Answer:
The no. of electrons is [tex]7.59\times 10^{21}[/tex]
Solution:
According to the question:
The rate at which the charge is delivered is given by:
[tex]\frac{dQ}{dt} = - 0.75[/tex]
Now,
[tex]\int_{0}^{Q}dQ = - 0.75\int_{0}^{27 min} dt[/tex]
[tex]Q = -0.75t|_{0}^{27 min}[/tex]
[tex]Q= -0.75\times 27\times 60 = - 1215 C[/tex]
No. of electrons, n can be calculated from the following relation:
Q = ne
where
e = electronic charge =[tex]1.6\times 10^{- 19} C[/tex]
Thus
[tex]n = \frac{Q}{e}[/tex]
[tex]n= \frac{1215}{1.6\times 10^{- 19}}[/tex]
[tex]n = 7.59\times 10^{21}[/tex]
