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 \[a=-2\] ...(i) \[b+d=1\] ...(ii) \[c+2d=1\] ...(iii) \[a-c-d=-2\] ...(iv) By equation (i), (ii), (iii) & (iv) we get, \[a=-2,\]\[b=0,\]\[c=-1,\]\[d=1\] \[\therefore \]\[[{{\mu }_{0}}]=\left[ \frac{h}{c{{e}^{2}}} \right]\]
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