Q. The ionization energy of the ground state hydrogen atom is $2.18 \times 10^{- 18}$ J. The energy of an electron in its second orbit would be
A) $- 2.67 \times 10^{- 18}$ J
B) $- 5.45 \times 10^{- 19}$ J
C) $- 3.58 \times 10^{- 18}$ J
D) $- 4.68 \times 10^{- 19}$ J
Explanation
(b)
Sine, $E \propto - \frac{1}{n^{2}}$
The energy of an electron in the second orbit will be
$E_{2} = \frac{E_{1}}{4} = \frac{( - 2.18 \times 10^{- 18}J)}{4}$
$= - 5.45 \times 10^{- 19}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.