Question
Physics Question on Dimensional Analysis
The dimensions of 4πε0hc′e2, where e,ε0,h and c and c are electronic charge, electric permittivity Planck's constant and velocity of light in vacuum respectively
A
[M0L0T0]
B
[ML0T0]
C
[M0LT0]
D
[M0L0T]
Answer
[M0L0T0]
Explanation
Solution
[e]=AT,ε0=[M−1L−3T4A2] [h]=[ML2T−1] and [c]=[LT−1] ∴[4πε0hce2] =[M−1L−3T4A2×ML2T−1×LT−1A2T2] =[M0L0T0]