Question

If the capacitance of a nano capacitor is measured in terms of a unit \['\mu '\] made by combining the electronic charge \['e'\] , bohr radius \['{a_0}'\] , Planck’s constant \['h'\] and speed of light \['c'\] then which of the following relation is possible?
A) \[\mu = \dfrac{{{e^2}c}}{{h{a_0}}}\]
B) \[\mu = \dfrac{{{e^2}h}}{{c{a_0}}}\]
C) \[\mu = \dfrac{{hc}}{{{e^2}{a_0}}}\]
D) \[\mu = \dfrac{{{e^2}{a_0}}}{{hc}}\]

Answer 1

Hint: Check the dimensions of the both sides of the given equations. The one which has dimensions on left hand and right hand equal will be the correct option.

Complete step by step solution:
We will solve this formula with the help of dimensional analysis . If a relation is correct, then the dimensions on the right hand side will be equal to the dimensions on the left hand side . All the physical quantities in physics can be expressed in terms of some sort of combinations of base quantities ( length, mass, time being the most common).

Dimensions of any quantity in physics are the powers to which the fundamental ( base) quantities can be raised to represent that quantity completely .
 Now let us assume that given four quantities are dimensionally comparable and are related as follows :
$[\mu ] = {[e]^w}{[{a_0}]^x}{[h]^y}{[c]^z}$ ……….(i)
 where $w,x,y,z$ are the powers of the to which these quantities are raised.
$
 [{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}] = {[AT]^w}{[L]^x}{[M{L^2}{T^{ - 1}}]^y}{[L{T^{ - 1}}]^z} \\
 [{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}] = [{M^y}{L^{x + 2y + z}}{T^{w - y - z}}{A^w}] \\
    \\
 $
Comparing both sides we get-
$
  w = 2 \\
  y = - 1 \\
 $
$
  x + 2y + z = - 2 \\
  w - y - z = 4 \\
 $
By solving above equations we get: $w = 2,x = 1,y = - 1,z = - 1$
Equation (i) now becomes –
$
  [\mu ] = {[e]^2}{[{a_0}]^1}{[h]^{ - 1}}{[c]^{ - 1}} \\
  \mu = \dfrac{{{e^2}{a_0}}}{{hc}} \\
 $
We have got the answer.

Hence , the correct option is (D).

Note: We have to keep in mind that while writing dimensional formula we need to write it only in terms of fundamental units and not derived units. This will not work if instead we write the derived units.