Question

When a hydrogen atom, initially at rest emits a photon resulting in transition $n=5\to n=1,$its recoil speed is about :

• A. ${10}^{-4}m/s$
• B. $2\times {10}^{-2}m/s$
• C. $4.3m/s$
• D. $3.8\times {10}^{-2}m/s$

Answer 1

Answer: (C)

$4.3m/s$

Given,

$n_1=1$

$n_2=5$

$m=1.6\times {10}^{-27}kg$

Momentum, $P=mv=\frac{h}{\lambda }$

recoil speed, $v=\frac{h}{m\lambda }$. . . . . . . . .(1)

we know that,

$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{1^2}-\frac{1}{5^2})$

$\lambda =9.48\times {10}^{-8}m$

From equation (1),

recoil speed, $v=\frac{6.67\times {10}^{-34}}{1.6\times {10}^{-27}\times 9.48\times {10}^{-8}}=4.3m/s$

The correct option is C.