Question

The ionisation potential of $H$-atom is $13.6V$. When it is excited from ground state by monochromatic radiations of $970.6\stackrel{{}_{\circ }}{A}$, the number of emission lines will be (according to Bohr's theory)

• A. $10$
• B. $8$
• C. $6$
• D. $4$

Answer 1

The correct option is C $6$

The wavelength for the corresponding series of the hydrogen spectrum is given by:

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\Rightarrow \frac{1}{970.6\times {10}^{-10}}=1.097\times {10}^7[\frac{1}{1^2}-\frac{1}{n_2^2}]$

$\Rightarrow n_2=4$

$\therefore$ Number of emission line, $N=\frac{n(n-1)}{2}=\frac{4\times 3}{2}=6$