Question

Hydrogen atom in ground state is excited by a monochromatic radiation of $\lambda =975\stackrel{\text{o}}{\text{A}}$. Number of spectral lines in the resultant spectrum emitted will be

• A. $3$
• B. $2$
• C. $6$
• D. $10$

Answer 1

The correct option is C $6$

Given $\lambda =975\times {10}^{-10}\text{m}$
Energy of emitted radiation is
$E=\frac{hc}{\lambda }$
$E=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}\times 1.6\times {10}^{-19}}$

$E=12.75\text{eV}$

It jumps to $n=4$ because the energy difference between the $n=1$ and $n=4$ is $12.75\text{eV}$
So, number of spectral lines emitted will be
$\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6$