Question

An isolated hydrogen atom emits a photon of energy $9eV$.

The momentum of recoil atom is:

• A. $4.8\times {10}^{-27}kg-m/s$
• B. $-4.8\times {10}^{-27}kg-m/s$
• C. $-6.8\times {10}^{-27}kg-m/s$
• D. $7.8\times {10}^{-27}kg-m/s$

Answer 1

The correct option is B $-4.8\times {10}^{-27}kg-m/s$

$\text{We Know,}$

$\text{According to De-Broglie equation,}$
$p=\frac{h}{\lambda }$ p=momentum.

$p=\frac{h\times c}{\lambda \times c}=\frac{\text{energy}}{c}$

$p=\frac{9\times 1.6\times {10}^{-19}}{3\times {10}^8}$

$p=\frac{14.4\times {10}^{-27}}{3}$

$p=4.8\times {10}^{-27}$

Now,
Taking photoelectron and atom a system, there is no external force acting ,hence momentum is conserved.
$p_{atom}+p_{electron}=0$
$p_{atom}=-4.8\times {10}^{-27}$

Hence option $\text{B}$ is correct.