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 $n^{t h}$ orbit is given by
$m _{ e } vr = n \cdot \frac{ h }{2\pi } ; n =1.2 .3 \ldots \ldots$
$n \frac{h}{2 \pi}=14 \frac{h}{11} \Rightarrow n=8$
$P . E=-2 \times 13.6 \frac{Z^{2}}{n^{2}} e V$
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 $L i^{2+}, Z=3$