If the capacitance of a nanocapacitor is measured in terms of a unit u , made by combining the electronic…

If the capacitance of a nanocapacitor is measured in terms of a unit u,  made by combining the electronic charge e, Bohr radius a0, Planck's constant h and speed of light c then
  1. u=e2a0hc
  2. u=hce2a0
  3. u=e2cha0
  4. u=e2hca0

Solution

∵     Capacitance C=Q∆V 

Also, hcλ= hca0= Energy. 

∴     C= Q∆V= Q Q∆V Q

∵     W=q∆V    ⇒     Q ∆V= Energy

∴     C= Q2Energy= Q2 a0hc

∴     Capacitance= Q2 a0hc

⇒     u= e2 a0hc

Asked in: JEE Main 2015 (10 Apr Online)

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