Question

What is the angular velocity $\left(\omega \right)$ of an electron occupying second orbit of $L{i}^{2+}$ ion?

• A. $\frac{8{\pi }^{3}m{e}^4}{h^3}{K}^2$
• B. $\frac{8{\pi }^{3}m{e}^4}{9h^3}{K}^2$
• C. $\frac{64}{9}\times \frac{{\pi }^{3}m{e}^4}{h^3}{K}^2$
• D. $\frac{9{\pi }^{3}m{e}^4}{h^3}{K}^2$

Answer 1

Answer: (D)

$\frac{9{\pi }^{3}m{e}^4}{h^3}{K}^2$

$v=r\omega$ where $r_n=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}Z.K}$
and $v_n=\frac{2\pi .Z.e^2.K}{n.h}$
$\therefore \frac{2\pi Z{e}^2.K}{n.h}=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{4}Z.K}\times \omega$;
$\omega =\frac{8{\pi }^{3}m{e}^4.Z^2.K^2}{n^3.h^3}$
$=\frac{9{\pi }^{3}m{e}^4.K^2}{h^3}(\because n=2andZ=3)$