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

  • Thread starter Thread starter kashmirekat
  • Start date Start date
  • Tags Tags
    Atomic Estimate
AI Thread Summary
The discussion centers on the need to multiply carbon isotope mass numbers (12, 13, 14) by 1.67x10^-27 kg to estimate their atomic masses for a mass spectrometry problem. It is clarified that while atomic mass units (amu) are commonly used, the conversion to kilograms is necessary for calculations involving electric and magnetic fields. The user confirms that since the equation requires mass in kilograms, they should use the conversion factor provided. The importance of understanding the relationship between mass, charge, and the resulting motion of ions in a mass spectrometer is emphasized. Ultimately, the conversion to kilograms is essential for accurate calculations in this context.
kashmirekat
Messages
30
Reaction score
1
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.
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top