Atomic Mass of Rare Carbon Isotope: Solving the Puzzle

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Homework Help Overview

The problem involves determining the atomic mass of a rare carbon isotope using data from a mass spectrometer. The context includes the comparison of two isotopes of carbon, focusing on their differing masses and the relationship between their movement in a magnetic field.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using a ratio based on the radii of the circles formed by the isotopes in the mass spectrometer. There is an exploration of the relationship between force, mass, and radius, with attempts to clarify the proportionality between radius and mass.

Discussion Status

Some participants have provided insights into the underlying physics principles, such as the relationship between radius and mass. However, there remains confusion regarding the final calculation and the implications of the proportionality established.

Contextual Notes

Participants are working under the assumption that the velocities and charges of the ions are equal, and that the magnetic field is constant, which influences their reasoning about the relationship between radius and mass.

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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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