Chapter 1 - Problem 6, 7, 8

Problem 6
A 100-keV photon collides with an electron at rest. It is scattered through $90^o$. What is its energy after the collision? What is the kinetic energy in eV of the electron after the collision, and what is the direction of its recoil?

Solution:
Let $h\nu$ be the incident photon energy, the final photon energy and p the outgoing electron momentum. Energy conservation reads
\[h\nu+mc^2=h\nu^\prime+\sqrt{p^2c^2+m^2c^4}\]
We write the equation for momentum conservation, assuming that the initial photon moves in the x -direction and the final photon in the y-direction. When multiplied by $c$ it read
\[\mathbf {i}(h\nu)=\mathbf{ j}(h\nu^\prime)+(\mathbf {i}p_xc+{\mathbf j}p_yc)\]
Hence $p_xc=h\nu~;p_yc=-h\nu^\prime$. We use this to rewrite the energy conservation equation as follows:
\[(h\nu+mc^2-h\nu^\prime)^2=m^2c^4+c^2(p_x^2+p_y^2)=m^2c^4+(h\nu)^2+(h\nu^\prime)^2\]
From this we get
\[h\nu^\prime=h\nu\left(\frac{mc^2}{h\nu+mc^2}\right)\]
We may use this to calculate the kinetic energy of the electron
$\begin{aligned}
K&=h\nu-h\nu^\prime=h\nu\left(1-\frac{mc^2}{h\nu+mc^2}\right)=h\nu\frac{h\nu}{h\nu+mc^2}\\
&=\frac{(100~\mbox{keV})^2}{100~\mbox{keV}+510~\mbox{keV}}=16.4~\mbox{keV}
\end{aligned}$
Also
\[\mathbf{p}c=\mathbf{ i}(100~\mbox{keV})+\mathbf{ i}(-83.6~\mbox{keV})\]
which gives the direction of the recoiling electron.

Problem 7
An electron of energy 100 MeV collides with a photon of wavelength $3\times 10^6~\mbox{nm}$ (corresponding to universal background of blackbody radiation). What is the maximum energy loss suffered by the electron?

Solution:
The photon energy is
$\begin{aligned}
h\nu&=\frac{hc}{\lambda}=\frac{(6.63\times 10^{-34}~\mbox{Js})(3\times 10^{8}~\mbox{m/s})}{3\times 10^6 \times 10^{-9}~\mbox{m}}=6.63\times 10^{-17}~\mbox{J}\\
&=\frac{6.63\times 10^{-17}~\mbox{J}}{1.60\times 10^{-19}~\mbox{J/eV}}=4.14\times 10^{-4}~\mbox{MeV}
\end{aligned}$
The momentum conservation for collinear motion (the collision is head on for maximum energy loss), when squared, reads
\[\left(\frac{h\nu}{c}\right)^2+p^2+2\left(\frac{h\nu}{c}\right)p\eta_i=\left(\frac{h\nu^\prime}{c}\right)^2+p^\prime^2+2\left(\frac{h\nu^\prime}{c}\right)p^\prime\eta_f\]
Here $eta_i=\pm 1$, with the upper sign corresponding to the photon and the electron moving in the same/opposite direction, and similarly for $\eta_f$. When this is multiplied by $c^2$ we get
\[(h\nu)^2+(pc)^2+2(h\nu)pc\eta_i=(h\nu^\prime)^2+(p^\prime c)^2+2(h\nu^\prime)p^\prime c\eta_f\]
The square of the energy conservation equation, with E expressed in terms of momentum and mass reads
\[(h\nu)^2+(pc)^2+m^2c^4+2Eh\nu=(h\nu^\prime)^2+(p^\prime c)^2+m^2c^4+2E^\primeh\nu^\prime\]
After we cancel the mass terms and subtracting, we get
\[h\nu(E-\eta_i pc)=h\nu^\prime(E^\prime-\eta_f p^\prime c)\]
From this can calculate $h\nu^\prime$ and rewrite the energy conservation law in the form
\[E-E^\prime=h\nu\left(\frac{E-\eta_i pc}{E^\prime-\eta_f p^\prime c}-1\right)\]
The energy loss is largest if $\eta_i=-1~;\eta_f=1$. Assuming that the final electron momentum is not very close to zero, we can write $E+pc=2E$ and $$E^\prime-p^\prime c=\frac{(mc^2)^2}{2E^\prime}$$ so that
\[E-E^\prime=h\nu\left(\frac{2E\times 2E^\prime}{(mc^2)^2}\right)\]
It follows that $$\frac{1}{E^\prime}=\frac{1}{E}+16h\nu$$ with everything expressed in MeV. This leads to $E^\prime = (100/1.64)=61~\mbox{MeV}$ and the energy loss is 39 MeV.

Problem 8
A beam of X rays is scattered by electrons at rest. What is the energy of the X rays if the wavelength of the X rays scattered at $60^o$ to the beam axis is $0.035 \AA$?

Solution:
We have $\lambda^\prime = 0.035\times 10^{-10}~\mbox{m}$, to be inserted into
$\begin{aligned}\lambda^\prime -\lambda&=\frac{h}{m_ec}(1-\cos 60^o)=\frac{h}{2m_ec}=\frac{6.63\times 10^{-34}~\mbox{Js}}{2\times (0.9\times 10^{-30}~\mbox{kg})(3\times 10^{8}~\mbox{m/s})}\\&=1.23\times 10^{-12}~\mbox{m}\end{aligned}$
Therefore
\[\lambda=(3.50-1.23)\times 10^{-12}~\mbox{m}=2.3\times 10^{-12}~\mbox{m}.\]
The energy of the X-ray photon is therefore
\[h\nu=\frac{hc}{\lambda}=\frac{(6.63 \times 10^{-34}~\mbox{Js})(3\times 10^{8}~\mbox{m/s})}{(2.3\times 10^{-12}~\mbox{m})(1.6\times 10^{-19}~\mbox{J/eV})}=5.4\times 10^{5}~\mbox{eV}.\]