A free iron atom emits K_α X-rays having an energy of 6.4 keV ....

By conservation of momentum, the recoil momentum of the atom is equal to the momentum of the emitted X-ray photon.

$p = \dfrac{E}{c}$

The recoil kinetic energy of the atom is given by:

$K = \dfrac{p^2}{2m} = \dfrac{E^2}{2mc^2}$

Given:

$E = 6.4 \text{ keV} = 6.4 \times 10^3 \text{ eV}$

$m = 9.3 \times 10^{-26} \text{ kg}$

$c = 3 \times 10^8 \text{ m/s}$

First, we find the rest mass energy of the iron atom in eV:

$mc^2 = \dfrac{9.3 \times 10^{-26} \times (3 \times 10^8)^2}{1.6 \times 10^{-19}} \text{ eV} = 52.3125 \times 10^9 \text{ eV}$

Substituting the values into the kinetic energy formula:

$K = \dfrac{(6.4 \times 10^3)^2}{2 \times 52.3125 \times 10^9} \text{ eV}$

$K = \dfrac{40.96 \times 10^6}{104.625 \times 10^9} \text{ eV}$

$K \approx 3.9 \times 10^{-4} \text{ eV}$