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 $n^{th}$ state and the ground state respectively. Let $\Lambda_n$ be the wavelength of the emitted photon in the transition from the $n^{th}$ state to the ground state. For large $n,$ ($A, B$ are constants)

• A. $\Lambda_n \approx A + \frac{B}{\lambda^2_n}$
• B. $\Lambda_n \approx A + B \lambda_n $
• C. $\Lambda_n^2 \approx A + B \lambda_n^2$
• D. $\Lambda_n^2 \approx \lambda$

Answer 1

The Correct Option is A

$P_{n}=\frac{h}{\lambda_{n}}, P_{g}=\frac{h}{\lambda_{g}}$
$k=\frac{P^{2}}{2 m}=\frac{h^{2}}{2 m \lambda^{2}}, E=-k=-\frac{h^{2}}{2 m \lambda^{2}}$
$E_{n}=-\frac{h^{2}}{2 m \lambda_{n}^{2}}, E_{g}=-\frac{h^{2}}{2 m \lambda_{g}^{2}}$
$E_{n}-E_{g}=\frac{h^{2}}{2 m}\left(\frac{1}{\lambda_{g}^{2}}-\frac{1}{\lambda_{n}^{2}}\right)=\frac{h c}{\Lambda_{n}}$
$\frac{h^{2}}{2 m}\left(\frac{\lambda_{n}^{2}-\lambda_{g}^{2}}{\lambda_{g}^{2} \lambda_{n}^{2}}\right)=\frac{h c}{\Lambda_{n}}$
$\Lambda_{n}=\frac{2 m c}{h}\left(\frac{\lambda_{g}^{2} \lambda_{n}^{2}}{\lambda_{n}^{2}-\lambda_{g}^{2}}\right)$
$\Lambda_{n}=\frac{2 m c \lambda_{g}^{2}}{h} \frac{\lambda_{n}^{2}}{\lambda_{n}^{2}\left(1-\frac{\lambda_{g}^{2}}{\lambda_{n}^{2}}\right)}$
$=\frac{2 m c \lambda_{g}^{2}}{h}\left[1-\frac{\lambda_{g}^{2}}{\lambda_{n}^{2}}\right]^{-1}$
$=\frac{2 m c \lambda_{g}^{2}}{h}\left[1+\frac{\lambda_{g}^{2}}{\lambda_{n}^{2}}\right]$
$=\frac{2 m c \lambda_{g}^{2}}{h}+\left(\frac{2 m c \lambda_{g}^{4}}{h}\right) \frac{1}{\lambda_{n}^{2}} $
$=A+\frac{B}{\lambda_{n}^{2}} $
$A= \frac{2 m c \lambda_{g}^{2}}{h}, B=\frac{2 m c \lambda_{g}^{4}}{h} $