Alternate Answer
$[\text{L} = \frac{\text{nh}}{2\pi}$ i.e. angular momentum of orbiting electron is quantised.] According to de Broglie hypothesis Linear momentum $(\text{P}) = \frac{\text{h}}{\lambda}$ And for circular orbit $\text{L} =\text{r}_{n}\text{p}$ where $' \text{r}_{n} '$ is the radius of quantized orbits. $ = \frac{\text{rh}}{\lambda}$ Also $\text{L} = \frac{\text{nh}}{2\pi}$ $\therefore\frac{\text{rh}}{\lambda} = \frac{\text{nh}}{2\pi}$ $\Rightarrow2\pi\text{r}_{n} = \text{n}\lambda$ $\therefore$ Circumference of permitted orbits are integral multiples of the wavelength $\lambda$ $\text{E}_{c} - \text{E}_{B} = \frac{\text{hc}}{\lambda}_{1}$ . . . . (i) $\text{E}_{B} - \text{E}_{A} =\frac{\text{hc}}{\lambda}_{2}$ . . . . (ii) $\text{E}_{c} - \text{E}_{A} = \frac{\text{hc}}{\lambda}_{a}$ . . . . . (iii) Adding (i) & (ii) $\text{E}_{c} - \text{E}_{A} = \frac{\text{hc}}{\lambda}_{1} + \frac{\text{hc}}{\lambda}_{2}$ . . . . . (vi) Using equation (iii) and (iv) $\frac{\text{hc}}{\lambda}_{3} = \frac{\text{hc}}{\lambda}_{1} + \frac{\text{hc}}{\lambda}_{2} = > \frac{1}{\lambda}_{3} = \frac{1}{\lambda}_{1} + \frac{1}{\lambda}_{2}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
