Q. An electron of a hydrogen like atom, having $Z\ =\ 4$, jumps from $4^{th}$ energy state to $2^{nd}$ energy state, The energy released in this process, will be: (Given $Rch\ =\ 13.6\ eV$) Where $R\ =$ Rydberg constant $c\ =$ Speed of light in vacuum $h\ =$ Planck's constant
A) $13.6\ eV$
B) $10.5\ eV$
C) $3.4\ eV$
D) $40.8\ eV$
Explanation
The correct answer is Option D: $40.8\ eV$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.