Question

calculate the frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 3 to n = 1 energy level based on the bohr theory

Answer 1

Answer:

$\nu=2.92\times 10^{15}\ Hz$

Explanation:

The expression for the energy of an electron in the nth orbit is:-

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

Given, $n_i=3\ and\ n_f=1$

$\Delta E=2.18\times 10^{-18}(\frac{1}{3^2} - \dfrac{1}{1^2})\ J$

$\Delta E=-1.94\times 10^{-18}\ J$  (negative sign indicates energy release)

Also, $E=h\times \nu$

Where,  

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

$1.94\times 10^{-18}=6.626\times 10^{-34}\times \nu$

$\nu=2.92\times 10^{15}\ Hz$