Question

If the capacitance of a nano-capacitor is measured in terms of a unit ${}^{\prime }{u}^{\prime }$ made by combining the electronic charge ${}^{\prime }{e}^{\prime }$, Bohr radius ${}^{\prime }{a}_0^{\prime }$, Plank's constant ${}^{\prime }{h}^{\prime }$ and speed of light ${}^{\prime }{c}^{\prime }$ then:

• A. $u=\frac{e^{2}c}{h{a}_0}$
• B. $u=\frac{hc}{e^2{a}_0}$
• C. $u=\frac{e^2{a}_0}{hc}$
• D. $u=\frac{e^{2}h}{c{a}_0}$

Answer 1

Answer: (C)

$u=\frac{e^2{a}_0}{hc}$

$\text{Solution}$

Let unit u related with e, $a_0$, h and c as follows.

$\left[u\right]={\left[e\right]}^a{\left[a_0\right]}^b{\left[h\right]}^c{\left[C\right]}^d$

Using dimensional method,

$\left[M^{-1}{L}^{-2}{T}^{+4}{A}^{+2}\right]={\left[A^1{T}^1\right]}^a{\left[L\right]}^b{\left[ML2T^{-1}\right]}^c{\left[L{T}^{-1}\right]}^d$

$\left[M^{-1}{L}^{-2}{T}^{+4}{A}^{+2}\right]=\left[M^c{L}^{b+2c+d}{T}^{a-c-d}{A}^a\right]$

a = 2, b=1, c = -1, d = -1

$\therefore u=\frac{e^2{a}_0}{hc}$

$\text{Hence C is the correct option}$