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pc2-brazil
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Homework Statement
A deuteron moves through the magnetic field of a cyclotron, in an orbit of 50 cm radius. Due to the slight collision with a target, the deuteron divides into a proton and a neutron, with a negligible loss of kinetic energy. Discuss the subsequent movements of each particle. Suppose that, in the break, the energy of the deuteron is evenly divided between the proton and the neutron.
Homework Equations
Radius of an orbit of a particle of charge |q| inside a magnetic field B (the velocity is perpendicular to the vector B):
r = \frac{mv}{|q|B}
The Attempt at a Solution
The proton has charge +e and the neutron is neutral. If the mass of the deuteron is 2m, then its kinetic energy is 2mv²/2 = mv². The mass of the proton and the neutron is m (half the mass of the deuteron). The kinetic energy of each particle (proton and neutron) after the collision will be half the energy of the original particle; thus, it will be mv²/2. So, the velocity of the neutron and the proton will be v.
Then, if the radius of the deuteron's orbit is r, the proton will follow an orbit of radius r/2 and the neutron will follow a straight line, since it has no charge.
If the division of the deuteron happens in a very small time interval, I think that the linear momentum can be considered approximately constant; then, I would be able to apply linear momentum conservation. If I call m the mass of the proton, the initial momentum is 2mv and the final momentum of each particle is mv.
So, the velocities of the neutron and the proton will have the same direction, so that the magnitude of the sum of the linear momenta of the two particles is 2mv (equal to the initial linear momentum).
Is this correct?
Thank you in advance.