The angular momentum of an electron in a stationary state of $\mathrm{Li}^{2+}(\mathrm{Z}=3)$ is $\frac{3…
The angular momentum of an electron in a stationary state of $\mathrm{Li}^{2+}(\mathrm{Z}=3)$ is $\frac{3 h}{\pi}$. The radius and energy of that stationary state are respectively
$3.174 Å,-5.45 \times 10^{-19} \mathrm{~J}$
$6.348 Å,-5.45 \times 10^{-19} \mathrm{~J}$
$6.348 Å,+5.45 \times 10^{-18} \mathrm{~J}$
$2.116 Å,-5.45 \times 10^{-19} \mathrm{~J}$
Solution
Angular momentum $=\frac{3 \mathrm{~h}}{\pi} \Rightarrow \mathrm{~m}_{\mathrm{e}} \mathrm{vr}= \frac{\mathrm{nh}}{2 \pi}$
$\frac{3 \mathrm{~h}}{\pi}=\mathrm{n} \cdot \frac{\mathrm{h}}{2 \pi} \Rightarrow \mathrm{n}=6$
For the Radius $\left(\mathrm{r}_{\mathrm{n}}\right)=\mathrm{a}_0 \times \frac{\mathrm{n}^2}{2}$
$=52.9 \mathrm{pm} \times \frac{(6)^2}{3}$
$=634.8 \times 10^{-12} \mathrm{~m} . \Rightarrow \mathrm{r}_{\mathrm{n}}=6.348 Å$
Energy of stationary state $\left(E_n\right)=-R_H \times\left(\frac{1}{n^2}\right)$
For $\mathrm{Li}^{+2}=(n=2)$
$E_2=-2.18 \times 10^{-18} \mathrm{~J} \times \frac{1}{2^2}$
$=-0.545 \times 10^{-18} \mathrm{~J}=-5.45 \times 10^{-19} \mathrm{~J}$