Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate H- (the hydride ion, which has one proton and two electrons) to an energy of 5 MeV to 20 MeV. This ion has a mass very close to that of a proton because the electron mass is negligible—about 1/2000 of the proton's mass. A typical magnetic field in such cyclotrons is 1.9 T. (a) What is the speed of a 5.0-MeV H-? (b) If the H- has energy 5.0 MeV and B = 1.9 T, what is the radius of this ion's circular orbit?

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Accepted Answer

Hey, everyone. So this problem is working with magnetic fields and charged particles, let's see what they are giving us and what they're asking from us. So we have a mass spec that's generating charged particles with a given mass and a given charge, the particles are accelerated through an electric field and gain an energy that's also given. And then the particles pass through a uniform magnetic field of a given magnitude, find the speed and the radius of the charged particles. So it is a two part question, but you'll see as we work through this, that both parts are very straightforward. So the first thing we can look at for part one when we were asked for the speed and were given the energy is to recall that the kinetic energy of a charged particle is just given as K equals one half M V squared. So we are solving for V R speed, we can rearrange that equation to B two K over M. And then looking at our, what they give us in the problem at our givens our energy, our kinetic energy is given as 0.4 and that's mega electron volts. So I'm going to rewrite that as times 10 to the sixth electron volts over our mass of 3.2 times 10 to the negative 25 kg. Now, the one tricky part here is to recognize that we still need to do some work with our units because we need to get our energy in terms of jewels. So we can recall that the conversion factor between electron volts and jewels is 1.6 times 10 to the negative 19 jewels per one electron volt. And then that cancels out. And then we are good to solve that out. And that equals 6.32 times to the fifth meters per second. So that is the answer for part one. And when we look at our potential choices off the bat, we can eliminate C and D. So for part two, they're asking for the radius of the circular path of the charged particles. And we can recall that the radius of charged particles in a uniform magnetic field is given by our equals M V over the absolute value of Q, the charge times the magnitude of the magnetic field. And again, as we look at each of these terms, we can see that we have everything we need to solve for radius, very straightforward. So the mass was already given to us in the problem. We recognize that that's 3.2 times 10 to the negative kg the speed, the velocity we just saw for So 6.3, 2 times 10 to the fifth, our charge Q. And the problem was given to us as plus two E. And so what we need to recall there is that E is a given Constant and that is 1. 02 times 10 to the -19 Coghlan's. And then our magnetic field B was also given to us in the problem as one Tesla. So we can plug all of that in to our equation. And we will have what we need to find the radius. So 3.2 times 10 to the minus 25 kg is our mass times velocity 6. times 10 to 5 m per second. Over our Charge two times 1.602 times 10 to the - columns times are magnitude of our magnetic field. One tesla plug that into our calculators and we are left with Radius of 6.631 m. So that is the answer to part two. So we go back up to our choices and we see that, that aligns with choice B. So B is the correct answer for this problem. And that's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Kinetic Energy and Relativistic Effects

Magnetic Force and Circular Motion

Mass-Energy Equivalence