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

  • A
    $u = \frac{e^2 h}{a_0}$
  • B
    $u = \frac{hc}{e^2 a_0}$
  • C
    $u = \frac{e^2 c}{h a_0}$
  • D
    $u = \frac{e^2 a_0}{hc}$

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