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The angular momentum of an electron moving in an orbit of hydrogen atom is $1.5\left(\dfrac{h}{\pi}\right)$. The energy in the same orbit is nearly.
A
$-1.5$ eV
B
$-1.6$ eV
C
$-1.3$ eV
D
$-1.4$ eV
Detailed Solution
Given $mvr=1.5\dfrac{h}{\pi}$. Comparing with $mvr=n\dfrac{h}{2\pi}$: $\dfrac{n}{2}=1.5\Rightarrow n=3$. $E_3=\dfrac{-13.6}{3^2}\ eV\approx-1.5$ eV.