JEE Main2015PhysicsUnits and DimensionsActual
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
Options
- Au = e 2 a 0 h c
- Bu = h c e 2 a 0
- Cu = e 2 c h a 0
- Du = e 2 h c a 0
Correct answer
A. u = e 2 a 0 h c
Step-by-step solution
∵           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 ∴