Question

A beam of 16 MeV deuterons from a cyclotron strikes a copper block. The beam is equivalent to current of 15 μA. (a) At what rate do deuterons strike the block? (b) At what rate is thermal energy produced in the block

Answer 1

(a) 9.4 X 10¹³s⁻¹

(b) 240 W

Step-by-step explanation:

Given-

Current, i = 15μA

Electrons, e = 16 MeV

(a) The rate at which the deuterons strike the block, r = ?

We know,

`i = (dq)/(dt) = e(dN)/(dt)`

`(dN)/(dt) = (i)/(e) \\\\(dN)/(dt) = (15 X 10^-^6 A)/(1.6 X 10^-^1^9C)`

`(dN)/(dt) = 9.4 X 10^1^3 s^-^1`

Therefore, the rate at which the deuterons strike the block is 9.4 X 10¹³s⁻¹

(b) Rate at which thermal energy is produced in the block

Now that we have
`(d_(N) )/(d_(t) )`, we can use it with the equation of power to solve for the thermal energy production rate.

`P = (dN)/(dt) U\\\\P = (9.4 X 10^1^3 s^-^1) (16MeV) ((1.6 X 10^-^1^3 J)/(1MeV) )\\\\P = 240W`

Therefore, the rate at which thermal energy is produced in the block is 240 W.