Question

If the capacitance of a nanocapacitor is measured in terms of a unit $'u'$ made by combining the electronic charge $V$, Bohr radius $'a_0'$, Planck's constant $'h'$ and speed of light $'c'$ then :

• A. $u = \frac{e^{2}c}{ha_{0}}$
• B. $u = \frac{e^{2}h}{ca_{0}}$
• C. $u = \frac{e^{2}a_{0}}{hc}$
• D. $u = \frac{hc}{e^{2}a_{0}}$

Answer 1

Answer: (C)

$u = \frac{e^{2}a_{0}}{hc}$

Answer 2

$C=\frac{q}{\Delta V} ; E=h f=\frac{h c}{\lambda}$
Working in dimensions
$[C]=\left[\frac{q}{\Delta V}\right]\left[\frac{q}{q}\right]=\left[\frac{q^{2}}{q \Delta V}\right]=\left[\frac{q^{2}}{E}\right]=\left[\frac{q^{2} a_{0}}{h c}\right]$