When an electron in a hydrogen atom jumps from the third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes

The De Broglie wavelength is \(\lambda=\frac{h}{p}\), where \(h=\) Planck's constant and \(p=\) momentum of the electron. Also, momentum \(p=m v\) and kinetic energy \(E_K=\frac{1}{2} m v^2\) From the above equation, we can relate momentum \(p\) and kinetic energy \(E_K\) as, \(p=\sqrt{2 m E_K}\)
Now the ratio of wavelength of hydrogen atom when the atom jumps from third excited state to ground state viz \((n=4\) to \(n=1\) )
\(\frac{\lambda_1}{\lambda_2}=\frac{\frac{h}{p_1}}{\frac{h}{p_2}}=\frac{p_2}{p_1}=\sqrt{\frac{E_{K 2}}{E_{K 1}}}\)
Also, we know \(E_{K n}=\frac{-13.6 Z^2}{n^2}\)
Using equation (1),
\(\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{1^2}}{\sqrt{4^2}}\)
So, the ratio of the wavelength of ground state \(\{n=1\}\) to \(3^{\text {rd }}\) excited state \(\{n=4\}\) is \(\frac{1}{4}\).