Question

Electrons of energies $10.20eV$ and $12.09eV$ can cause radiation to be emitted from hydrogen atom. The principal quantum number of the orbit to which electron belongs and the wavelength of the radiation emitted if it drops back to the ground state in each case is:

• A. $n=2$ and $3,\lambda =1216\mathring{A},1020\mathring{A}$
• B. $n=2$ and $4,\lambda =1216\mathring{A},1020\mathring{A}$
• C. $n=2$ and $3,\lambda =1516\mathring{A},1020\mathring{A}$
• D. $n=2$ and $4,\lambda =1516\mathring{A},1020\mathring{A}$

Answer 1

The correct option is A $n=2$ and $3,\lambda =1216\mathring{A},1020\mathring{A}$

The electron is excited from $n=1$ (ground state) to the excited state
$n=?$ (higher energy level)
During this process of excitation, electrons acquire energy of $10.2eV$ as $12.09eV$
We have $E_{n_2}-E_{n_1}=\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$
$i)10.20=13.6(\frac{1}{1^2}-\frac{1}{n^2})ev$
$=13.6-\frac{13.6}{n^2}\Rightarrow n^2=\frac{13.6}{3.41}=4\Rightarrow n=2$
$ii)12.09=13.6(\frac{1}{1^2}-\frac{1}{n^2})$
$\Rightarrow \frac{13.6}{n^2}=1.52\Rightarrow n^2=9\Rightarrow n=3$