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) \[\frac{9{{\pi }^{3}}m{{e}^{4}}}{{{h}^{3}}}{{K}^{2}}\]
Correct Answer: D
Solution :
[d] \[{{v}_{n}}={{r}_{n}}\omega \] where \[{{r}_{n}}=\frac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}m{{e}^{2}}Z.K}\] and \[{{v}_{n}}=\frac{2\pi \cdot Z\cdot {{e}^{2}}\cdot K}{n\cdot h\cdot }\] \[\therefore =\frac{2\pi Z{{e}^{2}}\cdot K}{n\cdot h\cdot }=\frac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}m{{e}^{2}}Z\cdot K}\times \omega ;\] \[\omega =\frac{8{{\pi }^{3}}m{{e}^{4}}\cdot {{Z}^{2}}\cdot {{K}^{2}}}{{{n}^{3}}\cdot {{h}^{3}}}\] \[=\frac{9{{\pi }^{3}}m{{e}^{4}}\cdot {{K}^{2}}}{{{h}^{3}}}(\therefore n=2and\,Z=3)\]
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