Q. The ionization energy of the electron in the hydrogen atom in its ground state is $13.6\ eV.$ The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between
A) $n = 3$ to $n = 2$ states
B) $n = 3$ to $n = 1$ states
C) $n = 2$ to $n = 1$ states
D) $n = 4$ to $n = 3$ states
Explanation
(d) Number of wavelength $= \frac{n(n - 1)}{2}$, where $n =$ No. of orbit from which transition takes place
$\therefore 6 = \frac{n(n - 1)}{2} \Rightarrow n = 4$
In all given options wavelength of emitted radiation’s will be maximum for transition $n = 4$ to $n = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.