A cyclotron produces a beam of protons that have 60 MeV of energy, The protons c

Question

A cyclotron produces a beam of protons that have 60 MeV of energy, The protons collide with a 50 g copper disc in a vacuum chamber and come to rest. If the target gains 3.5 10^-6 Coulombs of charge per second from collisions, how many protons per second is the cyclotron emitting? At what rate is energy being deposited into the copper target? Use the result from and Q = mc Delta T to calculate how long it will take for the target's temperature to increase by 300 degrees Celsius. Q is measured in Joules, Delta T represent the change in temperature, m is the mass of the target and C is the specific heat of copper (c = 0.385 J/g degree C). Ignore radiative losses as the target heats.

Explanation / Answer

a)

Charge increase rate , Q/t = 3.5*10^-6 C/s

So, number of protons per second , n = 3.5*10^-6/(1.6*10^-19)

= 2.19*10^13 <-------answer

b)

So, rate of energy deposition:

dU/dt = n*60*10^6*1.6*10^-19

= 2.19*10^13*60*10^6*1.6*10^-19

= 210.2 W <-----answer

c)

Q = m*C*T

So, Power , P = Q/t

So, P = m*C*T/t = 210.2

So, 50*0.385*300/t = 210.2

So, t = 27.5 s <------answer

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