# Homework Help: Mass of Atom in a Mass Spectrometer

1. May 9, 2010

### JonHO

1. The problem statement, all variables and given/known data
A mass spectrometer yields the following data for a beam of doubly ionized atoms of a single element.

B = 8.0*10^-2 T
q = 2(1.60*10^-19 C)
r = 0.077 m
V = 156 V

I am asked to solve for Mass (m) in kg.

2. Relevant equations
q/m=((2V)/((B^2)*(r^2)))

I am only partially sure that this is the correct equation. It is the equation for the "Charge-to-Mass Ratio of an Ion in a Mass Spectrometer".

3. The attempt at a solution
I tried to rearange the equation, but my Algebra skills aren't very good. This is what I ended up with:

m=q/(2V)/((B^2)*(r^2))

I tried to work this out with my calculator, but my answer was incorrect.

The number I got was very small: ~3.89*10^-28

EDIT:

Tried to use this instead: m=(q(B^2)(r^2))/(2V)

Still, my answer was far too small, at ~3.89*10^-26

My comprehension of both Physics and Algebra is limited at best. I would really appreciate some help. Thank you.

Last edited: May 9, 2010
2. May 10, 2010

### ehild

Show how do you calculate. And do not forget the units. What do you think, how much is the mass of an atom in kg-s?

$$\frac{q}{m}=\frac{2V}{r^2 B^2}$$

Plug in the numbers and simplify.

ehild

Last edited: May 10, 2010
3. May 10, 2010

### Jokerhelper

I was wondering how you guys obtained that formula.

4. May 10, 2010

### ehild

The ion is accelerated by an electric field. Flying through a potential difference V, it gains kinetic energy equal to qV

1/2 mv^2=qV.

It enters into the chamber of the mass spectrometer, where the magnetic field is perpendicular to the velocity. The magnetic force is

F=qvB.

This is a constant force, normal to the velocity, so the ion will move along a circle of radius R.
The centripetal force for the circular motion is equal to the force of the magnetic field:

(*) mv^2/R = qvB

but you know that mv^2= 2qV, so

2qV=RqvB --->v=2V/(RB).

(*) can be written as q/m=v/(RB). Plug in the expression for v.

ehild

5. May 10, 2010

### Jokerhelper

oh ok that makes sense. I hadn't thought of kinetic energy/work relationship as a way to substitute for speed. thanks!