Question

A proton beam in an accelerator carries a current of 130 μa. if the beam is incident on a target, how many protons strike the target in a period of 17.0 s?

Answer 1

The current intensity is the product between the total charge that flows through a certain point (in our case, the target) in a time interval $\delta t$:
$I= \frac{Q}{\Delta t}$
We know the current, $I=130 \mu A=130 \cdot 10^{-6} A$, and the time interval, $\Delta t=17 s$, so we can find the total charge:
$Q=I \Delta t= 2.21 \cdot 10^{-3}C$

The total charge Q is the product between the number of protons N and the charge of each protons, e, which is $e=1.6 \cdot 10^{-19}C$:
$Q=Ne$
we can  re-write the equation solving for N, so we can find the number of protons striking the target in 17 s:
$N= \frac{Q}{e}= \frac{2.21 \cdot 10^{-3}C}{1.6 \cdot 10^{-19}C} =1.38 \cdot 10^{16}$