Cyclotron with 1500 V between the two dees

[gsquare567]

Homework Statement

A cyclotron is designed to accelerate deuterium nuclei. (Deuterium has one proton and one neutron in its nucleus.)
[ANSWERED] (a) If the cyclotron uses a 2.0-T magnetic field, at what frequency should the dee voltage be alternated?
[ANSWERED] (b) If the vacuum chamber has a diameter of 0.90 m, what is the maximum kinetic energy of the deuterons?
(c) If the magnitude of the potential difference between the dees is 1500 V, how many orbits do the deuterons complete before achieving the energy from part (b)?

Homework Equations

frequency of particles in a cyclotron = (q * B) / (2 * PI * m), where q is the charge of the particle, B is the magnetic field of the cyclotron, and m is the mass of the particle

radius of a particle's circular motion in a cyclotron = (m * v) / (q * B)

The Attempt at a Solution

I have (what I think are) solutions for (a) and (b):
(a) f = qB/(2PIm) = eB/(2PIm) = 1.52 * 10^7 revolutions per second
(b) r = 0.45m = m * v-max / (qB)
v-max = 4.31 * 10^7 m/s
KE-max = 1/2mv-max^2 = 3.10 * 10^-12 J
(c)
...
I have no idea how to use the voltage to determine their speed or KE, all i know is:
electric potential = qV
but i don't know how, or if that helps.

 

[Delphi51]

The particle gains energy qV each time it crosses from one dee to the other. Would that be twice per orbit? Set n*2qV = KEmax from (b) and solve for n.

 

[gsquare567]

Thanks so much! I didn't know that it gains qV every time. I guess that's because the cyclotron is doing qV work on the particle when it is passed from one dee to the other?

 

[Delphi51]

Yes, you could work it out that way. Or use the definition of electric potential difference as the "energy per charge".

 

[rwisz]

Hello,

I would like to clarify that the units for frequency in a cyclotron are revolutions per second, not just revolutions. So the correct answer for part (a) would be 1.52 x 10^7 revolutions per second or Hz.

For part (b), your solution is correct. The maximum kinetic energy of the deuterons can be calculated using the formula KE = 1/2mv^2, where m is the mass of the deuteron and v is the maximum speed calculated in your solution.

For part (c), we can use the formula for the radius of a particle's circular motion in a cyclotron to determine the number of orbits completed by the deuterons before reaching the maximum kinetic energy. The formula is r = mv/qB, where r is the radius of the circular motion, m is the mass of the particle, v is the speed, q is the charge of the particle, and B is the magnetic field. We can rearrange this formula to solve for the number of orbits, which is given by n = (mv)/(qB^2). Plugging in the values given in the problem, we get n = (3.34 x 10^-27 kg * 4.31 x 10^7 m/s)/(1.6 x 10^-19 C * (2 T)^2) = 4.84 x 10^5 orbits. Therefore, the deuterons complete approximately 484,000 orbits before reaching the maximum kinetic energy calculated in part (b).

To use the voltage to determine the speed or KE, we can use the formula for the electric potential energy, which is given by PE = qV. We know the electric potential difference between the dees is 1500 V and the charge of a deuteron is 1.6 x 10^-19 C. Therefore, the potential energy gained by a deuteron in one orbit is PE = (1.6 x 10^-19 C)(1500 V) = 2.4 x 10^-16 J. This potential energy is converted to kinetic energy, so we can set it equal to the kinetic energy calculated in part (b) to solve for the speed v. This would give us v = sqrt(2PE/m) = 4.31 x 10^7 m/s, which is the same value calculated in part (b).

I hope