Question

A hydrogen atom in its ground state is irradiated by light of wavelength 970 $\mathring{A}$. Taking $\frac{hc}{e}=1.237\times {10}^{-6}eV$ and the ground state energy of hydrogen atom as $-13.6eV$, the number of lines present in the emission spectrum is

Answer 1

The electron in the ground state of H-atom jumps to the $n^{th}$ state after absorbing the radiation.
Wavelength of the radiation, $\lambda =970\mathring{A}=970\times {10}^{-10}m$
Energy gained by the electron, $E^{\prime }=\frac{hc}{\lambda }eV=\frac{1.237\times {10}^{-6}}{970\times {10}^{-10}}=12.75eV$
Energy of the hydrogen atom in the ground state $=-13.6eV$
Thus the energy of the atom $E_n$ in the state with principal quantum number $n$, $E_n=-13.6+12.75=-0.85eV$
Using $E_n=\frac{-13.6}{n^2}eV$
$\therefore$ $-0.85=\frac{-13.6}{n^2}$
$⟹n=4$
Number of (emission) spectral lines, $N={}^n{C}_2=\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6$ lines