Q. The principal quantum number of H-atom orbital, if the electron energy is $- 3.4$ eV, will be
A) 1
B) 2
C) 3
D) Zero
Explanation
(b)
$E_{n} = \frac{- 13.6 \times Z^{2}}{n^{2}}$eV
For H atom, $Z = 1,$
$- 3.4 = \frac{- 13.6 \times (1)^{2}}{n^{2}}$
$\Rightarrow \ n^{2} = 4$
$\therefore n = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.