A) \[\frac{ch}{2\pi \varepsilon _{0}^{2}}\]
B) \[\frac{{{e}^{2}}}{2\pi {{\varepsilon }_{0}}Gm_{e}^{2}}\left( {{\operatorname{m}}_{e}}=\,mass\,of\,electron \right)\]
C) \[\frac{{{\mu }_{0}}{{\varepsilon }_{0}}}{{{c}^{2}}}\frac{G}{h{{e}^{2}}}\]
D) \[\frac{2\pi \sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}{c{{e}^{2}}}\frac{h}{G}\]
Correct Answer: B
Solution :
[b]| The dimensional formulae of \[e=[{{M}^{0}}{{L}^{0}}{{T}^{1}}{{A}^{1}}]\] |
| \[{{\varepsilon }_{0}}=[{{M}^{-1}}{{L}^{3}}{{T}^{4}}{{A}^{2}}]\] |
| \[G=[{{M}^{1}}{{L}^{0}}{{T}^{0}}]\] and \[{{m}_{e}}=[{{M}^{1}}{{L}^{0}}{{T}^{0}}]\] |
| Now, \[\frac{{{e}^{2}}}{2\pi {{\varepsilon }_{0}}Gm_{e}^{2}}\]\[=\frac{{{[{{M}^{0}}{{L}^{0}}{{T}^{1}}{{A}^{1}}]}^{2}}}{2\pi [{{M}^{-1}}{{L}^{3}}{{T}^{4}}{{A}^{2}}][{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]{{[{{M}^{1}}{{L}^{0}}{{T}^{0}}]}^{2}}}\]\[=\frac{[{{T}^{2}}{{A}^{2}}]}{2\pi [{{M}^{-1-1+2}}{{L}^{-3+3}}{{T}^{4-2}}{{A}^{2}}]}\]\[=\frac{[{{T}^{2}}{{A}^{2}}]}{2\pi [{{M}^{0}}{{L}^{0}}{{T}^{2}}{{A}^{2}}]}=\frac{1}{2\pi }\] |
| \[\because \frac{1}{2\pi }\]is dimensionless thus the combination \[\frac{{{e}^{2}}}{2\pi {{\varepsilon }_{0}}Gm_{e}^{2}}\]would have the same value in different systems of units. |
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