Question

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

• A. 3
• B. 2
• C. 6
• D. 10

Answer 1

Answer: (C)

Energy of the photon, $E=\frac{hc}{\lambda }$
$E=\frac{6.63\times 10^{-34}\times 3\times 10^8}{975\times 10^{-10}}J$
$=\frac{6.63\times 10^{-34}\times 3\times 10^8}{975\times 10^{-10}\times 1.6\times 10^{-19}}eV=12.75eV$
After absorbing a photon of energy $12.75eV$, the electron will reach to third excited state of energy $-0.85eV$,
since energy difference corresponding
to $n=1$ and $n=4$ is $12.75eV$.
$\therefore$ Number of spectral lines emitted
$=\frac{(n)(n-1)}{2}=\frac{(4)(4-1)}{2}=6$