Why Multiply Carbon Isotope Mass Numbers by 1.67x10^-27 kg?

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Homework Help Overview

The discussion revolves around the use of carbon isotopes with mass numbers 12, 13, and 14 in the context of a mass spectrometry problem. Participants are examining the necessity of converting atomic mass units to kilograms using a provided conversion factor of 1.67x10^-27 kg.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning whether the mass numbers provided are already in atomic mass units and if the conversion to kilograms is necessary. There is also discussion about the implications of using different units for calculations in the context of the mass spectrometry equation.

Discussion Status

Some participants have provided insights regarding the need for unit conversion and the relevance of the conversion factor. There is an ongoing exploration of the implications of using kilograms versus grams in the calculations, with no explicit consensus reached yet.

Contextual Notes

The original poster indicates a lack of complete information regarding the overall problem context, which may affect the clarity of the unit conversion necessity. The discussion also touches on historical references to atomic mass units and their approximations.

kashmirekat
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Here's a verbatim section of the problem:
" The source contains carbon isotopes of mass numbers 12, 13, 14 from a long-dead piece of a tree. (To estimate atomic masses, multiply by 1.67x10^-27kg). "

My question is isn't the 12, 13, & 14 in the atomic mass units already (Carbon-12 has an amu of 12)? Why would I need to multiply? And if I do need to multiply, should I leave it in kg or convert to g?

Thank you.
 
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I'm afraid you haven't given enough of the problem statement for us to know what exactly you are going to be doing with the carbon masses so that we can know if you really need to convert to some unit such as kg. Since whoever wrote the question took the effort to provide a conversion factor for you, it is pretty likely that you really do need to make use of the conversion factor.

To an approximation, an atomic mass unit is kinda sort of the mass of a neutron or the mass of a proton. This glosses over the fact that the proton's mass is a little different than the neutron's mass. Also, there are issues of binding energy when protons and neutrons bind together to form a nucleus.

Way back when I was a kid, I am pretty sure they liked to call either the mass of a certain isotope of oxygen a perfect 16.00000 amu, or else maybe it was that the mass of the carbon-12 isotope was taken to be a perfect 12.00000 amu, I can't remember which.
 
I was afraid that would not be enough info, the entire question is as follows:

Suppose the e field between the e plates in the mass spec is 2.48x10^4 v/m and the magnetic fields B=B'=0.68T. The source contains carbon isotopes of mass #s 12, 13, 14 from a dead tree. (estimate masses by x by 1.67x10^-27) How far apart are the lines formed by the singly charged ions of each type on the photographic film?

I thought I would need to multiply since they bothered mentioning it. However, I do not know if I should convert to grams or leave it in kg. The eq I used is r=(mE)/(qBB')

Thanks
 
Nevermind, Telsa is in kg so for it to cancel I have to use kg.

Thanks.
 
Ah yes.

One Tesla is one kilogram[/color] per second squared per ampere.
 

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