Q. 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) $\frac{16}{25}\lambda$
B) $\frac{9}{16}\lambda$
C) $\frac{20}{7}\lambda$
D) $\frac{20}{13}\lambda$
Explanation
[Solution] $\frac{\lambda'}{\lambda} = \frac{1/4 - 1/9}{1/9 - 1/16} = \frac{5/36}{7/144} = \frac{20}{7}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.