Question

The dimension of the quantity $\frac{1}{{\in }_0}\frac{e^2}{hc}$ is (e = charge of electron, h = Planck's constant and c = velocity of light)

• A. $\left[M^{-1}{L}^{-3}{T}^{2}A\right]$
• B. $\left[M^0{L}^0{T}^0{A}^0\right]$
• C. $\left[M{L}^{-3}{T}^4{A}^{-2}\right]$
• D. $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

Answer 1

The correct option is B $\left[M^0{L}^0{T}^0{A}^0\right]$

The dimension of the physical quantity,

$\left[c\right]=\left[M^0{L}^1{T}^{-1}\right]$

$\left[e\right]=\left[A^1{T}^1\right]$

$\left[h\right]=\left[M^1{L}^2{T}^{-1}\right]$

$\left[{\epsilon }_0\right]=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

The dimension of the quantity $=\frac{e^2}{{\epsilon }_{0}hc}$

$=\frac{\left[e\right].\left[e\right]}{\left[{\epsilon }_0\right].\left[h\right].\left[c\right]}$

$=\frac{\left[AT\right].\left[AT\right]}{\left[M^{-1}{L}^{-3}{T}^4{A}^2\right].\left[M^1{L}^2{T}^{-1}\right].\left[L^1{T}^{-1}\right]}$

$=\left[M^0{L}^0{T}^0{A}^0\right]$

Hence, it is a dimensionless quantity.