Answer:
The maximum no. of electrons- [tex]2.25\times 10^{22}[/tex]
Solution:
As per the question:
Maximum rate of transfer of charge, I = 1.0 C/s
Time, t = 1.0 h = 3600 s
Rate of transfer of charge is current, I
Also,
[tex]I = \frac{Q}{t}[/tex]
Q = ne
where
n = no. of electrons
Q = charge in coulomb
I = current
Thus
Q = It
Thus the charge flow in 1. 0 h:
[tex]Q = 1.0\times 3600 = 3600\ C[/tex]
Maximum number of electrons, n is given by:
[tex]n = \frac{Q}{e}[/tex]
where
e = charge on an electron = [tex]1.6\times 10^{- 19}\ C[/tex]
Thus
[tex]n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}[/tex]
