Q. The dimensions of $e^{2}/4\pi\varepsilon_{0}hc$, where $e,\ \varepsilon_{0},\ h$ and $c$ are electronic charge, electric permittivity, Planck’s constant and velocity of light in vacuum respectively
A) $\lbrack M^{0}L^{0}T^{0}\rbrack$
B) $\lbrack M^{1}L^{0}T^{0}\rbrack$
C) $\lbrack M^{0}L^{1}T^{0}\rbrack$
D) $\lbrack M^{0}L^{0}T^{1}\rbrack$
Explanation
$\lbrack e\rbrack = \lbrack AT\rbrack,\ \in_{0} = \left\lbrack M^{- 1}L^{- 3}T^{4}A^{2} \right\rbrack,\ \lbrack h\rbrack = \lbrack ML^{2}T^{- 1}\rbrack$ And $\lbrack c\rbrack = \lbrack LT^{- 1}\rbrack$ $\therefore\left\lbrack \frac{e^{2}}{4\pi\epsilon_{0}hc} \right\rbrack = \left\lbrack \frac{A^{2}T^{2}}{M^{- 1}L^{- 3}T^{4}A^{2} \times ML^{2}T^{- 1} \times LT^{- 1}} \right\rbrack$ $= \lbrack M^{0}L^{0}T^{0}\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.