Question:
The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?
Correct Option: , 2
Solution:
(2) According to Bohr's Theory the wavelength of the radiation emitted from hydrogen atom is given by
$\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
$\because \quad Z=3$
$\therefore \frac{1}{\lambda}=9 R\left(1-\frac{1}{9}\right)$
$\Rightarrow \lambda=\frac{1}{8 R}=\frac{1}{8 \times 10973731.6}\left(\because R=10973731.6 \mathrm{~m}^{-1}\right)$
$\Rightarrow \lambda=11.39 \mathrm{~nm}$