Q. Energy of an electron in $n^{th}$ orbit of hydrogen atom is $\left( k = \frac{1}{4\pi\varepsilon_{0}} \right)$
A) $- \frac{2\pi^{2}k^{2}me^{4}}{n^{2}h^{2}}$
B) $- \frac{4\pi^{2}mke^{2}}{n^{2}h^{2}}$
C) $- \frac{n^{2}h^{2}}{2\pi\ k\ me^{4}}$
D) $- \frac{n^{2}h^{2}}{4\pi^{2}k\ me^{2}}$
Explanation
The correct answer is Option A: $- \frac{2\pi^{2}k^{2}me^{4}}{n^{2}h^{2}}$.
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.