Question

If the capacitance of a nano-capacitor is measured in terms of $'u'$ made by combining the electronic charge $'e'$, Bohr radius $'{{a}_{o}}'$, Planck's constant $'h'$ and speed of light $'c'$ then:
$A.u=\dfrac{{{e}^{2}}c}{h{{a}_{o}}}$
$B.u=\dfrac{hc}{{{e}^{2}}{{a}_{o}}}$
$C.u=\dfrac{{{e}^{2}}{{a}_{o}}}{hc}$
$D.u=\dfrac{{{e}^{2}}h}{c{{a}_{o}}}$

Answer 1

Hint: We have to apply the concept of capacitance and its unit to find the correct option. Capacitance can be defined as the ability of an object to store electric charge. SI unit of capacitance is farad but in this case we will find the unit in terms of the given parameters. We will use the concept of dimensional analysis to get the correct answer.
Formula used:
We will use the dimensional formula for the above mentioned parameters to get the correct answer. We are using the following formula:-
${{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{A}^{2}}=$${{\left[ {{A}^{1}}{{T}^{1}} \right]}^{a}}\times {{\left[ {{L}^{1}} \right]}^{b}}\times {{\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]}^{c}}\times {{\left[ L{{T}^{-1}} \right]}^{d}}$

Complete step by step solution:
We will analyse each and every parameters with their dimensional formula to get the correct answer. We will first write the dimensional formula of every parameter.
Dimensional formulae for different parameters are as follows:-
Capacitance,$u$ has dimensional formula of ${{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{A}^{2}}$.
For charge,$e$ it is ${{A}^{1}}{{T}^{1}}$.
For Bohr radius,${{a}_{o}}$ equals ${{L}^{1}}$.
For Plank’s constant,$h$ it is${{M}^{1}}{{L}^{2}}{{T}^{-1}}$.
For speed of light,$c$ it is ${{L}^{1}}{{T}^{-1}}$
Equating the dimensional formula of $u$ in terms of $e,{{a}_{o}},h$ and $c$ we get
${{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{A}^{2}}=$${{\left[ {{A}^{1}}{{T}^{1}} \right]}^{a}}\times {{\left[ {{L}^{1}} \right]}^{b}}\times {{\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]}^{c}}\times {{\left[ L{{T}^{-1}} \right]}^{d}}$…………..$(i)$
Where $a,b,c$ and $d$ are required values of exponents.
Now equating $(i)$ term wise we get
${{M}^{-1}}={{M}^{c}}$………………… $(ii)$
${{L}^{-2}}={{L}^{b+2c+d}}$……………… $(iii)$
${{T}^{4}}={{T}^{a-c-d}}$………………….. $(iv)$
${{A}^{2}}={{A}^{a}}$……………….. $(v)$
From the equation $(ii)$
$c=-1$…………. $(vi)$
From equation $(ii)$ and $(iii)$ we get
$-2=b+2c+d$
$-2=b-2+d$
$b=-d$…………….. $(vii)$
From the equation $(v)$ we get
$a=2$………………. $(viii)$
From$(iv),(vi)$ and $(viii)$ we get
$4=2+1-d$
$d=-1$…………. $(ix)$
From $(vii)$ we get
$b=1$……………… $(x)$
Putting these the values of $a,b,c$ and $d$ in $(i)$ we get
${{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{A}^{2}}=$${{\left[ {{A}^{1}}{{T}^{1}} \right]}^{2}}\times {{\left[ {{L}^{1}} \right]}^{1}}\times {{\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]}^{-1}}\times {{\left[ L{{T}^{-1}} \right]}^{-1}}$……………… $(xi)$
Putting the symbols in place of dimensions in equation $(xi)$ we get
$u=\dfrac{{{e}^{2}}{{a}_{o}}}{hc}$

Hence, option $(C)$ is correct among the given options.

Note:
We have to take care about the correct dimensional formula. We should use dimensional analysis for correct results. Our calculations should be accurate. We can also solve this problem by using units in terms of their mathematical formulae in place of dimensional units.