If the capacitance of a nanocapacitor is measured in terms of a unit \(u\) made by combining the electric charge \(e,\) Bohr radius \(a_0,\) Planck's constant \(h\) and speed of light \(c\) then

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(D) \(u\, = \,\frac{{{e^2}{a_0}}}{{hc}}\)

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The unit of capacitance is \(\frac {Coulomb}{V} = \frac {Coulomb}{\text{work done per unit charge}}\)=\(\frac{\text { Coulomb }}{\frac{ W }{\text { Coulomb }}}=\frac{\text { Coulomb }^{2}}{ kgm ^{2} s ^{-2}}= C ^{2} kg ^{-1} m ^{-2} s ^{2}\)

Checking units of each option:

\(\mu=\frac{ e ^{2} c }{\text { ha }} \Rightarrow=\frac{\text { Coulomb }^{2} ms ^{-1}}{ kgm ^{2} s ^{-1} m }= C ^{2} kg ^{-1} m ^{-2} s ^{0}\)

\(\mu=\frac{ e ^{2} h }{ ca _{\circ}} \Rightarrow=\frac{\text { Coulomb }^{2} kgm ^{2} s ^{-1}}{ ms ^{-1} m }= C ^{2} kg ^{1} m ^{0} s ^{0}\)

\(\mu=\frac{\text { hc }}{ e ^{2} a _{\circ}} \Rightarrow=\frac{ kgm ^{2} s ^{-1} ms ^{-1}}{\text { Coulomb }^{2} m }= C ^{-2} kg ^{1} m ^{1} s ^{-2}\)

\(\mu=\frac{ e ^{2} a _{\circ}}{ hc } \Rightarrow=\frac{\text { Coulomb }^{2} m }{ kgm ^{2} s ^{-1} ms ^{-1}}= C ^{2} kg ^{-1} m ^{-2} s ^{-2}\)

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