Atomic Mass of Rare Carbon Isotope: Solving the Puzzle

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SUMMARY

The discussion centers on calculating the atomic mass of a rare carbon isotope using mass spectrometry principles. The most abundant carbon isotope has an atomic mass of 12.0000 u, and the rarer isotope's radius in the mass spectrometer is 6.24 cm compared to 6.00 cm for the more abundant isotope. By applying the relationship between radius, mass, and charge in a magnetic field, specifically the equation r = mv/qB, participants confirm that the radius is directly proportional to mass, allowing for the calculation of the rarer isotope's atomic mass through a ratio method.

PREREQUISITES
  • Understanding of mass spectrometry principles
  • Familiarity with isotopes and atomic mass concepts
  • Knowledge of the relationship between force, mass, and radius in circular motion
  • Basic algebra for solving ratios and equations
NEXT STEPS
  • Explore the principles of mass spectrometry in detail
  • Learn about the calculation of atomic masses of isotopes
  • Study the effects of magnetic fields on charged particles
  • Investigate the derivation of the centripetal force equation in relation to mass spectrometry
USEFUL FOR

Students in chemistry or physics, educators teaching isotope behavior, and researchers involved in mass spectrometry applications will benefit from this discussion.

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Homework Statement


Natural carbon consists of two different isotopes (excluding 14C, which is present in only trace amounts). The isotopes have different masses, which is due to different numbers of neutrons in the nucleus; however, the number of protons in the same, and subsequently the chemical properties are the same. The most abundant isotope has an atomic mass of 12.0000 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 6.00 cm, while the rarer isotope moved in a circle of radius 6.24 cm. What is the atomic mass of the rarer isotope? (The ions are accelerated through the same potential difference before entering the magnetic field.)


Homework Equations





The Attempt at a Solution



would you just do a ratio of some sort i.e 6/6.24 =12/x??
 
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Well, yes that ratio ends up giving the correct answer but the key to why is finding the expression that gives the radius. We equate F= qvB to the centripetal force F = \frac{mv^2}{r}, and solve for r, giving r = \frac{mv}{qB}.

We can deduce their velocities and charges are the same from the given information, and may assume B is constant as well. Hence, r is directly proportional to m, enabling you to use that ratio formula.
 
Gib Z said:
Well, yes that ratio ends up giving the correct answer but the key to why is finding the expression that gives the radius. We equate F= qvB to the centripetal force F = \frac{mv^2}{r}, and solve for r, giving r = \frac{mv}{qB}.

We can deduce their velocities and charges are the same from the given information, and may assume B is constant as well. Hence, r is directly proportional to m, enabling you to use that ratio formula.

This logically all makes sense to me, but I am still confused as to how to get the final answer. Does that mean that r = m?
 
It doesn't mean r = m, we have a formula relating those two and there's more things ! It means r is *proportional* to m, meaning if there is an increase or decrease in r by a certain factor, the same will happen for m.
 

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