A) \[\frac{{{h}^{2}}}{{{\pi }^{2}}{{m}^{2}}{{r}^{3}}}\]
B) \[\frac{{{h}^{2}}}{8{{\pi }^{2}}{{m}^{2}}{{r}^{3}}}\]
C) \[\frac{{{h}^{2}}}{4{{\pi }^{2}}{{m}^{2}}{{r}^{3}}}\]
D) \[\frac{{{h}^{2}}}{4\pi {{m}^{2}}{{r}^{3}}}\]
Correct Answer: C
Solution :
[c] Speed of electron in first orbit (n=1) of hydrogen atom (z=1), \[v=\frac{{{e}^{2}}}{2{{\varepsilon }_{0}}h}\] \[r=\frac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi m{{e}^{2}}}\Rightarrow {{\varepsilon }_{0}}=\frac{r\pi m{{e}^{2}}}{{{h}^{2}}}\] Acceleration of electron, \[\frac{{{v}^{2}}}{r}=\frac{{{e}^{4}}}{4\varepsilon _{0}^{2}{{h}^{2}}}\times \frac{\pi m{{e}^{2}}}{{{h}^{2}}{{\varepsilon }_{0}}}\] \[\frac{{{e}^{4}}\times \pi m{{e}^{2}}}{4{{h}^{4}}\varepsilon _{0}^{3}}\] Eliminating \[{{\varepsilon }_{0}}\] \[=\frac{{{e}^{4}}\pi m{{e}^{2}}{{h}^{6}}}{4{{h}^{4}}{{r}^{3}}{{\pi }^{3}}{{m}^{3}}{{e}^{6}}}=\frac{{{h}^{2}}}{4{{\pi }^{2}}{{m}^{2}}{{r}^{3}}}\]
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