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If the capacitance of a nanocapacitor is...

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=(e^(2)c)/(ha_(0))`

B

`u=(e^(2)h)/(ca_(0))`

C

`u=(e^(2)a_(0))/(hc)`

D

`u=(hc)/(e^(2)a_(0))`

Text Solution

Verified by Experts

The correct Answer is:
C
Here, capacitance `C=ke^(x)a_(0)^(y)h^(z)c^(a)`
`[C]=[M^(-1)L^(-2)A^(2)T^(4)]`
`[e]=[AT],[a_(0)]=[L]`
`[c]=[L^(1)T^(-1)],[h]=[M^(1)L^(2)T^(-1)]`
`:.[M^(-1)L^(-2)A^(2)T^(4)]=[AT]^(x)[L]^(y)[M^(1)L^(2)T^(-1)]^(z)[L^(1)T^(-1)]^(a)`
Comparing both sides
`x=2,z=-1,y+2z+a-2,x-z-a=4`
On solving these eqns. we get
x=2, y=1, z=-1, a=-1
Also, [C]=u,so `u=(e^(2)a_(a))/(hc)`
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