Question

The potential energy of an electron present in the second orbit of $L{i}^{2+}$ ion will be:

• A. $-61.2eV$
• B. $-7.65eV$
• C. $-1.51eV$
• D. $13.6eV$

Answer 1

Answer: (A)

$-61.2eV$

The potential energy of an electron in the $n^{th}$ orbit is given by:

$P.E.=-\frac{27.2z^2}{n^2}$ eV

As it is lithium-ion $‘z^{\prime }=3$ and as it is 2nd orbit, $n=2$.

Putting $z=3,n=2$,

$P.E.=-\frac{27.2\times 3^2}{2^2}$

$P.E=-61.2eV$

Hence, option $A$ is correct.