Physics - Modern Physics Question with Solution | TestHub

Let the three energy levels be ${\mathrm{E}}_1,{\mathrm{E}}_2$, and ${\mathrm{E}}_3$. The wavelength ${\mathrm{\lambda }}_1,{\mathrm{\lambda }}_2$ and ${\mathrm{\lambda }}_3$ of the spectra lines corresponding to the three energy transitions are depicted as shown in figure.

$\mathrm{E}=\mathrm{hf}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ or $\mathrm{E}\propto \frac{1}{\mathrm{\lambda }}$
(given ${\mathrm{\lambda }}_1<{\mathrm{\lambda }}_2<{\mathrm{\lambda }}_3$)
Thus, for the three wavelengths, we have
${\mathrm{E}}_3-{\mathrm{E}}_2=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_3}\dots \left(\mathrm{i}\right)$
${\mathrm{E}}_2-{\mathrm{E}}_1=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}\dots \left(\mathrm{ii}\right)$
${\mathrm{E}}_3-{\mathrm{E}}_1=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}\dots \left(\mathrm{iv}\right)$
Now, ${\mathrm{E}}_3-{\mathrm{E}}_1=\left({\mathrm{E}}_3-{\mathrm{E}}_2\right)+\left({\mathrm{E}}_2-{\mathrm{E}}_1\right)$

$\Rightarrow \frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_3}+\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2}\Rightarrow \frac{1}{{\mathrm{\lambda }}_1}=\frac{1}{{\mathrm{\lambda }}_2}+\frac{1}{{\mathrm{\lambda }}_3}$