imone/ARB huggingface dataset:
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.

If the $z$-component of angular momentum, $L_{z}$, were measured, what is the probability that the result would be $L_{z}=+\hbar$ ?

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 $l=1$.The $z$-component of the angular momentum is

$$
\begin{aligned}
\left\langle\psi^{*}\left|L_{z}\right| \psi\right\rangle=& K^{\prime 2}\left[\frac{1}{2} \cdot \frac{8 \pi}{3} \cdot \hbar\left(Y_{1}^{1}\right)^{2}+\frac{1}{2} \cdot \frac{8 \pi}{3}(-\hbar)\left(Y_{1}^{-1}\right)^{2}\right.\\
&\left.+4 \cdot \frac{4 \pi}{3}(0)\left(Y_{1}^{0}\right)^{2}\right] \\
=& \frac{1}{8 \pi}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}(+\hbar)+\frac{1}{2} \cdot \frac{8 \pi}{3}(-\hbar)\right]=0 .
\end{aligned}
$$The probability of finding $L_{z}=+\hbar$ is

$$
\begin{aligned}
P &=\left|\left\langle L_{z}=+\hbar \mid \psi(\theta, \phi)\right\rangle\right|^{2} \\
&=\frac{1}{8 \pi} \cdot \frac{1}{2} \cdot \frac{8 \pi}{3}=\frac{1}{6} .
\end{aligned}
$$
Question: A student near a railroad track hears a train's whistle when the train is coming directly toward him and then when it is going directly away. The two observed frequencies are $250$ and $200 \mathrm{~Hz}$. Assume the speed of sound in air to be $360 \mathrm{~m} / \mathrm{s}$. What is the train's speed?
Answer: Let $\nu_{0}, \nu_{1}, \nu_{2}$ be respectively the frequency of the whistle emitted by the train, and the frequencies heard by the student when the train is coming and when it is moving away. The Doppler effect has it that

$$
\begin{aligned}
&\nu_{1}=\left(\frac{c}{c-v}\right) \nu_{0}, \\
&\nu_{2}=\left(\frac{c}{c+v}\right) \nu_{0},
\end{aligned}
$$

where $c$ is the speed of sound and $v$ is the speed of the train, and thus

$$
\frac{\nu_{1}}{\nu_{2}}=\frac{c+v}{c-v} .
$$

Putting in the data, we have

$$
1.25=\frac{360+v}{360-v},
$$

or

$$
\frac{2.25}{0.25}=\frac{720}{2 v},
$$

and thus

$$
v=\frac{360}{9}=40 \mathrm{~m} / \mathrm{s} .
$$
Question: An opaque sheet has a hole in it of $0.5 \mathrm{~mm}$ radius. If plane waves of light $(\lambda=5000 \AA)$ fall on the sheet, find the maximum distance from this sheet at which a screen must be placed so that the light will be focused to a bright spot.
Answer: The maximum distance $r$ of the screen from the hole for which the light will be focused to a bright spot is that for which the area of the hole corresponds to the first Fresnel zone only, and is given by

$$
\rho_{1}^{2}=\lambda r,
$$

where $\rho$ is the radius of the hole. Hence

$$
r=\frac{\rho_{1}^{2}}{\lambda}=\frac{0.5^{2}}{5000 \times 10^{-7}}=500 \mathrm{~mm} .
$$

Without the opaque screen, the number of the Fresnel zones will approach infinity. The total amplitude is then

hence

$$
\begin{aligned}
A &=A_{1}-A_{2}+A_{3}-\ldots \ldots \\
&=\frac{A_{1}}{2}+\left(\frac{A_{1}}{2}-A_{2}+\frac{A_{3}}{2}\right)+\left(\frac{A_{3}}{2}-A_{4}+\frac{A_{5}}{2}\right)+\ldots \\
& \approx \frac{A_{1}}{2}
\end{aligned}
$$

$$
I^{\prime} \sim A_{1}^{2} / 4 \text {, i.e., } I^{\prime}=I_{1} / 4
$$

Thus the intensity is now $1 / 4$ that before the opaque sheet is removed.
Question: Assume a visible photon of $3 \mathrm{eV}$ energy is absorbed in one of the cones (light sensors) in your eye and stimulates an action potential that produces a $0.07$ volt potential on an optic nerve of $10^{-9} \mathrm{~F}$ capacitance.

Calculate the charge needed.
Answer: $Q=V C=0.07 \times 10^{-9}=7 \times 10^{-11}$ Coulomb.
Question: Imagine the universe to be a spherical cavity, with a radius of $10^{28} \mathrm{~cm}$ and impenetrable walls.

If the temperature were $0 \mathrm{~K}$, and the universe contained $10^{80}$ electrons in a Fermi distribution, calculate the Fermi momentum of the electrons.
Answer: The number of photons in the angular frequency range from $\omega$ to $\omega+d \omega$ is

$$
d N=\frac{V}{\pi^{2} c^{3}} \frac{\omega^{2} d \omega}{e^{\beta \hbar \omega}-1}, \quad \beta=\frac{1}{k T} .
$$

The total number of photons is

$$
\begin{aligned}
N &=\frac{V}{\pi^{2} c^{3}} \int_{0}^{\infty} \frac{\omega^{2} d \omega}{e^{\beta \omega h / 2 \pi}-1}=\frac{V}{\pi^{2} c^{3}} \frac{1}{(\beta \hbar)^{3}} \int_{0}^{\infty} \frac{x^{2} d x}{e^{x}-1} \\
&=\frac{V}{\pi^{2}}\left(\frac{k T}{\hbar c}\right)^{3} \cdot 2 \sum_{n=1}^{\infty} \frac{1}{n^{3}} \approx \frac{2 \times 1.2}{\pi^{2}} \cdot V\left(\frac{k T}{\hbar c}\right)^{3} \\
&=\frac{2.4}{\pi^{2}} \cdot \frac{4}{3} \pi \cdot\left(10^{28}\right)^{3} \cdot\left(\frac{1.38 \times 10^{-16} \times 3}{1.05 \times 10^{-27} \times 3 \times 10^{10}}\right)^{3} \\
& \approx 2.5 \times 10^{87} .
\end{aligned}
$$

The total energy is

$$
\begin{aligned}
E &=\frac{V \hbar}{\pi^{2} c^{3}} \int_{0}^{\infty} \frac{\omega^{3} d \omega}{e^{\beta \omega h / 2 \pi}-1}=\frac{\pi^{2} k^{4}}{15(h c / 2 \pi)^{3}} V T^{4} \\
& \approx 2.6 \times 10^{72} \mathrm{ergs} .
\end{aligned}
$$

The Fermi momentum of the electrons is

$$
p_{F}=\hbar\left(3 \pi^{2} \frac{N}{V}\right)^{1 / 3}=2 \times 10^{-26} \mathrm{~g} \cdot \mathrm{cm} / \mathrm{s} .
$$
Question: In a simplified model of a relativistic nucleus-nucleus collision, a nucleus of rest mass $m_{1}$ and speed $\beta_{1}$ collides head-on with a target nucleus of mass $m_{2}$ at rest. The composite system recoils at speed $\beta_{0}$ and with center of mass energy $\varepsilon_{0}$. Assume no new particles are created.

Calculate $\beta_{0}$ for a ${ }^{40} \mathrm{Ar}$ nucleus impinging at $\beta_{1}=0.8$ on a ${ }^{238} \mathrm{U}$ nucleus.
Answer: As implied by the question the velocity of light is taken to be one for convenience.

For a system, $E^{2}-p^{2}$ is invariant under Lorentz transformation. In the laboratory frame $\Sigma$, writing $\gamma_{1}=\frac{1}{\sqrt{1-\beta_{1}^{2}}}$,

$$
E^{2}-p^{2}=\left(m_{1} \gamma_{1}+m_{2}\right)^{2}-\left(m_{1} \gamma_{1} \beta_{1}\right)^{2} .
$$

In the center of mass frame $\Sigma^{\prime}, E^{\prime 2}-p^{\prime 2}=\varepsilon_{0}^{2}$. Hence

$$
\begin{aligned}
\varepsilon_{0}^{2} &=\left(m_{1} \gamma_{1}+m_{2}\right)^{2}-\left(m_{1} \gamma_{1} \beta_{1}\right)^{2} \\
&=m_{1}^{2} \gamma_{1}^{2}\left(1-\beta_{1}^{2}\right)+2 m_{1} m_{2} \gamma_{1}+m_{2}^{2} \\
&=m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma_{1},
\end{aligned}
$$

or

$$
\varepsilon_{0}=\sqrt{m_{1}^{2}+m_{2}^{2}+\frac{2 m_{1} m_{2}}{\sqrt{1-\beta_{1}^{2}}}} .
$$

In the laboratory, the system of $m_{1}, m_{2}$ has total momentum $m_{1} \gamma_{1} \beta_{1}$ and total energy $m_{1} \gamma_{1}+m_{2}$. These are conserved quantities so that after the collision the composite system will move with velocity

$$
\beta_{0}=\frac{m_{1} \gamma_{1} \beta_{1}}{m_{1} \gamma_{1}+m_{2}}=\frac{m_{1} \beta_{1}}{m_{1}+m_{2} \sqrt{1-\beta_{1}^{2}}} .
$$

The masses are approximately

$$
\begin{aligned}
&m_{1}=40 \times 0.94=37.6 \mathrm{GeV}, \\
&m_{2}=238 \times 0.94=223.7 \mathrm{GeV} .
\end{aligned}
$$

Then

$$
\begin{aligned}
\varepsilon_{0} &=\sqrt{37.6^{2}+223.7^{2}+\frac{2 \times 37.6 \times 223.7}{\sqrt{1-0.64}}} \\
&=282 \mathrm{GeV}=2.82 \times 10^{5} \mathrm{MeV}, \\
\beta_{0} &=\frac{37.6 \times 0.8}{37.6+223.7 \times \sqrt{1-0.64}}=0.175
\end{aligned}
$$
Question: In a region of empty space, the magnetic field (in Gaussian units) is described by

$$
\mathrm{B}=B_{0} e^{a x} \hat{\mathbf{e}}_{z} \sin w
$$

where $w=k y-\omega t$.

Find the speed of propagation $v$ of this field.
Answer: Express $B$ as $\operatorname{Im}\left(B_{0} e^{a x} e^{i w}\right) \hat{\mathbf{e}}_{z}$.

Using Maxwell's equation

$$
\nabla \times \mathbf{B}=\frac{1}{c} \frac{\partial \mathrm{E}}{\partial t}
$$

and the definition $k=\frac{\omega}{c}$ for empty space, we obtain

$$
\mathbf{E}=\frac{i c}{\omega} \nabla \times \mathbf{B}=\frac{i}{k}\left|\begin{array}{ccc}
\mathbf{e}_{x} & \mathbf{e}_{y} & \mathbf{e}_{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & 0 \\
0 & 0 & B_{z}
\end{array}\right|
$$

where $\frac{\partial}{\partial z}=0$ as $\mathbf{B}$ does not depend on $z$.

Hence

$$
\begin{gathered}
E_{x}=\operatorname{Im}\left(\frac{i}{k} B_{0} e^{a x} i k e^{i w}\right)=-B_{0} e^{a x} \sin w \\
E_{y}=\operatorname{Im}\left(-\frac{i}{k} B_{0} a e^{a x} e^{i w}\right)=-\frac{a c}{\omega} B_{0} e^{a x} \cos w \\
E_{z}=0
\end{gathered}
$$If the wave form remains unchanged during propagation, we have

$$
d w=k d y-w d t=0,
$$

or $\frac{d y}{d t}=\frac{w}{k}=c$. Hence the wave propagates along the $y$ direction with a speed $v=c$.
Question: A meteorite of mass $1.6 \times 10^{3} \mathrm{~kg}$ moves about the earth in a circular orbit at an altitude of $4.2 \times 10^{6} \mathrm{~m}$ above the surface. It suddenly makes a head-on collision with another meteorite that is much lighter, and loses $2.0 \%$ of its kinetic energy without changing its direction of motion or its total mass.

Find the meteorite's distance of closest approach to the earth after the collision.
Answer: The laws of conservation of mechanical energy and conservation of angular momentum apply to the motion of the heavy meteorite after its collision.
For the initial circular motion, $E<0$, so after the collision we still have $E<0$. After it loses $2.0 \%$ of its kinetic energy, the heavy meteorite will move in an elliptic orbit.

From

$$
\frac{m v^{2}}{r}=\frac{G m M}{r^{2}},
$$

we obtain the meteorite's kinetic energy before collision:

$$
\begin{aligned}
\frac{1}{2} m v^{2} &=\frac{G m M}{2 r}=\frac{m g R^{2}}{2 r} \\
&=\frac{m \times 9.8 \times 10^{3} \times 6400^{2}}{2(6400+4200)}=1.89 \times 10^{7} \mathrm{~m} \text { Joules },
\end{aligned}
$$

where $m$ is the mass of the meteorite in kg. The potential energy of the meteorite before collision is

$$
-\frac{G m M}{r}=-m v^{2}=-3.78 \times 10^{7} m \text { Joules } .
$$

During the collision, the heavy meteorite's potential energy remains constant, while its kinetic energy is suddenly reduced to

$$
1.89 \times 10^{7} \mathrm{~m} \times 98 \%=1.85 \times 10^{7} \mathrm{~m} \text { Joules. }
$$

Hence the total mechanical energy of the meteorite after the collision is

$$
E=(1.85-3.78) \times 10^{7} \mathrm{~m}=-1.93 \times 10^{7} \mathrm{~m} \text { Joules } .
$$

From

$$
E=\frac{-G m M}{2 a}=\frac{-m R^{2} g}{2 a},
$$

we obtain the major axis of the ellipse as

$$
\begin{aligned}
2 a &=\frac{R^{2} g}{1.93 \times 10^{7}}=\frac{\left(6400 \times 10^{3}\right)^{2} \times 9.8}{1.93 \times 10^{7}} \\
&=2.08 \times 10^{7} \mathrm{~m}=2.08 \times 10^{4} \mathrm{~km} .
\end{aligned}
$$

As after the collision, the velocity of the heavy meteorite is still perpendicular to the radius vector from the center of the earth, the meteorite is at the apogee of the elliptic orbit. Then the distance of the apogee from the center of the earth is $6400+4200=10600 \mathrm{~km}$ and the distance of the perigee from the center of the earth is

$$
r_{\min }=20800-10600=10200 \mathrm{~km} .
$$

Thus the meteorite's distance of closest approach to the earth after the collision is $10200-6400=3800 \mathrm{~km}$.

From the above calculations, we see that it is unnecessary to know the mass of the meteorite. Whatever the mass of the meteorite, the answer is the same as long as the conditions remain unchanged.
Question: X-rays are reflected from a crystal by Bragg reflection. If the density of the crystal which is of an accurately known structure is measured with an rms error of 3 parts in $10^{4}$, and if the angle the incident and reflected rays make with the crystal plane is $6^{\circ}$ and is measured with an rms error of $3.4$ minutes of arc, then what is the rms error in the determination of the X-ray wavelength?
Answer: For simplicity consider a crystal whose primitive cell is simple cubic with edge $d$ (to be multiplied by a factor of about one for the primitive cells of other crystal structures). For first order reflection, $n=1$ and Bragg's law gives

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

Differentiating, we have

$$
\left|\frac{\Delta \lambda}{\lambda}\right|=\left|\frac{\Delta d}{d}\right|+\cot \theta|\Delta \theta| .
$$

The volume of a unit cell is

$$
d^{3}=\frac{M}{\rho N_{0}},
$$

where $M$ is the molar weight and $\rho$ the mass density of the crystal, and $N_{0}$ is Avogadro's number. This differentiates to give 

$$
\left|\frac{\Delta d}{d}\right|=\frac{1}{3}\left|\frac{\Delta \rho}{\rho}\right| .
$$

Thus

$$
\left|\frac{\Delta \lambda}{\lambda}\right|=\frac{1}{3}\left|\frac{\Delta \rho}{\rho}\right|+\cot \theta \cdot|\Delta \theta|,
$$

and, in terms of rms errors,

$$
\frac{\sigma_{\lambda}}{\lambda}=\left[\left(\frac{1}{3} \frac{\sigma_{\rho}}{\rho}\right)^{2}+\left(\sigma_{\theta} \cot \theta\right)^{2}\right]^{1 / 2} \text {. }
$$

As $\frac{\sigma_{\rho}}{\rho}=3 \times 10^{-4}, \sigma_{\theta} \cot \theta=\frac{\sigma_{\theta} \cos \theta}{\sin \theta} \approx \frac{\sigma_{\theta}}{\theta}=\frac{3.4}{60 \times 6}=9.4 \times 10^{-3}$,

$$
\frac{\sigma_{\lambda}}{\lambda}=\sqrt{10^{-8}+\left(9.4 \times 10^{-3}\right)^{2}}=9.4 \times 10^{-3} .
$$
Question: Find the angular separation in seconds of arc of the closest two stars resolvable by the following reflecting telescope: $8 \mathrm{~cm}$ objective, $1.5$ meter focal length, $80$X eyepiece. Assume a wavelength of $6000 \AA$. ($1 \AA=10^{-8}$ cm).
Answer: The angular resolving power of the telescope is

$$
\Delta \theta_{1}=1.22 \frac{\lambda}{D}=1.22 \times \frac{6000 \times 10^{-8}}{8} \approx 2^{\prime \prime} .
$$

The resolving power of the human eye is about 1 minute of arc. Using an eyepiece of magnification $80 \mathrm{X}$, the eye's resolving power is $\Delta \theta_{2}=1^{\prime} / 80<$ $2^{\prime \prime}$. Therefore, the resolvable angular separation of stars using the telescope is $\Delta \theta=\max \left(\Delta \theta_{1}, \Delta \theta_{2}\right)=2^{\prime \prime}$.
Question: A self-luminous object of height $h$ is $40 \mathrm{~cm}$ to the left of a converging lens with a focal length of $10 \mathrm{~cm}$. A second converging lens with a focal length of $20 \mathrm{~cm}$ is $30 \mathrm{~cm}$ to the right of the first lens.

Calculate the ratio of the height of the final image to the height $h$ of the object.
Answer: From $\frac{1}{f_{1}}=\frac{1}{u_{1}}+\frac{1}{v_{1}}$, where $f_{1}=10 \mathrm{~cm}, u_{1}=40 \mathrm{~cm}$, we obtain $v_{1}=13 \frac{1}{3} \mathrm{~cm}$.

From $\frac{1}{f_{2}}=\frac{1}{u_{2}}+\frac{1}{v_{2}}$, where $f_{2}=20 \mathrm{~cm}, u_{2}=\left(30-\frac{10}{3}\right) \mathrm{cm}$, we obtain $v_{2}=-100 \mathrm{~cm}$, i.e., the final image is $100 \mathrm{~cm}$ to the left of the second lens.The ratio of the final image height to the height $h$ of the object can be calculated from

$$
m=m_{2} \times m_{1}=\left(\frac{v_{2}}{u_{2}}\right)\left(\frac{v_{1}}{u_{1}}\right)=-2 .
$$

The minus sign signifies an inverted image.
Question: A currently important technique for precisely measuring the lifetimes of states of multiply ionized atoms consists of exciting a beam of the desired ions with a laser tuned to a resonance wavelength of the ion under study, and measuring the emission intensity from the ion beam as a function of down- stream distance. When a particular ion beam with a resonance wavelength at $4885 \AA$ is excited with a $4880 \AA$ argon-ion laser output, the intensity is found to drop by a factor of two, $10 \mathrm{~mm}$ down-stream from the excitation point. If ion beam velocity is $v / c=5 \times 10^{-3}$, calculate the angle the laser beam must make with the normal to the ion beam to achieve resonance and the lifetime of the excited state of the ion.

Answer: Take two coordinate frames $\Sigma, \Sigma^{\prime} . \Sigma$ is the laboratory frame in which the laser source $L$ is at rest and the ions move with velocity $\beta c$ along the $x$-axis as shown in Fig. 1.46. $\Sigma^{\prime}$ is the rest frame of an ion in the beam and moves with velocity $\beta c$ relative to $\Sigma$ along the $x$ direction. Consider the transformation of the 4 -vector $\left(\mathbf{k}, \frac{\omega}{c}\right)$. We have

$$
\frac{\omega^{\prime}}{c}=\gamma\left(\frac{\omega}{c}-\beta k \sin \theta\right),
$$

where $\gamma=\left(1-\beta^{2}\right)^{-\frac{1}{2}}, k=\frac{\omega}{c}$. Hence 

$$
\begin{aligned}
\sin \theta &=\frac{1}{\beta}\left(1-\frac{\lambda}{\lambda^{\prime}} \sqrt{1-\beta^{2}}\right) \\
& \approx \frac{1}{\beta}\left[1-\frac{\lambda}{\lambda^{\prime}}\left(1-\frac{\beta^{2}}{2}\right)\right]
\end{aligned}
$$

With $\lambda=4885 \AA, \lambda^{\prime}=4880 \AA, \beta=5 \times 10^{-3}$, we find

$$
\sin \theta=0.2074, \quad \text { or } \quad \theta=11.97^{\circ} \text {. }
$$

As the excited ions move with velocity $\beta c$, on account of time dilation the laboratory lifetime of an ion is $\gamma \tau$, where $\tau$ is the proper lifetime. Hence the excited ion beam will decay according to

$$
I=I_{0} \exp \left(-\frac{t}{\gamma \tau}\right)
$$

giving

$$
\begin{aligned}
\tau &=\frac{t}{\gamma \ln \left(\frac{I_{0}}{I}\right)}=\frac{x}{\gamma \beta c \ln \left(\frac{I_{0}}{I}\right)} \\
& \approx\left(1-\frac{\beta^{2}}{2}\right) \frac{x}{\beta c \ln \left(\frac{I_{0}}{I}\right)}=9.6 \times 10^{-9} \mathrm{~s} .
\end{aligned}
$$
Question: A one-dimensional square well of infinite depth and $1 \AA$ width contains 3 electrons. The potential well is described by $V=0$ for $0 \leq x \leq 1 \AA$ and $V=+\infty$ for $x<0$ and $x>1 \AA$. For a temperature of $T=0 \mathrm{~K}$, the average energy of the 3 electrons is $E=12.4 \mathrm{eV}$ in the approximation that one neglects the Coulomb interaction between electrons. In the same approximation and for $T=0 \mathrm{~K}$, what is the average energy for 4 electrons in this potential well?
Answer: For a one-dimensional potential well the energy levels are given by

$$
E_{n}=E_{1} n^{2},
$$

where $E_{1}$ is the ground state energy and $n=1,2, \ldots$. Pauli's exclusion principle and the lowest energy principle require that two of the three electrons are in the energy level $E_{1}$ and the third one is in the energy level $E_{2}$. Thus $12 \cdot 4 \times 3=2 E_{1}+4 E_{1}$, giving $E_{1}=6 \cdot 2 \mathrm{eV}$. For the case of four electrons, two are in $E_{1}$ and the other two in $E_{2}$, and so the average energy is

(Note: the correct value of $E_{1}$ is

$$
E=\frac{1}{4}\left(2 E_{1}+2 E_{2}\right)=\frac{5}{2} E_{1}=15.5 \mathrm{eV} .
$$

$$
\begin{aligned}
\frac{\pi^{2} \hbar^{2}}{2 m a^{2}} &=\frac{1}{2 m c^{2}}\left(\frac{\pi \hbar c}{a}\right)^{2}=\frac{1}{1.02 \times 10^{6}}\left(\frac{\pi \times 6.58 \times 10^{-16} \times 3 \times 10^{10}}{10^{-8}}\right)^{2} \\
&=37.7 \mathrm{eV} .)
\end{aligned}
$$
Question: Evaluate

$$
I=\int_{0}^{\infty} \frac{d x}{4+x^{4}} .
$$
Answer: As

$$
y=1+2 x+3 x^{2}+4 x^{3}+\ldots+n x^{n-1}+\ldots
$$

we have

$$
x y=x+2 x^{2}+3 x^{3}+4 x^{4}+\ldots+n x^{n}+\ldots
$$

and thus

$$
y-x y=1+x+x^{2}+x^{3}+\ldots+x^{n-1}+\ldots=\frac{1}{1-x}
$$

since $|x|<1$. Hence

$$
y=\frac{1}{(1-x)^{2}} .
$$

The mean value of $x$ is

$$
\bar{x}=\frac{\int_{0}^{\infty} x f(x) d x}{\int_{0}^{\infty} f(x) d x}=\frac{\int_{0}^{\infty} x^{2} e^{-x / \lambda} d x}{\int_{0}^{\infty} x e^{-x / \lambda} d x}=\frac{2 \lambda \int_{0}^{\infty} x e^{-x / \lambda} d x}{\int_{0}^{\infty} x e^{-x / \lambda} d x}=2 \lambda,
$$

where integration by parts has been used for the numerator and $\lambda>0$ has been assumed . The most probable value of $x, x_{m}$, is given by

$$
\left(\frac{d t}{d x}\right)_{x_{m}}=0,
$$

whence $x_{m}=\lambda$.

Note that if $\lambda<0, f(x)$ is infinite at $\infty$ and the mean and most probable values are also infinite.

Fig. $3.14$Write

$$
\begin{aligned}
I &=\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4}+\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4} \\
&=\frac{1}{2} \int_{-\infty}^{0} \frac{d x}{x^{4}+4}+\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4}=\frac{1}{2} \int_{\infty}^{\infty} \frac{d x}{x^{4}+4}
\end{aligned}
$$

and consider

$$
\oint_{C} \frac{d z}{z^{4}+4}
$$

where $C$ is the closed contour of Fig. $3.14$ consisting of the line from $-R$ to $R$ and the semicircle $\Gamma$, traversed in anticlockwise sense, in the complex plane. As

$$
\frac{1}{z^{4}+4}=\frac{1}{\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right)\left(z-z_{4}\right)},
$$

where

$$
z_{1}=1+i, \quad z_{2}=-1+i, \quad z_{3}=-1-i, \quad z_{4}=1-i
$$

the residue theorem gives

$$
\begin{aligned}
&\oint_{C} \frac{d z}{z^{4}+4} \\
&=2 \pi i\left[\frac{1}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)\left(z_{1}-z_{4}\right)}+\frac{1}{\left(z_{2}-z_{1}\right)\left(z_{2}-z_{3}\right)\left(z_{2}-z_{4}\right)}\right] \\
&=2 \pi i\left[\frac{1}{2(2+2 i) \cdot 2 i}-\frac{1}{2 \cdot 2 i(-2+2 i)}\right]=\frac{\pi}{4} .
\end{aligned}
$$

But

$$
\oint_{C} \frac{d z}{z^{4}+4}=\int_{-R}^{R} \frac{d x}{x^{4}+4}+\int_{\Gamma} \frac{d z}{z^{4}+4} \rightarrow \int_{-\infty}^{\infty} \frac{d x}{x^{4}+4}
$$

as $R \rightarrow \infty$. Hence

$$
I=\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8} .
$$