Question

The angular momentum of an electron in a certain orbit of $L{i}^{+2}$ ion is $3.16\times {10}^{-34}{\text{kg m}}^2{\text{sec}}^{-1}$ . What will be the potential energy of electron in that orbit?
Given : $h=6.62\times {10}^{-34}\text{J sec}\pi =3.143.16\times 3.14\approx 9.93$

• A. $-13.6\text{eV}$
• B. $-27.2\text{eV}$
• C. $+13.6\text{eV}$
• D. $-53.4\text{eV}$

Answer 1

The correct option is B $-27.2\text{eV}$

$\text{Angular momentum}=\frac{nh}{2\pi }=3.15\times {10}^{-34}{\text{kg m}}^2{\text{sec}}^{-1}\Rightarrow n=3$
$\text{Total energy of electron}=-13.6\left(\frac{Z^2}{n^2}\right)=-13.6\left(\frac{3^2}{3^2}\right)=-13.6\text{eV}$
$\text{P.E.}=2\times \text{(Total energy)}=-27.2\text{eV}$