Question

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $\lambda$. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

• A. $\frac{20}{13}\lambda$
• B. $\frac{16}{25}\lambda$
• C. $\frac{9}{16}\lambda$
• D. $\frac{20}{7}\lambda$

Answer 1

The correct option is D $\frac{20}{7}\lambda$

Wavelength of emitted photon:

$\frac{1}{\lambda }=R(\frac{1}{n_2^2}-\frac{1}{n_1^2})$

Transition from $n_1=3$ to $n_2=2$:

$\therefore$ $\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R}{36}$ .............(1)

Transition from $n_1=4$ to $n_3=2$:

$\therefore$ $\frac{1}{{\lambda }^{\prime }}=R(\frac{1}{3^2}-\frac{1}{4^2})=\frac{7R}{144}$ .............(2)

Dividing (1) by (2) we get:

$\frac{{\lambda }^{\prime }}{\lambda }=\frac{5R/36}{7R/144}=\frac{20}{7}$ $⟹{\lambda }^{\prime }=\frac{20}{7}\lambda$