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

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, u = e 2 a 0 h c, u = h c e 2 a 0, u = e 2 c h a 0, u = e 2 h c a 0

MARKS App · Accepted Answer

∵ Capacitance C = Q ∆ V Also , h c λ = h c a 0 = Energy . ∴ C = Q ∆ V = Q Q ∆ V Q ∵ W = q ∆ V ⇒ Q ∆ V = Energy ∴ C = Q 2 Energy = Q 2 a 0 h c ∴ Capacitance = Q 2 a 0 h c ⇒ u = e 2 a 0 h c

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