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 ($m$ is the mass of the electron, $R$, Rydberg constant and $h$ Planck's constant)

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

Answer 1

Answer: (A)

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