Question

In a hydrogen atom, in transition of electron a photon of energy $2.55$ eV is emitted, then calculate change in wavelength of the electron.

Answer 1

Let the energy of photon is

$\Delta E=2.55eV=2.55\times 1.602\times {10}^{-19}J$

The wavelength of the photon of light is

$\lambda =\frac{hc}{\Delta E}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{2.55\times 1.602\times {10}^{-19}m}=4861A^0$

This corresponds to the second line Balmar series.

The transition is $n_2=4\to n_1=2$

The change in the wavelength of the electron is

$\Delta \lambda =\frac{h}{m{u}_4}-\frac{h}{m{u}_2}=\frac{h}{m}(u_4-u_2)=\frac{6.626\times {10}^{-34}}{9.1085\times {10}^{-31}}\left(\frac{4-2}{2.19\times {10}^m/s}\right)=6.64A^0$.