Question

If the electron in hydrogen atom jumps from the third orbit to second orbit, the wavelength of the emitted radiation is given by

• A. $\lambda =\frac{36}{5R}$
• B. $\lambda =\frac{5R}{36}$
• C. $\lambda =\frac{5}{R}$
• D. $\lambda =\frac{R}{6}$

Answer 1

Answer: (A)

The wavelength $\left(\lambda \right)$ of emitted photon when electrons makes a transition from a higher $n_1^{th}$ orbit to a lower $n_2^{th}$ orbit, is given by
$\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
where $R$ is the Rydberg constant.
Given, $n_1=3$ and $n_2=2$
$\Rightarrow \frac{1}{\lambda }=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$
$\Rightarrow \frac{1}{\lambda }=\frac{5R}{36}$
So, $\lambda =\frac{36}{5R}$