Question

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground lend. The velocity that the atom of mass $m$ acquired as a result of photon emission will be-
$($Assume $R$ is Rydberg constant and $h$ is Planck's constant$)$

• A. $\frac{24hR}{25m}$
• B. $\frac{25hR}{24m}$
• C. $\frac{25m}{24hR}$
• D. $\frac{24m}{25hR}$

Answer 1

The correct option is A $\frac{24hR}{25m}$

Using Rydberg's formula for Hydrogen atom,

$\frac{1}{\lambda }=R[\frac{1}{n_f^2}-\frac{1}{n_i^2}]$

Where, $n_i=\text{Quantum no. of higher energy}{n}_f=\text{Quantum no. of lower energy}$

$\frac{1}{\lambda }=R[\frac{1}{1^2}-\frac{1}{5^2}]$

$\Rightarrow \frac{1}{\lambda }=\frac{24R}{25}$

From the de-Broglie wavelength relation,

$\lambda =\frac{h}{p}=\frac{h}{mv}$

$\therefore v=\frac{h}{m\lambda }=\frac{24hR}{25m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.