Question

When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de-Broglie wave length associated with the electron change? Justify your answer.

Answer 1

De-Broglie wavelength associated with a moving charge particle having a K.E. 'K' can be given as

$\lambda =\frac{h}{}\frac{h}{\sqrt{mK}}[K=\frac{1}{2}m{v}^2=\frac{p^2}{2m}]$ ...(i)

The kinetic energy of the electron in any orbit of hydrogen atom can be given as

$K=-E=-\left(\frac{13.6}{n^2}eV\right)=\frac{13.6}{n^2}eV$ ...(ii)

Let $K_1$ and $K_4$ be the K.E. of the electron in ground state and third excited state, where $n_1=1$ shows ground state and $n_2=4$ shows third excited state.

Using the concept of equations (i) and (ii), we have

$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{K_4}{K_1}}=\sqrt{\frac{n_1^2}{n_2^2}}$

$\frac{{\lambda }_1}{{\lambda }_4}=\sqrt{\frac{1^2}{4^2}}=\frac{1}{4}$

$\Rightarrow {\lambda }_1=\frac{{\lambda }_4}{4}$
i.e., the wavelength in the ground state will decrease.