Question

The angular momentum of electron in $L{i}^{2+}$ is found to be $14\left(\frac{h}{11}\right)$. Calculated the potential energy (in $eV$) of system.

• A. $13.6\times {\left(\frac{3}{8}\right)}^2$
• B. $-13.6\times {\left(\frac{3}{8}\right)}^2$
• C. $-2\times 13.6\times {\left(\frac{3}{8}\right)}^2$
• D. $-2\times 13.6\times {\left(\frac{8}{3}\right)}^2$

Answer 1

Answer: (C)

The angular momentum of an electron in an $n^{th}$ orbit is given by
$m_{e}vr=n\cdot \frac{h}{2\pi };n=\mathrm{1.2.3}\dots \dots$
$n\frac{h}{2\pi }=14\frac{h}{11}\Rightarrow n=8$
$P.E=-2\times 13.6\frac{Z^2}{n^2}eV$
Potential energy $=-2\times 13.6\times {\left(\frac{Z}{n}\right)}^2$
Potential energy $=-2\times 13.6\times {\left(\frac{3}{8}\right)}^2$
$\because$ for $L{i}^{2+},Z=3$