imone/ARB huggingface dataset:
Question: A beam of electrons is fired in free space over a distance of $10^{4} \mathrm{~km}$. If the transversal width of the initial packet is $1 \mu \mathrm{m}$, what will be the minimum spread upon arrival if the kinetic energy of the electrons is $13.6 \mathrm{eV}$ ?
Answer: The initial wave packet of an electron in the beam has uncertainties $\Delta p_{x}, \Delta x$ in a transverse direction related by the uncertainty relation

$$
\Delta p_{x} \cdot \Delta x \geq \frac{\hbar}{2} .
$$

If it has momentum $p$, the angle of divergence is

$$
\theta=\frac{\Delta p_{x}}{p} \gtrsim \frac{1}{2 \Delta x} \frac{\hbar}{p}=\frac{1}{2 \Delta x} \frac{\hbar c}{p c} .
$$

As

$$
\begin{gathered}
p=\sqrt{2 m T}=\frac{1}{c} \sqrt{2 m c^{2} T}=\frac{1}{c} \sqrt{2 \times 0.51 \times 10^{6} \times 13.6}=3.72 \times 10^{3} \mathrm{eV} / \mathrm{c}, \\
\Delta x \approx 10^{-6} \mathrm{~m},
\end{gathered}
$$

we have

$$
\theta \gtrsim \frac{6.6 \times 10^{-16} \times 3 \times 10^{8}}{2 \times 10^{-6} \times 3.72 \times 10^{3}}=2.66 \times 10^{-5} \mathrm{radian}
$$

and the minimum transverse spread is

$$
L \theta_{\min }=266 \mathrm{~m} .
$$
Question: Assume that the earth's magnetic field is caused by a small current loop located at the center of the earth. Given that the field near the pole is $0.8$ gauss, that the radius of the earth is $R=6 \times 10^{6} \mathrm{~m}$, and that $\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$, use the Biot-Savart law to calculate the strength of the mangnetic moment of the small current loop.
Answer: Assume that the axis of the current loop, of small radius a, coincides with the axis of rotation of the earth, which is taken to be the $z$-axis as shown in Fig. 2.8. The contribution of a current element $I d I$ to the magnetic induction $B$ at an axial point $z$ is, according to the Biot-Savart law,

$$
d \mathrm{~B}=\frac{\mu_{0}}{4 \pi} \frac{I d \rrbracket \times \mathrm{r}}{r^{3}} .
$$

$d \mathrm{~B}$ is in the plane containing the $z$-axis and $\mathbf{r}$ and is perpendicular to r. Summing up the contributions of all current elements of the loop, by symmetry the resultant $B$ will be along the $z$-axis, i.e.,

$$
\begin{aligned}
&B=B_{z} \mathbf{e}_{z}, \text { or } \\
&d B_{z}=d B \cdot \frac{a}{r} .
\end{aligned}
$$

At the pole, $z=R$. As $R \gg a, r=\sqrt{R^{2}+a^{2}} \approx R$ and

$$
\begin{aligned}
B_{z} &=\frac{\mu_{0}}{4 \pi} \frac{I a}{R^{3}} \oint d l=\frac{\mu_{0}}{4 \pi} \frac{I a}{R^{3}} \cdot 2 \pi a \\
&=\frac{\mu_{0}}{2 \pi} \frac{I S}{R^{3}},
\end{aligned}
$$

where $S=\pi a^{2}$ is the area of the current loop. 

Fig. $2.8$

The magnetic moment of the loop is $\mathbf{m}=I S \mathbf{e}_{z}$, thus

$$
m=\frac{2 \pi R^{3}}{\mu_{0}} B_{z} .
$$

Using the given data $R=6 \times 10^{6} \mathrm{~m}, B_{z}=0.8 \mathrm{Gs}$, we obtain

$$
m \approx 8.64 \times 10^{-26} \mathrm{Am}^{2} \text {. }
$$
Question: Consider simple models for the earth's atmosphere. Neglect winds, convection, etc, and neglect variation in gravity.

Assume that the atmosphere is isothermal $\left(\right.$ at $\left.0^{\circ} \mathrm{C}\right)$. By using an expression for the distribution of molecules with height, estimate roughly the height below which half the molecules lie.
Answer: The molecular number density at height $h$ is denoted by $n(h)$. From the condition of mechanical equilibrium $d p=-n m g d h$ and the equation of state $p=n k T$, we find

$$
\frac{1}{p} d p=-\frac{m g}{k T} d h .
$$

Thus $n(h)=n_{0} \exp (-m g h / k T)$. Let $\int_{0}^{H} n(h) d h / \int_{0}^{\infty} n(h) d h=\frac{1}{2}$, then

$$
H=\frac{k T}{m g} \ln 2=\frac{R T}{N_{0} m g} \ln 2 .
$$

The average molecular weight of the atmosphere is 30 . We have

$$
H=\frac{8.31 \times 10^{7} \times 273}{30 \times 980} \times \ln 2 \approx 8 \times 10^{5} \mathrm{~cm}=8 \mathrm{~km} .
$$
Question: A parabolic mirror of small relative aperture ($10 \mathrm{~cm}$ dia, $500 \mathrm{~cm}$ focal length) is used to photograph stars. Roughly what is the diameter of the image "blob" for a star on axis for visible light (take $\lambda=5000 \AA$)?
Answer: For photographing stars, the dominant limitation of resolution is the Fraunhofer diffraction by the aperture of the telescope. The diameter of the image "blob" (the Airy disk) of a star on axis is

$$
d=\frac{1.22 \lambda f}{D} .
$$

With $D=10 \mathrm{~cm}, f=500 \mathrm{~cm}$ and $\lambda=5000 \AA$ we have

$$
d=0.3 \times 10^{-2} \mathrm{~cm} .
$$

For a atar off axis, the ray makes an angle $\theta$ with the axis of the telescope. So the effective diameter of the aperture reduces to $D \cos \theta$, and the diameter of the Airy disk becomes

$$
d^{\prime}=\frac{1.22 \lambda f}{D \cos \theta} .
$$
Question: Suppose an electron is in a state described by the wave function

$$
\psi=\frac{1}{\sqrt{4 \pi}}\left(e^{i \phi} \sin \theta+\cos \theta\right) g(r)
$$

where

$$
\int_{0}^{\infty}|g(r)|^{2} r^{2} d r=1
$$

and $\phi, \theta$ are the azimuth and polar angles respectively.

What is the expectation value of $L_{z}$ ?
Answer: As

$$
Y_{10}=\sqrt{\frac{3}{4 \pi}} \cos \theta, Y_{1, \pm 1}=\mp \sqrt{\frac{3}{8 \pi}} \sin \theta e^{\pm i \phi},
$$

the wave function can be written as

$$
\psi=\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right) g(r)
$$

Hence the possible values of $L_{z}$ are $+\hbar, 0$.

Since

$$
\begin{aligned}
\int|\psi|^{2} d r &=\frac{1}{4 \pi} \int_{0}^{\infty}|g(r)|^{2} r^{2} d r \int_{0}^{\pi} d \theta \int_{0}^{2 \pi}(1+\cos \phi \sin 2 \theta) \sin \theta d \phi \\
&=\frac{1}{2} \int_{0}^{\pi} \sin \theta d \theta=1
\end{aligned}
$$

the given wave function is normalized. The probability density is then given by $P=|\psi|^{2}$. Thus the probability of $L_{z}=+\hbar$ is $\left(\sqrt{\frac{2}{3}}\right)^{2}$ or $2 / 3$ and that of $L_{z}=0$ is $\left(\frac{1}{\sqrt{3}}\right)^{2}$ or $1 / 3$.

$$
\begin{aligned}
\int \psi^{*} L_{z} \psi r^{2} \sin \theta d \theta d \phi d r=& \int\left[\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right)\right]^{*} \\
& \times \hat{L}_{z}\left[\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right)\right] \\
& \times|g(r)|^{2} r^{2} d r \sin \theta d \theta d \phi \\
=& \frac{2}{3} \hbar \int_{0}^{\pi} d \theta \int_{0}^{2 \pi} Y_{11}^{2} d \phi=\frac{2}{3} \hbar
\end{aligned}
$$