Question

A hydrogen atom initially in the ground level absorbs a photon and is excited to $n = 4$ level then the wavelength of photon is

• A. $790 \mathring{A}$
• B. $870 \mathring{A}$
• C. $970 \mathring{A}$
• D. $1070 \mathring{A}$

Answer 1

Answer: (C)

Here, $ n_1 = 1$, and $n_2 = 4$
Energy of photon absorbed, $E = E_2 - E_1$
Since, $E_n = - \frac{13.6}{n^2} eV $
Then, $E_2 - E_1 = -\frac{13.6}{(4)^2} - (-\frac{13.6}{(1)^2})$
$ = -\frac{13.6}{16} + 13.6 $
$ = \frac{13.6\times15}{16} eV $
$ = 12.75 eV $
$= 12.75 \times1.6 \times10^{-19} J$
$= 20.4 \times10^{-19} J$
$E_{2} -E_{1} = \frac{hc}{\lambda} $
$ \therefore \lambda = \frac{hc}{E_{2} -E_{1}}$
$ = \frac{6.6 \times10^{-34}\times3\times10^{8}}{20.4\times10^{-19}} $
$ = 9.70 \times10^{-8} m $
$ = 970 \times10^{-10} $
$ = 970 \mathring{A}$