Question

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be:

• A. $\frac{25m}{24hR}$
• B. $\frac{24m}{24hR}$
• C. $\frac{24hR}{25m}$
• D. $\frac{25m}{14hR}$

Answer 1

Answer: (C)

$\frac{24hR}{25m}$

According to Rydberg formula

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

Here, $n_f=1$ and $n_i=5$

Therefore,

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

$=R[\frac{1}{1}-\frac{1}{25}]=\frac{24}{25}R$

According to conservation of linear momentum, we get

Momentum of photon $=$ Momentum of atom

That is,

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

or

$v=\frac{h}{m\lambda }=\frac{h}{m}\left(\frac{24R}{25}\right)=\frac{24hR}{25m}$