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?

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 :

     ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, 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 .

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

= 22.90°

what was the answer?