A) \[\left( \frac{hc}{m{{e}^{2}}} \right)\]
B) \[\left( \frac{h}{m{{e}^{2}}} \right)\]
C) \[\left( \frac{h}{c{{e}^{2}}} \right)\]
D) \[\left( \frac{m{{c}^{2}}}{h{{e}^{2}}} \right)\]
Correct Answer: C
Solution :
\[\left[ e \right]=IT\] \[\left[ m \right]=M\] \[\left[ c \right]=L{{T}^{-1}}\] \[\left[ h \right]=M{{L}^{2}}{{T}^{-1}}\] \[\left[ {{\mu }_{0}} \right]=ML{{I}^{-2}}{{T}^{-3}}\] If \[{{\mu }_{0}}={{e}^{a}}{{m}^{b}}{{c}^{c}}{{h}^{d}}\] \[ML{{T}^{-2}}{{T}^{-3}}={{\left[ IT \right]}^{a}}{{\left[ M \right]}^{b}}{{\left[ L{{T}^{-1}} \right]}^{c}}{{\left[ M{{L}^{2}}{{T}^{-1}} \right]}^{d}}\] by equating powers, we get \[a=-2,b=0,c=-1,d=1\] \[\therefore \] \[\left[ {{\mu }_{0}} \right]=\left[ \frac{h}{c{{e}^{2}}} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
