Solveeit Logo

Question

Physics Question on Dimensional Analysis

The dimensions of e24πε0hc\frac{e^{2}}{4 \pi \varepsilon_{0} h c'}, where e,ε0,he, \varepsilon_{0}, h and cc and cc are electronic charge, electric permittivity Planck's constant and velocity of light in vacuum respectively

A

[M0L0T0]\left[ M ^{0} L ^{0} T ^{0}\right]

B

[ML0T0]\left[ ML ^{0} T ^{0}\right]

C

[M0LT0]\left[ M ^{0} LT ^{0}\right]

D

[M0L0T]\left[ M ^{0} L ^{0} T\right]

Answer

[M0L0T0]\left[ M ^{0} L ^{0} T ^{0}\right]

Explanation

Solution

[e]=AT,ε0=[M1L3T4A2][e]=A T, \varepsilon_{0}=\left[ M ^{-1} L ^{-3} T ^{4} A ^{2}\right] [h]=[ML2T1][h]=\left[ ML ^{2} T ^{-1}\right] and [c]=[LT1][c]=\left[ LT ^{-1}\right] [e24πε0hc]\therefore\left[\frac{e^{2}}{4 \pi \varepsilon_{0} h c}\right] =[A2T2M1L3T4A2×ML2T1×LT1]=\left[\frac{A^{2} T^{2}}{M^{-1} L^{-3} T^{4} A^{2} \times M L^{2} T^{-1} \times L T^{-1}}\right] =[M0L0T0]=\left[M^{0} L^{0} T^{0}\right]