Question

A beam having photon of energy 480 k eV is incident on a foil of aluminum. If the photon is scattered at 45°, then calculate the energy of the recoil electron. (a) 2.56x10-14 J (b) 1.65 x10-14 J (c) 3.28x10-14 J (d) 4.12 x 10-14 J​

Answer 1

Answer:

option (a) is the correct answer

Explanation:

Given :-    E = 480 Kev          θ = 45°

Recoil energy $T_{e}$ = E/ 1 - cosθ

$T_{e}$ = 480/1 - cos45° = 1638 Kev

1Kev = 1.602 × $10^{-16}$J

1638 × 1.602 × $10^{-16}$ = 2.57 × $10^{-14}$J