Question

Question 4
A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly

(a) ${10}^{20}$
(b) ${10}^{16}$​​​​​​​
(c) ${10}^{18}$​​​​​​​
(d) ${10}^{23}$​​​​​​​

Answer 1

(a) ${10}^{20}$
​​​​​​​
$1A=1C{s}^{-1}=6.24\times {10}^{18}$ electrons per second
In 16 seconds, number of electrons = $16\times 6.24\times {10}^{18}=99.84\times {10}^{18}=100\times {10}^{18}={10}^{20}$ (almost)