Q. The ionization potential of $H$-atom is $13.6V$. When it is excited from ground state by monochromatic radiations of $970.6\ \mathring{\mathrm{A}}$, the number of emission lines will be (according to Bohr’s theory)
A) 10
B) 8
C) 6
D) 4
Explanation
(c) $\frac{1}{\lambda} = R\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$\Rightarrow \frac{1}{970.6 \times 10^{- 10}} = 1.097 \times 10^{7}\left\lbrack \frac{1}{1^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack \Rightarrow n_{2} = 4$
$\therefore$ Number of emission lines $N = \frac{n(n - 1)}{2} = \frac{4 \times 3}{2} = 6$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.