A) \[[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\]
B) \[[M{{L}^{0}}{{T}^{0}}]\]
C) \[[{{M}^{0}}L{{T}^{0}}]\]
D) \[[{{M}^{0}}{{L}^{0}}T]\]
Correct Answer: A
Solution :
From \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{{{e}^{2}}}{{{r}^{2}}}\] \[\therefore \,\,\frac{{{e}^{2}}}{{{\varepsilon }_{0}}}=4\pi \,F\,{{r}^{2}}\] (dimensionally) \[\frac{{{e}^{2}}}{{{\varepsilon }_{0}}hc}=\frac{F{{r}^{2}}}{hc}=\frac{(ML{{T}^{-2}}){{L}^{2}}}{M{{L}^{2}}{{T}^{-1}}(L{{T}^{-1}}]}=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\],
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