Question

How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of electrons from an initial point where the electric potential is 6.00 V to a point where the potential is -5.00 V

Answer 1

Answer:

$1.06\times 10^6 J$

Explanation:

We are given that

$V_1=6 V$

$V_2=-5 V$

Charge on 1 electron, 1 e=$-1.6\times 10^{-19} C$

Avogadro's number, $N_A=6.02\times 10^{23}$

$q=ne$

$q=6.02\times 10^{23}\times 1.6\times 10^{-19}=-9.632\times 10^4 C$

$W=q\Delta V$

$W=q(V_2-V_1)$

$W=-9.632\times 10^4\times (-5-6)=1.06\times 10^6 J$

Hence, the work done=$1.06\times 10^6 J$