Question

What is dimensional formula of the quantity,$\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{{{e}^{2}}}{hc}$ where $e$ is the charge of electron, $h$ is the Planck’s constant, and $c$ is the velocity of light?
$\begin{align}
  & A.{{M}^{-1}}{{L}^{-3}}{{T}^{2}}A \\
 & B.{{M}^{0}}{{L}^{0}}{{T}^{0}}{{A}^{0}} \\
 & C.M{{L}^{3}}{{T}^{-4}}{{A}^{-2}} \\
 & D.{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \\
\end{align}$

Answer 1

Hint: Dimensional formula contains the basic physical quantities called as fundamental quantities in physics. This kind of notation helps us to understand the basic nature of a material, whether it’s a form of energy, force etc.

Complete step by step solution:
Dimensional analysis is the overview of the relationships in between different physical quantities by finding out their basic physical quantities and units of measurement and using these dimensional values for calculations or comparisons to be performed. Dimensional analysis is useful in various ways. It is primarily used to verify the consistency of a dimensional equation. To derive the relationship between physical quantities in a natural phenomenon and to vary units from one system to the other one. There are some limitations to this too. It will not be useful in deriving the relation or formula if a physical quantity is dependable upon more than three factors having dimensions. It may not help to derive a formula including trigonometric function, exponential function, and logarithmic function. It is not helpful in deriving a relation having more than one part in an equation. The most basic rule in dimensional analysis is the dimensional homogeneity. Only the physical quantities having the similar dimension can be compared, equated, or done any operations.
Here the given equation is
$\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{{{e}^{2}}}{hc}$
Where
Let us take
${{\varepsilon }_{0}}$Be permittivity,
\[\begin{align}
  & e=AT \\
 & h=M{{L}^{2}}{{T}^{-2}} \\
 & c=L{{T}^{-1}} \\
\end{align}\]
We know that the permittivity will be,
${{\varepsilon }_{0}}=\dfrac{{{q}_{1}}{{q}_{2}}}{F{{r}^{2}}}$
From this the dimension of ${{\varepsilon }_{0}}$ will be,
$\left[ {{\varepsilon }_{0}} \right]=\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]$
Hence we can write that.
\[\begin{align}
  & =\dfrac{{{\left( AT \right)}^{2}}}{\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]M{{L}^{2}}{{T}^{-2}}L{{T}^{-1}}} \\
 & ={{M}^{0}}{{L}^{0}}{{T}^{0}}{{A}^{0}} \\
\end{align}\]

So, the correct answer is “Option B”.

Note: Dimensions are the powers raised to the fundamental quantities to represent other physical quantities. Dimensional formula is the expression of the dimensions of a physical quantity which is represented as fundamental quantities.