Question

An electron of a stationary hydrogen atom makes a transition from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be ($m$ is the mass of the electron, $R$ is the Rydberg constant and $h$ is the 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 $\frac{24hR}{25m}$

According to Rydberg formula
$\frac{1}{\lambda }=R[\frac{1}{{n_f}^2}-\frac{1}{{n_i}^2}]$
Here, $n_f=1,n_i=5$
$\therefore \frac{1}{\lambda }=R[\frac{1}{1^2}-\frac{1}{5^2}]=R[\frac{1}{1}-\frac{1}{25}]$
$=\frac{24}{25}R$
According to conservation of linear momentum, we get
Momentum of photon = Momentum of atom
$\frac{h}{\lambda }=mv$
or $v=\frac{h}{m\lambda }=\frac{h}{m}\left(\frac{24R}{25}\right)=\frac{24hR}{25m}$