Question

An electron in a hydrogen atom first jumps from second excited state to first excited state and then from excited state to ground state. If the ratio of wavelength, momentum and energy of photons emitted in these two cases be $a, b$ and c respectively. Then show that

$$b=c=5 / 27$$

Answer 1

$E=\frac{h c}{\lambda}, \quad E=p c$ and $E \propto \frac{1}{n^{2}}$

$\begin{array}{llrl} & E \propto \frac{1}{\lambda} ; & & p \propto E \\\therefore & b=c \quad \text { and } & & a=\frac{1}{b}\end{array}$

By question

$\quad c=\frac{\frac{1}{2^{2}}-\frac{1}{3^{2}}}{1^{2}-\frac{1}{2^{2}}}=\frac{\frac{5}{36}}{3 / 4}=\frac{5}{36} \times \frac{4}{3}=\frac{5}{27}$

$b=c=5 / 27 \quad \text { and } \quad a=27 / 5$