Question

The ionisation energy of the hydrogen atom is given to be $13.6eV$. A photon falls on a hydrogen atom which is initially at the ground state and excites it to the $(n=4)$. Calculate the wavelength of the photon.

Answer 1

As energy,

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

$E_1=-13.6eV$

$E_4=\frac{13.6}{4^2}=-0.85eV$

So, energy of photon is

$\Delta E=E_4-E_1=-0.85-(-13.6)=12.75eV$

Wavelength corresponding to this energy

$\lambda =\frac{hc}{\Delta E}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{12.75\times 1.6\times {10}^{-19}}$

$=0.974\times {10}^{-7}$m $=974\mathring{A}$