Respuesta :
Answer:
e. 5.6 x 10^21 electrons.
Explanation:
In order to calculate the charge Q passing through the resistor for in a definite amount of time is
[tex]Q \ (\mathrm{coulombs}) \ = \ I \ (\mathrm{amperes}) \ \times \ t \ (\mathrm{seconds}) \\ \\ \-\hspace{2.22cm} = \ 3.0 \ \mathrm{A} \ \times \ 5.0 \ \mathrm{min} \ \times \ \displaystyle\frac{60 \ \mathrm{sec}}{1 \ \mathrm{min}} \\ \\ \-\hspace{2.22cm} = \ 900 \ \mathrm{C}[/tex]
Thus, using the law of quantization of electric charge, the number of electrons passing through the resistor during this time interval can be calculated.
[tex]Q \ = \ n \times e \\ \\ n \-\hspace{0.18cm} = \ \displaystyle\frac{Q}{e} \\ \\ n \-\hspace{0.18cm} = \displaystyle\frac{900 \ \mathrm{C}}{1.6 \times 10^{-19} \ \mathrm{C}} \\ \\ n \-\hspace{0.18cm} = 5.6 \times 10^{21} \ \ \mathrm{electrons}[/tex], where n denotes the number of electrons and e is the unit charge of an electron ([tex]1.6 \times 10^{-19} \ \mathrm{C}[/tex]).
