Question

The kinetic energy of the electron present in the ground state of $L{i}^{2+}$ ion is represented by:

• A. $\frac{3e^2}{8\pi {\in }_{0}r}$
• B. $-\frac{3e^2}{8\pi {\in }_{0}r}$
• C. $\frac{3e^2}{4\pi {\in }_{0}r}$
• D. $-\frac{3e^2}{4\pi {\in }_{0}r}$

Answer 1

Answer: (B)

$-\frac{3e^2}{8\pi {\in }_{0}r}$

$F_e=\frac{k{q}_1{q}_2}{r^2}$

$=\frac{-1}{4\lambda {ϵ}_o}\times \frac{3e\times e}{r^2}$

$q_1+3e$

$q_2=-e$

$F_e=\frac{m{v}^2}{r}=\frac{-3e^2}{4\pi {ϵ}_o{r}^2}$

$m{v}^2=\frac{-3e^2}{4\lambda {ϵ}_{o}r}$

$\frac{1}{2}m{v}^2=\frac{-3e^2}{8\pi {ϵ}_{o}r}$

$\therefore K.E.=\frac{-3e^2}{8\pi {ϵ}_{o}r}$