Question

A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam) is the recoiling electron found?

Answer 1

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$      ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ      ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of  λ .

$\cot\phi = (1+\frac{h}{m_{e}c\lambda })\tan\frac{\theta }{2}$

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for $m_{e}$, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ  and 134° for θ in the above equation.

$\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9} })\tan\frac{134 }{2}$

$\cot\phi=2.37$

$\phi$ = 22.90°