The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state.
Question:
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, $\lambda_{1} / \lambda_{2}$, of the photons emitted in this process is :
Correct Option: 1,
Solution:
(1) $\frac{1}=\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$
$=\frac{7 R}{16 \times 9}$
And $\frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$
$=\frac{5 R}{36}$
Now $\frac{\lambda_{1}}{\lambda_{2}}=\frac{(5 R / 36)}{7 R /(16 \times 9)}=\frac{20}{7}$