Question

Feedback A photon of wavelength 4.88 pm scatters at an angle of 147" from an initially stationary, unbound electron. What is the de Broglie wavelength of the electron after the photon has been scattered? de Broglie wavelength: 9.35 pm Incorrect

Answer 1

Given:

the wavelength of incident photon = $\lambda_{i}$

the angle of scattering =

0= 147°

Let the wavelength of the scattered photon be

X3 = 4.88pm = 4.88 * 10-12m

According to Compton's effect:

$\lambda_{s}-\lambda_{i}=\frac{h}{mc}(1-cos\theta)$

$\Rightarrow \lambda_{i}=\lambda_{s}-\frac{h}{mc}(1-cos\theta)$

  $=(4.88*10^{-12})-\frac{6.63*10^{-34}}{9.1*10^{-31}*3*10^{8}}(1-cos147^{\circ})$

= 0.41 pm [answer]

Let the momentum of the electron is p.

therefore, the energy of the electron is $E_{e}=\frac{p^{2}}{2m}$

Now, applying conservation of energy:

$E_{i}=E_{s}+E_{e}$ [where Ee = energy of electron]

$\Rightarrow E_{e}=E_{i}-E_{s}$

$\Rightarrow \frac{p^{2}}{2m}=\frac{hc}{\lambda_{i}}-\frac{hc}{\lambda_{\omega}}$

$\Rightarrow p=\sqrt{2m\left (\frac{hc}{\lambda_{i}}-\frac{hc}{\lambda_{s}} \right )}$

$=\sqrt{2*9.1*10^{-31}*6.63*10^{-34}*3*10^{8}\left (\frac{1}{0.41*10^{-12}}-\frac{1}{4.88*10^{-12}} \right )}$

  $=8.99*10^{-22}Ns$

According to de-Broglie's equation,

the wavelength of electric = $\frac{h}{p}=\frac{6.6*10^{-34}}{8.99*10^{-22}}=0.73*10^{-12}m = 0.73pm$ [answer]