Question

Suppose an atom of Osmium at rest emits an X-ray photon of energy 6.8 keV. Calculate the
“recoil” momentum and kinetic energy of the atom.

Answer 1

Since, system has initial momentum is zero and no external interaction is going on with the system,Momentum of the system is conserved, hence, by conservation law of momentum

$$p_{osmium}+p_{\gamma}=0$$

We have given that, emitted photon has energy is

$$8.6keV=8.6\times 10^3eV$$

Thus,

$$E_{\gamma}=8.6\times 10^3eV=\frac{hc}{\lambda}=\frac{h}{\lambda}c=p_{\gamma}c\\ \implies p_{\gamma}=\frac{E_{\gamma}}{c}=8.6keV/c$$

Thus,

$$p_{osmium}=-8.6keV/c$$

Kinetic energy is

$$K=\frac{p^2}{2m}$$

Hence,

$$K_{osmium}=2.088\times 10^{-4}eV$$