Answer:
[tex]4.31\times10^{15}particles[/tex]
Explanation:
Current (I) is the flow of charge (Q) per unit of time (t). The ampere unit is defined as the amount of charge that flows in one second.
[tex]Current=I=\dfrac{Q}{t}[/tex]
[tex]1\text{ }ampere=1\dfrac{coulomb}{second}[/tex]
You have the current, 138 µA, and the time, 25.0s, thus you can calculate the charge:
[tex]Q=I\times t=138\mu A\times \dfrac{10^{-6}A}{\mu A}\times 25.0s=0.00345C[/tex]
Now calculate the number of particles with a charge of +5eV that have a charge of 0.00345C
The charge of one electron is [tex]1.602\times 10^{-19}C[/tex]
Then, a positve charge equivalent to the magnitude of the charge of 5 electrons (one particle) is:
[tex]5\times 1.602\times 10^{-19}C=8.010\times 10^{-19}C[/tex]
Divide the total charge by the charge of a particle:
[tex]\dfrac{0.00345C}{8.010\times 10^{-19}C}=4.3071\times 10^{15}[/tex]
Rounding to 3 significant figures that is [tex]4.31\times 10^{15}particles[/tex]
