Question

Show that one ampere is equivalent to flow of `6.25xx10^(-18)` elementary electrons per second ? Charge on electron `=1.6 xx 10^(-19)C`.

Answer 1

Here, `I=1 A, t= 1s, e = 1.6xx10^(-19)C`
`I= ("ne")/(t)` or `n= (It)/(e) = (1 xx 1)/(1.6xx10^(-19))=6.25 xx 10^(18)`