What is the angular velocity $(\omega)$ of an electron occupying second orbit of $-\mathrm{Li}^{2+}$ ion?
- $\frac{8 \pi^{3} \mathrm{me}^{4}}{\mathrm{~h}^{3}} \mathrm{~K}^{2}$
- $\frac{8 \pi^{3} m e^{4}}{9 h^{3}} K^{2}$
- $\frac{64}{9} \times \frac{\pi^{3} \mathrm{me}^{4}}{\mathrm{~h}^{3}} \mathrm{~K}^{2}$
- $\frac{9 \pi^{3} \mathrm{me}^{4}}{\mathrm{~h}^{3}} \mathrm{~K}^{2}$
Solution
\mathrm{v}_{\mathrm{n}}=\mathrm{r}_{\mathrm{n}} \omega \text { where } \mathrm{r}_{\mathrm{n}}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{4 \pi^{2} \mathrm{me}^{2} \mathrm{Z} \cdot \mathrm{K}} \\
\text { and } \mathrm{v}_{\mathrm{n}}=\frac{2 \pi \cdot \mathrm{Z} \cdot \mathrm{e}^{2} \cdot \mathrm{K}}{\mathrm{n} \cdot \mathrm{h}}
\end{array}$
$\therefore$ $\frac{2 \pi \mathrm{Ze}^{2} \cdot \mathrm{K}}{\mathrm{n} \cdot \mathrm{h}}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{4 \pi^{2} \mathrm{me}^{2} \mathrm{Z} \cdot \mathrm{K}} \times \omega ;$
$\omega=\frac{8 \pi^{3} \mathrm{me}^{4} \cdot \mathrm{Z}^{2} \cdot \mathrm{K}^{2}}{\mathrm{n}^{3} \cdot \mathrm{h}^{3}}$
$=\frac{9 \pi^{3} \mathrm{me}^{4} \cdot \mathrm{K}^{2}}{\mathrm{~h}^{3}}(\because \mathrm{n}=2$ and $\mathrm{Z}=3)$
Asked in: JEE-TOPICTESTS-CHEMISTRY