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, $E_2-E_1=-\frac{13.6}{(4{)}^2}-(-\frac{13.6}{(1{)}^2})$
$=-\frac{13.6}{16}+13.6$
$=\frac{13.6\times 15}{16}eV$
$=12.75eV$
$=12.75\times 1.6\times 10^{-19}J$
$=20.4\times 10^{-19}J$
$E_2-E_1=\frac{hc}{\lambda }$
$\therefore \lambda =\frac{hc}{E_2-E_1}$
$=\frac{6.6\times 10^{-34}\times 3\times 10^8}{20.4\times 10^{-19}}$
$=9.70\times 10^{-8}m$
$=970\times 10^{-10}$
$=970\mathring{A}$