Atomic Mass in a Mass Spectrometer

[futurepocket]

Homework Statement

Natural carbon consists of two different isotopes. The isotopes have different masses, which is due to different numbers of neutrons in the nucleus; however, the number of protons is the same, and subsequently, the chemical properties are the same. The most abundant isotope has an atomic mass of 12.00 u. When natural carbon is placed in a mass spectrometer, two lines are formed on the photographic plate. The lines show that the more abundant isotope moved in a circle of radius 15.0 cm, while the rarer isotope moved in a circle of radius 15.6 cm. What is the atomic mass of the rarer isotope? (The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field)

Homework Equations

a(r) = v^2/r = F/m = qvB/m

The Attempt at a Solution

The magnetic field and the charge are constant. I realize that velocity is NOT constant since it passes through the same potential difference (but the two isotopes have two different masses). After some trial and error, I found the ratio:

r2/r1 = m2/m1

Which doesn't work because I get an answer of 12.48 while the answer is 13.0 u. So, basically I have to consider that the velocity matters as well, so you get:

r2/r1 = m2v2 / m1v1

But if r2/r1 is proportional to m2/m1, is it also proportional to v2/v1? I get the right answer if I do the following:

m2 = r2/r1 x m1 x r2/r1
m2 = 2(r2/r1) x m1

I consider the second r2/r1 is because of the velocity, but I don't have an explanation for the question. If someone could explain why this is done (if its the correct method), or explain the correct method if it is not, it would be appreciated.

Thanks!

 

[Dadface]

A charge q on a mass m accelerated through a pd V(in a vacuum) gains KE given by:
qV=mv squared/2.....(v= speed)
For both isotopes q is the same and V is the same.

 

[futurepocket]

I thought so too, but the textbook states that if the mass is different of the two charges, than, when they pass through a potential difference, the velocity is not the same. If I follow the logic you provided, I end up with the ratio:

r2/r1 = m2/m1
m2 = r2/r1 x m1
m2 = 12.48 u

When the actual answer is m2 = 13 u. I know they didn't round off to 13 u, so I suppose velocity has something to do with it? If not, what am I doing wrong?

 

[Dadface]

Yes,the velocities are not the same.From the equation above
m1v1 squared=m2v2 squared(m1 and m2are the two masses corresponding respectively to velocities v1 and v2).
On entering the magnetic field each isotope is deflected into a circular path of radius given by:
Bqv=mv squared/r therefore r=mv/Bq

 

[gneill]

You wrote:

v²/r = F/m = qvB/m

So

mv² = rqvB

E/2 = rqvB

Where E is the energy that the atom was given passing through the accelerator stage. Assuming its constant for all particles with the same charge, you can also replace the remaining v in terms of energy (it will depend upon the mass). Everything but m and r should be constants in what remains.

 

[futurepocket]

Okay, so I am still totally confused. I know r = mv / Bq, the magnetic field and the charge are constant but the mass and velocity are NOT constant, but we aren't given the velocity. So if I set up a ratio:

r2/r1 = m2v2/m1v1

and isolated for m2

(r2/r1 x m1v1) / v2 = m2
r2m1v1/r1v2 = m2

Do I just take the ratio of v1/v2 as the ratio of r1/r2 since they're proportional? Sorry for the misunderstanding, but I just can't seem to figure it out.

Also, if I express the other v in terms of E, it stays constant but I get it under a square root.

 

[futurepocket]

Actually, I think I just figured it out:

If I express them in terms of E and energy stays constant for both:

r2 / r1 = m2v2 / m1v1 = m2sqroot(2E/m2) / m1sqroot(2E/m1)
r2 / r1 x sqroot(m1) = sqroot(m2)
m2 = (r2 / r1 x sqroot(m1))^2
m2 = 13 u

Is that correct?

 

[Dadface]

Thats what I made it.I was surprised because I thought the other isotope would be carbon 14 and not carbon 13.

 

[futurepocket]

Great, thanks for all the help guys :)