A) \[\left( \frac{h}{m{{e}^{2}}} \right)\]
B) \[\left( \frac{hc}{m{{e}^{2}}} \right)\]
C) \[\left( \frac{h}{c{{e}^{2}}} \right)\]
D) \[\left( \frac{m{{c}^{2}}}{h{{e}^{2}}} \right)\]
Correct Answer: C
Solution :
[c] Let \[{{\mu }_{0}}\] related with e, m, c and h as follows. \[{{\mu }_{0}}=k{{e}^{a}}{{m}^{b}}{{c}^{c}}{{h}^{d}}\] \[[ML{{T}^{-2}}{{A}^{-2}}]={{[AT]}^{a}}{{[M]}^{b}}{{[L{{T}^{-1}}]}^{c}}{{[M{{L}^{2}}{{T}^{-1}}]}^{d}}\] \[-[{{M}^{b+d}}{{L}^{c+2d}}{{T}^{a-c-d}}{{A}^{a}}]\] On comparing both sides we get \[a=-2,b=0,c=-1,d=1\] \[\therefore \,\,\,[{{\mu }_{0}}]-\left[ \frac{h}{c{{e}^{2}}} \right]\]
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