What is the drift velocity of electrons in a silver wire of length `1m` having cross-section area `3.14 xx 10^(-6)m^(2)` and carriying a current of `10A`. Given atoms weight of weight of silver `= 108` density of silver `10.5 xx 10^(3) kg//m^(3)`, charge of electron `1.6 xx 10^(-19) C`, Avogadro's number `6.023 xx 10^(26)` per kg.atom

Explanation

Use the formula `l = "ne" v_(d) A`
First calculate n = no of electrons per unit volume , Iamgine the volume of silver to be `1 m^(3)`
Now its mass = density `xx` volume `= 10.5 xx 10^(3) kg `
Now, number of moles of silver ` = (10.5 xx 10^(3) kg)/(108 xx 10^(3) kg //mol)`
So, no of silver atoms in this will be got by mulitplying the number of moles by Avogadro.s number i.e., no of silver atoms = `(10.5 xx 10^(3) xx 6.023 xx 10^(23))/(108 xx 10^(-3))`
Now, since the valency of silver is one , we can assume each atom of silver contributes can electron. So finally
` n = (10.5 xx 10^(3) xx6.023 xx 10^(23))/(108 xx 10^(-3)) = 5.86 xx 10^(24) per m^(3)` ( give l = 10 A and `A= 3.14 xx 10^(-6) m^(2)` )
So, we have `v_(d) = (l)/(n.e.A)`
` = (10)/(5.86 xx 10^(29) xx 1.602 xx 10^(-19) xx 3.14 xx 10^(-6)) = 3.39 xx 10^(-4)` m/s.