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 :
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 :
Updated On: Jul 22, 2024
- $u = \frac{e^{2}c}{ha_{0}}$
- $u = \frac{e^{2}h}{ca_{0}}$
- $u = \frac{e^{2}a_{0}}{hc}$
- $u = \frac{hc}{e^{2}a_{0}}$
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The Correct Option is C
Solution and Explanation
$C=\frac{q}{\Delta V} ; E=h f=\frac{h c}{\lambda}$
Working in dimensions
$[C]=\left[\frac{q}{\Delta V}\right]\left[\frac{q}{q}\right]=\left[\frac{q^{2}}{q \Delta V}\right]=\left[\frac{q^{2}}{E}\right]=\left[\frac{q^{2} a_{0}}{h c}\right]$
Working in dimensions
$[C]=\left[\frac{q}{\Delta V}\right]\left[\frac{q}{q}\right]=\left[\frac{q^{2}}{q \Delta V}\right]=\left[\frac{q^{2}}{E}\right]=\left[\frac{q^{2} a_{0}}{h c}\right]$
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