Question

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

• A. $3.9\times {10}^{-4}\text{eV}$
• B. $5\times {10}^{-4}\text{eV}$
• C. $8.3\times {10}^{-4}\text{eV}$
• D. $2.3\times {10}^{-4}\text{eV}$

Answer 1

The correct option is A $3.9\times {10}^{-4}\text{eV}$

Given energy of photon,

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

$\therefore$ Momentum of photon,

$p=\frac{E}{c}=\frac{6.4\times {10}^3\times 1.6\times {10}^{-19}}{3\times {10}^8}$

$p=3.41\times {10}^{-24}\text{kg m/s}$

According to collision theory,

momentum of photon = momentum of atom

$\therefore$ Recoil $KE$ of atom will be,

$KE=\frac{p^2}{2m}=\frac{(3.41\times {10}^{-24}{)}^2}{\left(2\right)(9.3\times {10}^{-26})}$

$\Rightarrow KE=6.25\times {10}^{-23}\text{J}$

$\therefore KE=\frac{6.25\times {10}^{-23}}{1.6\times {10}^{-19}}=3.9\times {10}^{-4}\text{eV}$

Hence, option $\left(\text{A}\right)$ is correct.

Why this question ? Concept:- This question helps in solving the problems having multi concepts ( $\text{X}$-rays and conservation of momentum) .