Q. If the electron in hydrogen atom jumps from the third orbit to second orbit, the wavelength of the emitted radiation is given by
NTA AbhyasNTA Abhyas 2022
Solution:
The wavelength $\left(\lambda \right)$ of emitted photon when electrons makes a transition from a higher $n_{1}^{t h}$ orbit to a lower $n_{2}^{t h}$ 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 $n_{2}=2$
$\Rightarrow \frac{1}{\lambda }=R\left[\frac{1}{2^{2}} - \frac{1}{3^{2}}\right]$
$\Rightarrow \frac{1}{\lambda }=\frac{5 R}{36}$
So, $\lambda =\frac{36}{5 R}$
$\frac{1}{\lambda }=R\left[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right]$
where $R$ is the Rydberg constant.
Given, $n_{1}=3$ and $n_{2}=2$
$\Rightarrow \frac{1}{\lambda }=R\left[\frac{1}{2^{2}} - \frac{1}{3^{2}}\right]$
$\Rightarrow \frac{1}{\lambda }=\frac{5 R}{36}$
So, $\lambda =\frac{36}{5 R}$