A) \[\frac{ch}{2\pi \varepsilon _{o}^{2}}\]
B) \[\frac{{{e}^{2}}}{2\pi \varepsilon _{o}^{{}}Gm_{e}^{2}}\](\[{{m}_{e}}=\] mass of electron)
C) \[\frac{{{\mu }_{o}}{{\varepsilon }_{o}}}{{{c}^{2}}}\frac{G}{h{{e}^{2}}}\]
D) \[\frac{2\pi \sqrt{{{\mu }_{o}}{{\varepsilon }_{o}}}}{c{{e}^{2}}}\frac{h}{G}\]
Correct Answer: B
Solution :
| [b] The dimensional formulae of \[e=\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}}{{A}^{1}} \right]\] |
| \[{{\varepsilon }_{0}}=\left[ {{M}^{-1}}{{L}^{3}}{{T}^{4}}{{A}^{2}} \right]\] |
| \[G=\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]\]and\[{{m}_{e}}=\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]\] |
| Now, \[\frac{{{e}^{2}}}{2\pi {{\varepsilon }_{0}}Gm_{e}^{2}}\] |
| \[\frac{{{\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}}{{A}^{1}} \right]}^{2}}}{2\pi \left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]{{\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]}^{2}}}\] |
| \[=\frac{\left[ {{T}^{2}}{{A}^{2}} \right]}{2\pi \left[ {{M}^{-1-1+2}}{{L}^{-3+3}}{{T}^{4-2}}{{A}^{2}} \right]}\] |
| \[=\frac{\left[ {{T}^{2}}{{A}^{2}} \right]}{2\pi \left[ {{M}^{0}}{{L}^{0}}{{T}^{2}}{{A}^{2}} \right]}\] \[=\frac{1}{2\pi }\] |
| \[\because \]\[\frac{1}{2\pi }\]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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