Answer:
[tex]N = 1.2 \times 10^{23}[/tex]
Explanation:
As we know that the current through the battery is given as
[tex]i = (0.88 A) e^{-t/6}[/tex]
here from above equation we know that current will become zero when time elapsed is very large
so here we can say that charge will flow through the battery from t = 0 to t = infinite
now we have
[tex]q = \int i dt[/tex]
[tex]q = \int (0.88 A) e^{-t/6} dt[/tex]
[tex]q = 0.88\times -6(3600)\times (e^{-t/6})[/tex]
[tex]q = -1.90 \times 10^{4} (0 - 1)[/tex]
[tex]q = 1.90 \times 10^{4} C[/tex]
As we know that
[tex]q = Ne[/tex]
[tex]N = \frac{q}{e}[/tex]
[tex]N = \frac{1.90 \times 10^4}{1.6 \times 10^{-19}}[/tex]
[tex]N = 1.2 \times 10^{23}[/tex]
