Quantum Physics Problem

1. May 1, 2010

scoldham

1. The problem statement, all variables and given/known data

A particle of charge q and a mass m, moving with a constant speed v, perpendicular to a constant magnetic field B, follows a circular path. If in this case the angular momentum about the center of this circle is quantized so that $$mvr_n = 2nh$$, determine the allowed radii for the particle in terms of n, h, q, and B for n = 1,2,3,....

2. Relevant equations

$$F = qvBsin\vartheta$$

3. The attempt at a solution

As far as I can tell, this has something to do with relating magnetism to the quantum level. It is easy enough to calculate the radius at a given energy level by solving for $$r_n$$. But I do not understand how to relate the charge and the B field to the situation. The best I can come up with is the formula provided... I feel like there is some way it ties into the problem. Help greatly appreciated.

2. May 1, 2010

kuruman

Use the relevant equation you have provided to write Newton's Second Law, F = ma. What is the acceleration for circular motion?

3. May 2, 2010

scoldham

$$\frac{mv^2}{r_n} = qVB sin \vartheta$$

$$sin \vartheta = 0$$ as the angle of the particle with the B field is 90 degrees.

So, simplifying I get,

$$r_n = \frac{mv}{qB}$$

How do I tie in this equation with the above?

Last edited: May 2, 2010
4. May 2, 2010

thebigstar25

try to use your original equation mvr = 2nh again in the last equation to get red of mv ..

5. May 2, 2010

scoldham

I think I see it now.

$$mv = \frac{2nh}{r_n}$$

Subbing mv into equation from above $$r_n = \frac{mv}{qB}$$

I get

$$r_n = \frac{2nh}{r_nqB}$$

A bit more simplification yields $$r_n = \sqrt{\frac{2nh}{qB}}$$

Is that correct?

6. May 2, 2010

thebigstar25

well, it seems correct to me since you achieved what is required in the question which was asking to write r in terms of n, h, q, and B ..