Question

Hydrogen atom is excited from ground state to another state $\left(n\right)$. Number of spectral lines in emission spectrum is 10. If this excitation is done by a photon, then energy of the photon is

• A. $13eV$
• B. $12.1eV$
• C. $10.2eV$
• D. $4eV$

Answer 1

The correct option is A $13eV$

If excitation is done to principal quantum number (n) then no of spectral lines in emission spectrum $=\frac{n(n-1)}{2}$
$10=\frac{n(n-1)}{2}$
$n^2-n=20$
$n^2-n-20=0$
$n^2-5n+4n-20=0$
$n(n-5)+4(n-5)=0$
$n=5$

$E_1=-13.6eV$
$E_5=-\frac{13.6}{25}=-0.544eV$
$E_{\text{photon}}=E_5-E_1=-0.544+13.6\approx 13eV$