Question

The dimension of $\dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}hc}$ where $e,$ ${{\varepsilon }_{0}}$,$h$ and $c$ are the electronic charge, electric permittivity. Planck’s constant and velocity of the light in vacuum, respectively, is?
A) $[{{M}^{0}}{{L}^{0}}{{T}^{0}}]$
B) $[{{M}^{1}}{{L}^{0}}{{T}^{0}}]$
C) $[{{M}^{0}}{{L}^{1}}{{T}^{0}}]$
D) $[{{M}^{0}}{{L}^{0}}{{T}^{1}}]$

Answer 1

The dimension of $\dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}hc}$ is obtained by using the dimension of charge, electric permittivity. Planck’s constant and velocity of the light in vacuum .Light travels approximately 300,000kilometers per second in a vacuum and the speed of light in air and in water medium is different and the refractive index in vacuum is 1.3.

Complete step-by-step solution:
Electric charge is the product of electric current and the time. Electric charge is of two types and that is positive charge and negative charge. Positive charge are protons and negative charge are electrons
Electric charge (q or e) =current $\times$ time
The dimension of electric charge is
$e=[I]\times [T]$
$e=[{{M}^{0}}{{L}^{0}}{{T}^{1}}{{I}^{1}}]$
Electric permittivity in free space$({{\varepsilon }_{0}})$
It is the ability of a material to permit electric field which in turn relates to permittivity. Electric permittivity of free space is a constant of proportionality
${{\varepsilon }_{0}}=\dfrac{1}{{{c}^{2}}{{\mu }_{0}}}$
Dimension of ${{\varepsilon }_{0}}=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{I}^{2}}]$
Planck’s constant $(h)$
Energy of photon is given as
$E=hv$
$h=\dfrac{E}{v}$
We know the Dimension of $E=[{{M}^{0}}{{L}^{2}}{{T}^{-2}}]$
And also we known the Dimension of $v={{T}^{-1}}$
Thus Dimension of Planck’s constant is $[{{M}^{1}}{{L}^{2}}{{T}^{-1}}]$
Speed of light$(c)$
Is defined as the ratio of distance travelled by light in the vacuum to the time taken
Dimension of speed of light is
$c=[{{M}^{0}}{{L}^{1}}{{T}^{-1}}]$
The dimension of $\dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}hc}$ $\cdots \cdots (1)$
In equation $(1)$ substitute the dimensional formula of $e$ ${{\varepsilon }_{0}}$ $c$ and$h$
After substituting
$\begin{aligned} &\dfrac{{{e}^{2}}}{4\pi{{\varepsilon}_{0}}hc}=\dfrac{[{{I}^{2}}{{T}^{2}}]}{[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{I}^{2}}][{{M}^{1}}{{L}^{2}}{{T}^{-1}}][{{M}^{0}}{{L}^{1}}{{T}^{-1}}]} \\\ & \dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}hc}=[{{M}^{0}}{{L}^{0}}{{T}^{0}}] \\\ \end{aligned}$
So the correct option is A.

Note: Students take the dimensional formula of${{\varepsilon }_{0}}$ and don’t take the value of ${{\varepsilon }_{0}}$ that is$8.854\times {{10}^{-12}}c$. Don’t consider $4\pi$ because it is a value. The Planck’s constant is the quantum of electromagnetic action which in turn relates to a photon’s energy to its frequency and the value of Planck’s constant is$6.6260705\times {{10}^{-34}}J.s$.