An electron from various excited states of hydrogen atom emit radiation to come to the ground state.

Question

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $\lambda_n, \lambda_g$ be the de Broglie wavelength of the electron in the $n^{	ext {th }}$ state and the ground state respectively. Let $\Lambda_n$ be the wavelength of the emitted photon in the transition from the $n^{	ext {th }}$ state to the ground state. For large $n$, ( $A, B$ are constants)

Professor PC Thomas Classes · Answer

(a) : Momentum of electron in different states \ $ p_n=\frac{h}{\lambda_n}, p_g=\frac{h}{\lambda_g} $\  Kinetic energy, $K=\frac{p^2}{2 m}=\frac{h^2}{2 m \lambda^2}$ \ Total energy in an orbit of hydrogen atom, \ $ \begin{aligned} & E=-K=-\frac{h^2}{2 m \lambda^2} ; E_n-E_g=\frac{h^2}{2 m}\left(\frac{1}{\lambda_g^2}-\frac{1}{\lambda_n^2}ight) \ & \frac{h^2}{2 m}\left(\frac{\lambda_n^2-\lambda_g^2}{\lambda_g^2 \lambda_n^2}ight)=\frac{h c}{\Lambda_n} ; \Lambda_n=\frac{2 m c}{h}\left(\frac{\lambda_g^2 \lambda_n^2}{\lambda_n^2-\lambda_g^2}ight) \ & \Lambda_n=\frac{2 m c \lambda_g^2}{h}\left[1-\frac{\lambda_g^2}{\lambda_n^2}ight]^{-1}=\frac{2 m c \lambda_g^2}{h}\left[1+\frac{\lambda_g^2}{\lambda_n^2}ight] \quad\left(\because \lambda_g \ll \lambda_night) \ & =\frac{2 m c \lambda_g^2}{h}+\left(\frac{2 m c \lambda_g^4}{h}ight) \frac{1}{\lambda_n^2}=A+\frac{B}{\lambda_n^2} \end{aligned} $ \ where $A$ and $B$ are $A=\frac{2 m c \lambda_g^2}{h}, B=\frac{2 m c \lambda_g^4}{h}$