How Do You Calculate the Mass of Tritium in a Mass Spectrometer?

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Homework Help Overview

The problem involves calculating the mass of the hydrogen isotope tritium as it passes through a mass spectrometer, utilizing electric and magnetic fields. The context includes the application of concepts from electromagnetism and circular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the velocity of the isotope using the ratio of electric field to magnetic field. There are attempts to derive voltage and mass through various equations, with some participants expressing uncertainty about their calculations. Questions arise regarding the charge of the isotope and its implications for the calculations.

Discussion Status

Some participants have provided calculations and insights into the relationships between velocity, charge, and mass. There is ongoing exploration of the assumptions regarding the charge of the hydrogen isotope, with a few participants questioning the validity of their approaches and results.

Contextual Notes

Participants are working under the constraints of the problem statement and the definitions of physical quantities involved, such as charge and mass. There is a noted lack of consensus on certain values and assumptions, particularly regarding the charge of the isotope.

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Homework Statement



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?

Homework Equations



v (velocity) = E/B

where E = electric field intensity
B = magnetic field intensity

.5mv^2 = qV

V = kq/r


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
 
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Hold on, I redid my calculation and found the volts to be 5625
 
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
 
Any ideas on this one?
 
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
 
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.
 
You are correct. Hydrogen has only 1 electron to lose and therefore must be singly + charged.
 

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