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If the capacitance of a nanocapacitor is measured in terms of a unit $'u'$ made by combining the electronic charge $'e'$ , Bohr radius'a₀', Plancks constant 'h' and speed of light 'c' then :
Option: 1
$u= \frac{e^{2}c}{ha_{0}}$

Option: 2
$u= \frac{e^{2}h}{ca_{0}}$

Option: 3
$u= \frac{e^{2}a_{0}}{hc}$

Option: 4
$u= \frac{hc}{e^{2}a_{0}}$

Answers (1)

best_answer

Unit of capacitance is u

Since E= $\frac{q^{2}}{2c}$

$\therefore c=\frac{q^{2}}{2E}$

Here q=e, E= $\frac{hc}{\lambda }or\frac{hc}{a_o}$

$\therefore capacitance\: can\:be\: represented$

as u= $\frac{e^{2}}{(hc/a_{o})}=\frac{a_{o}e^{2}}{hc}$

Posted by

jitender.kumar

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