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= (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= (Q)/(e)= (2.21 \cdot 10^(-3)C)/(1.6 \cdot 10^(-19)C) =1.38 \cdot 10^(16)`