Q. The H atom electron dropped from $n = 3\ to\ n = 2$, then energy emitted is
A) 1.9 eV
B) 12 eV
C) 10.2 eV
D) 0.65 eV
Explanation
(a)
$E_{n} = \frac{13.6}{n^{2}}eV$
$E_{3} - E_{2} = 13.6\left( \frac{1}{(2)^{2}} - \frac{1}{(3)^{2}} \right)eV$
$E_{3} - E_{2} = 13.6\ \left( \frac{1}{4} - \frac{1}{9} \right)eV$
$E_{3} - E_{2} = 13.6 \times \left( \frac{5}{36} \right)eV$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.9\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.