Chapter 1 - Problem 9, 10, 11, 12, 13

Problem 9
A nitrogen nucleus (mass $\cong 14\times$ proton mass) emits a photon of energy 6.2 MeV. If the nucleus is initially at rest, what is the recoil energy of the nucleus in eV?

Solution:
With the nucleus initially at rest, the recoil momentum of the nucleus must be equal and opposite to that of the emitted photon. We therefore have its magnitude given by $p=h\nu/\lambda$, where $h\nu=6.2~\mbox{MeV}$. The recoil energy is
$\begin{aligned}
E&=\frac{p^2}{2M}=h\nu\frac{h\nu}{2Mc^2}=(6.2~\mbox{MeV})\frac{6.2~\mbox{MeV}}{2\times 14\times (940~\mbox{MeV})}\\
&=1.5\times 10^{-13}~\mbox{MeV}
\end{aligned}$

Problem 10
Consider a crystal with planar spacing 0.32 nm. What order of magnitude of energies would one need for (a) electrons, (b) helium nuclei (mass $\cong 4\times$ proton mass) to observe up to three interference maxima?

Solution:
The formula $\lambda=2a\sin\theta /n$ implies that $\lambda/\sin\theta\leq 2a/3$. Since $\lambda=h/p$ this leads to $p=\geq 3h/2a\sin\theta$, which implies that the kinetic energy obeys
\[K=\frac{p^2}{2m}\geq \frac{9h^2}{8ma^2\sin^2\theta}\]
Thus the minimum energy for electrons is
\[K=\frac{9(6.63\times 10^{-34}\mbox{Js})^2}{8(0.9\times 10^{-30}\mbox{kg})(0.32\times 10^{-9}\mbox{m})^2(1.6\times 10^{-19}\mbox{J/eV})}=3.35~\mbox{eV}\]
For Helium atoms the mass is $4(1.67\times 10^{-27}\mbox{kg})/(0.9\times 10^{-30}\mbox{kg})=7.42\times 10^3$ larger, so that
\[K=\frac{33.5~\mbox{eV}}{7.42\times 10^3}=4.5\times 10^{-3}~\mbox{eV}\]

Problem 11
The smallest separation resolvable by microscope is of the order of magnitude of the wavelength used. What energy electrons would one need in an electron microscope to resolve separations of (a) 15 nm, (b) 0.5 nm?

Solution:
We use $$K=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}$$ with $\lambda=15\times 10^{-9}~\mbox{m}$ to get
\[K=\frac{(6.63\times 10^{-34}~\mbox{Js})^2}{2(0.9\times 10^{-30}~\mbox{kg})(15\times 10^{-9}~\mbox{m})^2(1.6\times 10^{-19}~\mbox{J/eV})}=6.78\times 10^{-3}~\mbox{eV}\]
For $\lambda=0.5~\mbox{nm}$, the wavelength is 30 times smaller, so that the energy is 900 times larger. Thus $K =6.10~\mbox{eV}$.

Problem 12
If one assumes that in a stationary state of the hydrogen atom the electron fits into a circular orbit with an integral number of wavelengths, one can reproduce the results of the Bohr theory. Work this out.

Solution:
For a circular orbit of radius $r$, the circumference is $2\pi r$. If $n$ wavelength $\lambda$ are to fit into the orbit, we must have $2\pi r=n\lambda=nh/p$. We therefore get the condition
\[pr=n\frac{h}{2\pi}=n\hbar\]
which is just the condition that the angular momentum in a circular orbit is an integer in units of $\hbar$.

Problem 13
The distance between adjacent planes in a crystal are to be measured. If X rays of wavelength $0.5~\AA$ are detected at an angle of $5^o$, what is the spacing? At what angle will the second maximum occur?

Solution:
We have $a=n\lambda/2\sin\theta$. For $n=1~,\lambda=0.5\times 10^{-10}~\mbox{m}$ and $\theta=5^o$. We get $a=2.87\times 10^{-10}~\mbox{m}$. For $n=2$, we require $\sin\theta_2 = 2\sin\theta_1$. Since the angles are very small, $\theta_2 = 2\theta_1$. So that the angle is $10^o$.