Question

A hydrogen atom undergoes a transition form n=6 to n=2. Find the energy and frequency of the emitted photon

Answer 1

As energy of electron in hydrogen atom in n'th level is

$E_{n}=-\frac{13.6}{n^{2}}eV$

for n=2

  $E_{n}=-\frac{13.6}{2^{2}}eV$

for n=6

$E_{n}=-\frac{13.6}{6^{2}}eV$

energy of emitted photon is

  $\Delta E=-\frac{13.6}{6^{2}}eV-(-\frac{13.6}{2^{2}}eV)=3.022eV$

$\Delta E=3.022eV$

energy of emitted photon is 3.022eV.

frequency of emitted photon is

$\nu =\frac{\Delta E}{h}$ where h is plank's constant =4.136 * 10^-15eV-s

  $\nu =\frac{3.022eV}{4.136\times 10^{-15}eV-s}=0.7306576\times 10^{15}$

  $\nu =0.7306576\times 10^{15}$

frequency of photon emitted 0.7306576 * 10¹⁵ s^-1.