If the capacitance of a nanocapacitor is measured in terms of a unit '‘u'’ made by combining the electronic charge ‘e’, Bohr radius ‘$a_0$’, Planck’s constant ‘h’ and speed of light ‘c’ then :

Question

$\begin{array}{1 1} u= \frac{e^2 c}{ha_0} \\ u= \frac{e^2 h}{ca_0} \\u= \frac{e^2 a_0}{hc} \\ u= \frac{hc}{e^2a_0} \end{array} $