Question

In Compton scattering , A X-ray photon is found to have doubled its wavelength on being scattered by

90°

. Find the wavelength and energy of incident X-ray photon.​

Answer 1

Answer:

A photon of wavelength 6000 nm collides with an electron at rest. After scattering, the wavelength of the scattered photon is found to change by exactly one Compton wavelength.

Answer 2

Given:

Final wavelength $\lambda _{f}=2\lambda _{i}$

Scattering angle$\theta=90^{0}$

To find:

$\lambda _{i}= ?\\E=?$

Formula:

Compton scattering: $del\lambda=\lambda _{f}-\lambda _{i}=\frac{h(1-cos\theta)}{mc}$

Solution:

Step 1 of 3

Compton scattering is the phenomenon of shift in the wavelength of the incident radiation after getting scattered in a medium.

The incident wavelength is represented by $\lambda _{i}$ and the final wavelength after scattering is represented by $\lambda _{f}$.

In the formula, m is the mass of the electron and c is the velocity of light.

Step 2 of 3

$\lambda _{f}=2\lambda _{i}$

$\lambda _{f}-\lambda _{i}=\frac{h(1-cos 90^{0}) }{(9.1)(10^{-31} )(3)(10^{8}) }$

$cos 90^{0}=0 \\2\lambda _{i}-\lambda _{i}=\frac{h(1-0) }{(27.3)(10^{-23})} \\\lambda _{i}=\frac{(6.63)(10^{-34})}{(27.3)(10^{-23})} \\\lambda _{i}=0.2429(10^{-11})$

So, the incident wavelength of the X-ray photon is 0.2429×$10^{-11}$m

Step 3 of 3

Energy of the photon is always inversely proportional to the wavelength of radiation.

$E=\frac{hc}{\lambda _{i} }$

$E=\frac{(6.63)(10^{-34})(3)(10^{8})}{(0.2429)(10^{-11})} \\E=(81.88)(10^{-15})$

So, the energy of the incident photon is 81.88×$10^{-15}eV$

Answer:

The wavelength and energy of the incident X-ray photon is 0.2429×$10^{-11}$m and 81.88×$10^{-15}eV$, respectively