Steven W. answered • 09/28/16

Tutor

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Physics Ph.D., college instructor (calc- and algebra-based)

Hi Prashant!

The formula for the radius (r) of a circle traced out by a charged particle in a uniform magnetic field turns out to be:

r = mv/qB

where

m = mass of particle

v = speed of particle

q = (electric) charge of particle

B = strength of magnetic field

(we can derive this if you need to, but, for now, I will take it as a given)

If we rearrange this to put v on one side of the equation by itself, it becomes:

v = qBr/m

I take the energy of the particle to mean its kinetic energy (KE)

KE = (1/2)mv

^{2}Putting in the expression for v we got above, for a particle moving in a circle, this becomes:

KE = (1/2)m(qBr/m)

^{2}--> q^{2}B^{2}r^{2}/2mSo this is an expression for kinetic energy of a particle of mass m and charge q moving in a circular path of radius r in magnetic field B.

The charge and the mass of a particle are important factors in this energy expression. For the two particles in the problems, we can compare those quantities qualitatively (without numbers) by a knowledge of what an alpha particle is:

proton: mass = m, charge = e

alpha: mass = 4m, charge = 2e

This is because an alpha particle is a helium nucleus, containing two protons (each with mass m and charge e) and two neutrons (each with, effectively, mass m, and no electric charge).

For the proton, we can then write the KE expression for the circular path as:

KE

_{p}= e^{2}B^{2}r^{2}/2mFor the alpha particle, as I read the problem, we assume that the magnetic field B and path radius r are the same as for the proton. The kinetic energy for that particle then becomes:

KE

_{a}= (2e)^{2}B^{2}r^{2}/2(4m) = 4e^{2}B^{2}r^{2}/8m = e^{2}B^{2}r^{2}/2mThis is exactly the same expression as for the proton. So the alpha particle turns out to have the same kinetic energy, 8 eV, in this case. This comes from the specific circumstance of having increased the mass by a factor of 4, but the charge only by a factor of 2.

I hope this helps! If you want to discuss any more about this, or look at the details, just let me know.