A) \[\frac{nh}{16\,ke{{\pi }^{2}}\,{{m}^{3/2}}}\]
B) \[\frac{{{n}^{2}}{{h}^{2}}}{18\,{{k}^{2}}{{e}^{2}}{{\pi }^{2}}\,{{m}^{3}}}\]
C) \[\frac{{{n}^{3}}{{h}^{3}}}{18\,{{k}^{2}}{{e}^{2}}{{\pi }^{3}}\,{{m}^{4}}}\]
D) \[\frac{{{n}^{2}}{{h}^{2}}}{4\sqrt{2}\,ke{{\pi }^{2}}{{m}^{3/2}}}\]
Correct Answer: D
Solution :
[d] \[\frac{d[U(r)]}{dr}=\frac{4k{{e}^{2}}}{{{r}^{5}}}=force\] |
\[\frac{4k{{e}^{2}}}{{{r}^{5}}}=\frac{m{{v}^{2}}}{r}\] |
and \[mvr=\frac{nh}{2\pi }\] |
or \[r=\frac{nh}{2\pi mv}\Rightarrow \frac{1}{r}=\frac{2\pi mv}{nh}\] |
\[4k{{e}^{2}}\times \frac{1}{{{r}^{5}}}=\frac{m{{v}^{2}}}{r}\] |
\[2k{{e}^{2}}\times \frac{1}{{{r}^{4}}}=m{{v}^{2}}\] |
\[2k{{e}^{2}}\times \frac{16{{\pi }^{4}}{{m}^{4}}{{v}^{4}}}{{{n}^{4}}{{h}^{4}}}=m{{v}^{2}}\] |
\[{{v}^{2}}={{n}^{4}}{{h}^{4}}/32k{{e}^{2}}{{\pi }^{4}}{{m}^{3}}\] |
\[v=\frac{{{n}^{2}}{{h}^{2}}}{4\sqrt{2}ke{{\pi }^{2}}{{m}^{3/2}}}\] |
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