Q. Energy required for the electron excitation in $Li^{+ +}$ from the first to the third Bohr orbit is
A) $12.1\ eV$
B) $36.3\ eV$
C) $108.8\ eV$
D) $122.4\ eV$
Explanation
(c) $E_{1} = - \frac{13.6(3)^{2}}{(1)^{2}}$
$E_{3} = - \frac{13.6(3)^{2}}{(3)^{2}}$
$\therefore\mathrm{\Delta}E = E_{3} - E_{1} = 13.6(3)^{2}\left\lbrack 1 - \frac{1}{9} \right\rbrack = \frac{13.6 \times 9 \times 8}{9}$
$\Rightarrow \mathrm{\Delta}E = 108.8eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.