Q. The energy of the electron in first Bohr’s orbit is $-$13.6eV. The energy of the electron in its first excited state is
A) $- 3.4\ eV$
B) $- 27.8\ eV$
C) $- 6.8\ eV$
D) $- 10.2\ eV$
Explanation
(a)
For first excited state $n = 2$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{n} = \frac{E_{1}}{n^{2}}$
(Where, $E_{1}$= energy of first Bohr’s orbit)
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{2} = \frac{- 13.6}{(2)^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - 3.4eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.