Chapter 1 - Problem 3, 4, 5

Problem 3
Ultraviolet light of wavelength 350 nm falls on a potassium surface. The maximum energy of the photoelectrons is 1.6 eV. What is the work function of potassium?

Solution:
The relationship is
\[h\nu=K+W\]
where $K$ is the electron kinetic energy and $W$ is the work function. Here
\[h\nu=\frac{hc}{\lambda}=\frac{(6.626 \times 10^{-34}~\mbox{Js})(3 \times 10^{8}~\mbox{m/s})}{350 \times 10^{-9}~\mbox{m}}=5.68 \times 10^{-19}~\mbox{J}=3.55~\mbox{eV}\]
With $K=1.60~\mbox{eV}$, we get $W=1.95~\mbox{eV}$.

Problem 4
The maximum energy of photoelectrons from aluminum is 2.3 eV for radiation of 200 nm and 0.90 eV for radiation of 258 nm. Use these data to calculate Planck's constant and the work function of aluminum.

Solution:
We use
\[\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}=K_1-K_2\]
since $W$ cancels. From this we get
$\begin{aligned}h&=\frac{1}{c}\frac{\lambda_1\lambda_2}{\lambda_2-\lambda_1}(K_1-K_2)\\
&=\frac{(200 \times 10^{-9}~\mbox{m})(258 \times 10^{-9}~\mbox{m})}{(3 \times 10^{8}~\mbox{m/s})(58 \times 10^{-9}~\mbox{m})}\times (2.3-0.9)~\mbox{eV}\times (1.60 \times 10^{-19}~\mbox{J/eV})\\
&=6.64 \times 10^{-34}~\mbox{J.s} \end{aligned}$

Problem 5
A 100-MeV photon collides with a proton that is at rest. What is the maximum possible energy loss for the photon?

Solution:
The maximum energy loss for the photon occurs in a head-on collision, with the photon scattered backwards. Let the incident photon energy be $h\nu$, and the backward-scattered photon energy be $h\nu^\prime$. Let the energy of the recoiling proton be $E$. Then its recoil momentum is obtained from $E=\sqrt{p^2~c^2+m^2~c^4}$. The energy conservation equation reads
\[h\nu+mc^2=h\nu^\prime+E\]
and the momentum conservation equation reads
\[\frac{h\nu}{c}=- \frac{h\nu^\prime}{c} + p \]
that is
\[h\nu=-h\nu^\prime+ pc \]
We get $E+pc-mc^2=2h\nu$ from which it follows that
\[p^2~c^2+m^2~c^4=(2h\nu-pc+mc^2)^2\]
so that
\[pc=\frac{4h^2\nu^2+4h\nu mc^2}{4h\nu+2mc^2}\]
The energy loss for the photon is the kinetic energy of the proton
$K=E-mc^2$. Now $h\nu=100~\mbox{MeV}$ and $mc^2=938~\mbox{MeV}$,
so that
\[pc=182~\mbox{MeV}\]
and
\[E-mc^2=K=17.6~\mbox{MeV}.\]