Question

An electron from various excited states of a hydrogen atom emit radiation to come to the ground state. Let ${\lambda }_n,{\lambda }_9$ be the de Broglie wavelength of the electron in the $n^{\text{th}}$ state and the ground state respectively. Let ${\Lambda }_n$ be the wavelength of the emitted photon in transition from the $n^{\text{th}}$ state to the ground state. For large $n,(A,B$ are constants)

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

Answer 1

The correct option is C ${\wedge }_n\approx A+\frac{B}{{\lambda }_n^2}$

$$\begin{array}{c}{v}_n=\left(\frac{2\pi {Ke}^2}{h}\right)\frac{1}{n}\\ E_n=-\frac{1}{2}{mv}_n^2\\ {\lambda }_n=\frac{h}{{mv}_n}=\left(\frac{h^2}{m2\pi {Ke}^2}\right)n{\lambda }_n>>>{\lambda }_g\\ \frac{hc}{{\lambda }_n}=E_n-E_1=\frac{1}{2}m(v_1^2-v_n^2)\\ \frac{hc}{{\wedge }_n}=\frac{1}{2}m({\left(\frac{h}{m{\lambda }_g}\right)}^2-{\left(\frac{h}{m{\lambda }_n}\right)}^2)\\ \frac{hc}{{\wedge }_n}=\frac{h^2}{2m}(\frac{1}{{\lambda }_g^2}-\frac{1}{{\lambda }_n^2})=\frac{h^2}{2m{\lambda }_g^2}(1-\frac{{\lambda }_g^2}{{\lambda }_n^2})\\ \frac{{\wedge }_n}{hc}=\frac{2m{\lambda }_g^2}{h^2}\frac{1}{(1-\frac{{\lambda }_g^2}{{\lambda }_n^2})}\end{array}$$
Using binomial expansion,
${\Lambda }_n=\left(\frac{2mc{\lambda }_9^2}{h}\right)(1+\frac{\lambda g^2}{{\lambda }_n^2})\Rightarrow {\Lambda }_n=A+\frac{B}{{\lambda }_n^2}$
(Since $\frac{{\lambda }_g}{{\lambda }_n}<<1$)