Physics - Modern Physics Question with Solution | TestHub

The De Broglie wavelength is $\lambda =\frac{h}{p}$, where $h=$ Planck's constant and $p=$ momentum of the electron. Also, momentum $p=mv$ 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, $p=\sqrt{2m{E}_K}$ Now the ratio of wavelength of hydrogen atom when the atom jumps from third excited state to ground state viz $(n=4$ to $n=1$ ) $\frac{{\lambda }_1}{{\lambda }_2}=\frac{\frac{h}{p_1}}{\frac{h}{p_2}}=\frac{p_2}{p_1}=\sqrt{\frac{E_{K2}}{E_{K1}}}-\cdots (1)$ Also, we know $E_{Kn}=\frac{-13.6Z^2}{n^2}$ 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 $\frac{1}{4}$.