Physics - Modern Physics Question with Solution | TestHub

Let ${\lambda }_1$ be the wavelength when the electron jumps from the second orbit to the first, and ${\lambda }_2$ be the wavelength when the electron jumps from the third orbit to the first.

The energy of the emitted photon is proportional to the difference in energy levels: $E=hf=\frac{hc}{\lambda }$.

For the transition from $n=2$ to $n=1$:

$$\frac{1}{{\lambda }_1}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R\left(1-\frac{1}{4}\right)=\frac{3R}{4}$$

where $R$ is the Rydberg constant.

For the transition from $n=3$ to $n=1$:

$$\frac{1}{{\lambda }_2}=R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R\left(1-\frac{1}{9}\right)=\frac{8R}{9}$$

Therefore,

${\lambda }_1=\frac{4}{3R}$

${\lambda }_2=\frac{9}{8R}$

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{9}{8R}\cdot \frac{3R}{4}=\frac{27}{32}$

${\lambda }_2=\frac{27}{32}{\lambda }_1=\frac{27}{32}\lambda$