When an electron in a hydrogen atom jumps from the third excited state to the ground state, the de-Broglie…
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
$\frac{1}{32}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{16}$
Solution
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 $n=1$ )
$\frac{\lambda_1}{\lambda_2}=\frac{\frac{h}{p_1}}{\frac{h}{p_2}}=\frac{p_2}{p_1}=\sqrt{\frac{E_{K 2}}{E_{K 1}}}-\cdots(1)$
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 $\frac{1}{4}$.