# Mass Spec

1. Apr 6, 2004

### kashmirekat

I have two molecules CO and N2 with masses 28.0106u and 28.0134u respectively.

I need to determine the radius of curvature the spectrometer must have if the molecules are to be separated on film by 0.50mm.

I don't even know where to begin to solve this. The only equation relating to it in the book is m=qB'r/v, where r=mv/qB'. But I don't have v or q or B'. Agghhh.

Mco=28.0106u
Mn2=28.0134u
d=0.50mm
r=?

Would someone point me in the right direction? Thank you.

Last edited: Apr 7, 2004
2. Apr 6, 2004

### jdavel

kashmirekat,

Since the molecules have mass, Newton's 1st law says they'll travel at a constant speed, v, and in a constant direction unless there's a force applied to them. The force is from the magnetic field B. It's magnitude is given by qvB, and its direction is always perpendicular to v. This means the molecules will travel in a circle.

But how big is the circle? The acceleration of an object going in a circle turns out to be v^2/r. This makes sense because the smaller r is, the faster the particle is has to be changing is direction (hich is one form of acceleration) and the faster its going the faster it's making that turn.

Now we apply Newton's 2nd law that says F = ma. So write this equation for each of your molecules, and see if that helps you understand what's going on.

Also, you might check your book where it talks about mass spectrometers, but usually the way they work is to send all the molecules through a little device called a velocity selector so that everything coming out is going at the same speed. You can probably assume this even if your book doesn't say it (because I think you have to!).

Try messing around with all that for a while, and see what you come up with.

3. Apr 6, 2004

### kashmirekat

Okay, I understand that their is going to be a circular path due to the magnetic force (E field) and that since the masses of the molecules are nearly the same, their velocity is going to be the same too. However, I do not understand where the .5mm is integrated into the equation.

I'm sorry for not understanding, it just hasn't clicked yet as to how to work out this problem. So frustrating....

4. Apr 7, 2004

### Staff: Mentor

The circular paths have different radii, thus they will hit the detector at different positions. (Think of the particles as swinging around in a semicircle: distance from initial position = diameter of circle.) The position on the detector is a function of the mass. The question is: What must the radius be (of the larger path say) so that the difference in where they hit the detector is 0.5mm?

5. Apr 7, 2004

### kashmirekat

The answer to the problem involved taking the ratio of masses to the ratio of r. m/[delta]m=[delta]r/r . Where [delta]r = .5mm