Question

What is the wavelength (in nm) of a photon emitted during transition from the n = 3 state to the n = 1 state in the H atom?

Answer 1

Answer:

$\lambda=103\ nm$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,  

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

c is the speed of light having value $3\times 10^8\ m/s$

So,  

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

So,  

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

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

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times |(\frac{1}{3^2} - \dfrac{1}{1^2})}|\ m$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179|\left(\frac{1}{9}-\frac{1}{1}\right)|}\ m$

$\lambda=1.03\times 10^{-7}\ m$

1 m = 10⁻⁹ nm

$\lambda=103\ nm$