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.
$$\frac{{8{\pi ^3}m{e^4}}}{{9{h^3}}}{K^2}$$
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 :
$$\frac{{9{\pi ^3}m{e^4}}}{{{h^3}}}{K^2}$$
Solution :
$$\eqalign{
& {v_n} = {r_n}\omega \,\,{\text{where}}\,\,{r_n} = \frac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}Z \cdot K}} \cr
& {\text{and}}\,\,{v_n} = \frac{{2\pi \cdot Z \cdot {e^2} \cdot K}}{{n \cdot h}} \cr
& \therefore \,\,\frac{{2\pi Z{e^2} \cdot K}}{{n \cdot h}} = \frac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}\,Z \cdot K}} \times \omega \,; \cr
& \omega = \frac{{8{\pi ^3}m{e^4} \cdot {Z^2} \cdot {K^2}}}{{{n^3} \cdot {h^3}}} \cr
& = \frac{{9{\pi ^3}m{e^4} \cdot {K^2}}}{{{h^3}}}\left( {\because \,\,n = 2\,{\text{and}}\,Z = 3\,} \right) \cr} $$