamma(1+\beta \cos \theta) k
$$

As

$$
k_{x}^{\prime}=\gamma\left(k_{x}+\frac{\beta \omega}{\mathrm{c}}\right)=\gamma k(\cos \theta+\beta),
$$

the angle $\mathbf{k}^{\prime}$ makes with the $x^{\prime}$-axis is given by

$$
\cos \theta^{\prime}=\frac{k_{x}^{\prime}}{k^{\prime}}=\frac{\cos \theta+\beta}{1+\beta \cos \theta} .
$$

With $\beta=0.8, \theta=60^{\circ}$, we have

$$
\nu^{\prime}=\left(\frac{1+0.8 \cos 60^{\circ}}{\sqrt{1-0.8^{2}}}\right) \nu=\frac{1.4}{0.6} \nu=\frac{7}{3} \nu,
$$

$$
\cos \theta^{\prime}=\frac{0.5+0.8}{1+0.8 \times 0.5}=\frac{13}{14},
$$

giving $\theta^{\prime}=21.8^{\circ}$
Question: Suppose that the ionosphere is in thermal equilibrium at temperature $T$. It consists of electrons (mass $m$ ) and singly charged ions (mass $M$ ). If there were no large-scale electric field, the electrons and ions would have very different scale heights, $k T / m g$ and $k T / M g$ respectively. This would produce a large electric field, thereby modifying the density vs. height relation.

Taking into account the electrostatic field, find the numerical value of $E$ if the ions are protons.
(You may ignore the curvature of the Earth's surface. it may be helpful to know that the electric field turns out to be uniform.)
Answer: Use a coordinate system with origin at the Earth's surface and the $z$-axis vertically upward. The self-consistent electric field $\mathbf{E}$ is along the $z$ direction:

$$
E=-\frac{\partial V}{\partial z},
$$

or

$$
V=-E z
$$

assuming $E$ to be uniform. Then an electron and an ion at height $z$ will respectively have energies

$$
\begin{aligned}
&\varepsilon_{\mathrm{e}}=m g z+e E z \\
&\varepsilon_{\mathrm{i}}=M g z-e E z
\end{aligned}
$$

and number densities

$$
\begin{aligned}
&n_{\mathrm{e}}=n_{\mathrm{e} 0} e^{-\varepsilon_{\mathrm{e}} / k T}=n_{\mathrm{e} 0} e^{-(m g+e E) z / k T}, \\
&n_{\mathrm{i}}=n_{\mathrm{i} 0} e^{-\varepsilon_{\mathrm{i}} / k T}=n_{\mathrm{i} 0} e^{-(M g-e E) z / k T}
\end{aligned}
$$

as functions of height.Since

$$
\Delta \cdot \mathbf{E}=\frac{\rho}{\epsilon_{0}}=0
$$

for a uniform field, the net charge density $\rho=0$ everywhere. This means that

$$
-e n_{\mathrm{e}}+e n_{\mathrm{i}}=0,
$$

or

$$
n_{\mathrm{e} 0}=n_{\mathrm{i} 0}, \quad m g+e E=M g-e E .
$$

The last equation gives

$$
E=\frac{(M-m) g}{2 e} \approx \frac{M g}{2 e}
$$

as $M \gg m$.If the ions are protons, we have $M=1.672 \times 10^{-27} \mathrm{~kg}$, $e=$ $1.602 \times 10^{-19} \mathrm{C}$, and

$$
E=\frac{1.672 \times 10^{-27} \times 9.807}{2 \times 1.602 \times 10^{-19}}=5.12 \times 10^{-8} \mathrm{~V} \mathrm{~m}^{-2} .
$$
Question: An electron (mass $m$, charge $e$ ) moves in a plane perpendicular to a uniform magnetic field. If energy loss by radiation is neglected the orbit is a circle of some radius $R$. Let $E$ be the total electron energy, allowing for relativistic kinematics so that $E \gg m c^{2}$.

Compute the needed field induction $B$ numerically, in Gauss, for the case where $R=30$ meters, $E=2.5 \times 10^{9}$ electron-volts.
Answer: In uniform magnetic field $B$ the motion of an electron is described in Gaussian units by

$$
\frac{d p}{d t}=\frac{e}{c} \mathbf{v} \times \mathbf{B},
$$

where $p$ is the momentum of the electron,

$$
\mathbf{p}=m \boldsymbol{v},
$$

with $\gamma=\left(1-\beta^{2}\right)^{-\frac{1}{2}}, \beta=\frac{v}{c}$. Since $\frac{e}{c} \mathbf{v} \times \mathbf{B} \cdot \mathbf{v}=0$, the magnetic force does no work and the magnitude of the velocity does not change, i.e., $v$, and hence $\gamma$, are constant. For circular motion,

$$
\left|\frac{d v}{d t}\right|=\frac{v^{2}}{R} .
$$

Then

$$
m \gamma\left|\frac{d v}{d t}\right|=\frac{e}{c}|\mathbf{v} \times \mathbf{B}| .
$$

As $\mathbf{v}$ is normal to $B$, we have

$$
m \gamma \frac{v^{2}}{R}=\frac{e}{c} v B
$$

or

$$
B=\frac{p c}{e R} \text {. }
$$

With $E \gg m c^{2}, p c=\sqrt{E^{2}-m^{2} c^{4}} \approx E$ and

$$
B \approx \frac{E}{e R} \approx 0.28 \times 10^{4} \mathrm{Gs} .
$$
Question: Two particles with the same mass $m$ are emitted in the same direction, with momenta $5 m c$ and $10 m c$ respectively. As seen from the slower one, what is the velocity of the faster particle? ($c=$ speed of light.)
Answer: In the laboratory frame $K_{0}$, the slower particle has momentum

$$
m \gamma_{1} v_{1}=m \gamma_{1} \beta_{1} c=5 m c
$$

giving

$$
\gamma_{1} \beta_{1}=\sqrt{\gamma_{1}^{2}-1}=5
$$

or

$$
\gamma_{1}^{2}=26
$$

Hence

$$
\beta_{1}^{2}=1-\frac{1}{26}=\frac{25}{26}, \quad \text { or } \quad v_{1}=\sqrt{\frac{25}{26}} c \text {. }
$$

Similarly for the faster particle, the velocity is

$$
v_{2}=\sqrt{\frac{100}{101}} c
$$

Let $K_{1}, K_{2}$ be the rest frames of the slower and faster particles respectively. The transformation for velocity between $K_{0}$ and $K$, which moves with velocity $v$ in the $x$ direction relative to $K_{0}$, is 

$$
u_{x}^{\prime}=\frac{u_{x}-v}{1-\frac{u_{x} v}{c^{2}}} .
$$

Thus in $K_{1}$, the velocity of the faster particle is

$$
v_{2}^{\prime}=\frac{v_{2}-v_{1}}{1-\frac{u_{2} v_{1}}{c^{2}}}=\left(\frac{\sqrt{\frac{100}{101}}-\sqrt{\frac{25}{26}}}{1-\sqrt{\frac{100}{101} \cdot \frac{25}{26}}}\right) \mathrm{c}=0.595 \mathrm{c}
$$

In $K_{2}$, the velocity of the slower particle is

$$
v_{1}^{\prime}=\frac{v_{1}-v_{2}}{1-\frac{v_{1} v_{2}}{\mathrm{c}^{2}}}=-0.595 \mathrm{c} \text {. }
$$