Question

In the upper atmosphere, suppose an 0.800 nm X-ray photon scatters from a free electron. After the scattering event, the electron recoils at 1.40 x 100 (a) What is the Compton shift in the photon's wavelength? Leave your final result in picometers (10 points) (b) Through what angle is the photon scattered? Leave your result in degrees. (5 points)

Answer 1

1) given :

wavelength of scattered photon =

12 = 0.8 * 10 m

velocity of electron = v = $1.4*10^{6}m/s$

let wavelength of incident photon be $\lambda_{1}$ .

From conservation of energy :

$\frac{hc}{\lambda_{1}}=\frac{1}{2}mv^{2}+\frac{hc}{\lambda_{2}}$

$\Rightarrow \lambda_{1}=\frac{1}{\frac{mv^{2}}{2hc}+\frac{1}{\lambda_{2}}}=\frac{1}{\frac{9.1*10^{-31}*(1.4*10^{6})^{2}}{2*6.63*10^{-34}*3*10^{8}}+\frac{1}{0.8*10^{-9}}}=0.797*10^{-9}m$

= 0.797nm

a) the compton shift = difference in scattered and incident wavelength of photon = $\lambda_{2}-\lambda_{1} = 0.8-0.797 = 0.003nm$ [answer]

b) According to Compton effect :

Compton shift = $\lambda_{2}-\lambda_{1} = \frac{h}{mc}*(1-cos\phi)$ [ phi = angle of scattering of photon, m = mass of electron]

$cos\phi =1- \frac{mc(\lambda_{2}-\lambda_{1})}{h}$

$\Rightarrow \phi =cos^{-1}(1- \frac{mc(\lambda_{2}-\lambda_{1})}{h})=cos^{-1}(1- \frac{9.1*10^{-31}*3*10^{8}*0.003*10^{-9}}{6.63*10^{-34}})$

$=103.61^{\circ \:}$ [answer]