Question

2. (5 pts) X-rays with a wavelength of 2 = 18.0 pm are used in a Compton scattering experiment (i.e. the photons are scattered by essentially free electrons). a. What is the energy, E, of the incident photons in eV? b. If the Compton effect X-ray scatters at right angles to the incoming X-ray direction (0 = 90°), what is the wavelength of the Compton scattered X-ray? c. If the Compton effect X-ray scatters at right angles to the incoming X-ray direction (@= 90°), what kinetic energy, K, does the ejected electron have? d. At what X-ray scattering angle, 220, will the wavelength of the scattered X-ray be 20.0% longer than the wavelength of the incident X-ray (i.e. 12 =1.202, )?

Answer 1

given :

11 = 18pm = 18 * 10-12m

a) Energy of incident photon [answer]

hc 6.6 * 10-34 * 3 * 10% E1 = 1.1 * 10-14 J = 68750eV 11 18 * 10-12

b) According to Compton's Effect, the wavelength od scattered X-Ray is :

$\lambda_{2}=\lambda_{1}+(\frac{h}{m_{e}c})(1-cos\theta)$ [where $\theta$ = angle of scattering]

given, $\theta$ = 90o.

therefore, $\lambda_{2}=\lambda_{1}+(\frac{h}{m_{e}c})(1-cos90^{o}) = 18*10^{-12}+(\frac{6.6*10^{-34}}{9.1*10^{-31}*3*10^{8}})$

   [answer]

c) From conservation of energy :

$\frac{hc}{\lambda_{2}}+K = \frac{hc}{\lambda_{1}}$

$\Rightarrow K = hc\left ( \frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}} \right ) = 6.6*10^{-34}*3*10^{8}(\frac{1}{18*10^{-12}}-\frac{1}{20.42*10^{-12}})J$

= [answer]

d) if $\lambda_{2} = 1.2\lambda_{1}$

then, $\lambda_{2}=\lambda_{1}+(\frac{h}{m_{e}c})(1-cos\theta)$

$\Rightarrow cos \theta = 1-\frac{(\lambda_{2}-\lambda_{1})m_{e}c}{h} = 1-\frac{(20.42-18)*10^{-12}*9.1*10^{-31}*3*10^{8}}{6.6*10^{-34}}$

= =-0.001

$\Rightarrow \theta = cos^{-1}(-0.001) = 90.06^{o}$ [answer]