Question

A free atom of iron emits $K_a$ X-rays of energy 6.4 KeV. Calculate the recoil kinetic energy of the atom. Mass of the iron atom = $9.3\times {10}^{-26}kg$

Answer 1

Energy of photon = E = 6.4 KeV

So, momentum of photon = $p=\frac{E}{c}=\frac{6.4\times {10}^3\times 1.6\times {10}^{-19}J}{3\times {10}^{8}m/s}=3.41\times {10}^{-24}kgm/s$

According to the principle of conservation of momentum, momentum of photons = momentum of iron atom

Thus, momentum of iron atom = $3.4\times {10}^{-24}kgm/s$

Thus, recoil kinetic energy of iron atom = $K=\frac{p^2}{2m}=\frac{(3.41\times {10}^{-24}{)}^{2}k{g}^2{m}^{2/s^2}}{2\times 9.3\times {10}^{-26}kg}=0.6252\times {10}^{-22}J=\frac{0.6252\times {10}^{-22}eV}{1.6\times {10}^{-19}}=0.39075KeV$