# Homework Help: Measuring the mass of tritium

1. Jan 16, 2012

### bobsmith76

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

An isotope of hydrogen passes, without deflection, through a velocity selector that has an electric field of 2.40 × 10^5 N/C and a magnetic field of 0.400 T. It then enters a mass spectrometer that has an applied magnetic field of 0.494 T and consequently describes a circular path with a radius of 3.80 cm.
(a) What is the mass of the particle?

2. Relevant equations

v (velocity) = E/B

where E = electric field intensity
B = magnetic field intensity

.5mv^2 = qV

V = kq/r

3. The attempt at a solution

step 1. find the velocity

v = E/B

(2.4 * 10^5)/.4 = 6 * 10^5

step 2. find voltage

I tried to find q and V, then find the mass but I couldn't so I'm working backwards, starting with the answer and trying to find V

.5mv^2 = qV

.5* (5.01*10^-27)*(6*10^5)^2 = 1.6 * 10^-19 * V

step 3 get V by itself

(.5* (5.01*10^-27)*(6*10^5)^2)/(1.6 * 10^-19) = V

which is

1.44 * 10^20

step 4 look for an equation which will give me V (electric potential difference)

V = kq/r

((9*10^9)(1.6*10^-19)/.238 =

I forgot what that number is, but it did not add up to 1.44 * 10^20, so there's something I'm doing wrong. Let's try a different equation.

E (electric field intensity) = ΔV/Δd

step 5 E * d = V

The E with a magnetic field intensity of .494 is 2.9*10^5 so

2.9*10^5 * .238 =

I also forget what that number is but it also does not add up to around 10^20

2. Jan 16, 2012

### bobsmith76

Hold on, I redid my calculation and found the volts to be 5625

3. Jan 16, 2012

### bobsmith76

Well, I'm also there

step 5 E * d = V

The E with a magnetic field intensity of .494 is 2.9*10^5 so

2.9*10^5 * .238 = 69020, not that far away from 5625

4. Jan 16, 2012

### bobsmith76

Any ideas on this one?

5. Jan 16, 2012

### technician

you have the velocity of the isotope = 6 x 10^5 m/s as it enters the magnetic field.
The force on the isotope due to the magnetic field is Bqv where q will be 1.6 x 10^-19 C
this force equals the centripetal force = (mv^2)/r
equating Bqv = (mv^2)/r gives Bq =mv/r
if you substitute into this equation you should get m = 5 x 10^-27kg

6. Jan 16, 2012

### bobsmith76

The force on the isotope due to the magnetic field is Bqv where q will be 1.6 x 10^-19 C

But given the information in the question, how do I know that the charge of the isotope will be as you say? If it's because a hydrogen isotope's by definition has one charge then that will be my answer but i'm not sure that's true.

7. Jan 16, 2012

### technician

You are correct. Hydrogen has only 1 electron to lose and therefore must be singly + charged.