A potential difference of 3 V is applied across a conductor of resistance 1.5 Ω. The number of electrons flowing through it in one second would be: (Given charge of an electron, $e = 1.6 × 10^{-19} C$)

$1.25 × 10^{19}$

The correct answer is Option (3) → $1.25 × 10^{19}$

Using Ohm's law,

$I=\frac{V}{R}=\frac{3}{1.5}=2A$

and,

$I=\frac{Q}{t}$

$⇒Q=I.t=2C$

$∴n=\frac{Q}{e}=\frac{2}{1.6×10^{-19}}$

$=1.25×10^{19}$ electrons