Calculate the mass defect of a carbon-12 nucleus

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Homework Help Overview

The discussion revolves around calculating the mass defect of a carbon-12 nucleus, focusing on the relationship between mass and energy as described by Einstein's equation E=mc². Participants are examining the definitions and calculations related to mass defect and energy conversion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of mass defect and its implications, questioning the definitions and units involved. There are attempts to convert mass defect into energy and discussions about the precision of values used in calculations.

Discussion Status

The discussion is active, with participants providing feedback on calculations and questioning the precision of units. Some guidance has been offered regarding the conversion of units and the importance of using accurate values for constants, but no consensus has been reached on the final calculations.

Contextual Notes

Participants are navigating issues related to unit conversions and precision in calculations, with some expressing uncertainty about the appropriate values to use for constants like the speed of light.

amazingphysics2255
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Homework Statement
Carbon-12 nucleus, has a mass of 11.9967u. Given: proton mass = 1.00728 u;
neutron mass = 1.00866 u

Calculate the mass defect of carbon-12 nucleus.
Relevant Equations
N/a
My attempt: Proton 1.00728ux6+ Neutron 1.00866ux6=12.09564u

12.09564u-11.9967u=0.09894u

E=mc^2 E=0.09894x(3.00x10^8ms-^1)^2
 
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Well, I think you have the correct idea, you should also look at the exact definition of the mass defect, as I learn it is not exactly the final answer you give. In fact, it seems obvious that de mass defect should have dimensions of mass, not energy.
 
Gaussian97 said:
Well, I think you have the correct idea, you should also look at the exact definition of the mass defect, as I learn it is not exactly the final answer you give. In fact, it seems obvious that de mass defect should have dimensions of mass, not energy.
My final answer for the mass defect was 0.09894u. I was then going to calculate the Energy in the reaction. @Gaussian97
 
Last edited:
Ok, then I think it's right but, are you sure that you want the energy in these units?
 
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I hadn't solved the E=mc^2 but I think it equals 8.9046x10^15J
 
No it doesn't -- Units, Units, Units
 
Do I have to convert 0.09894u to kg ? how do I do that ?
 
so 1.64293721x10^-28x(3.00x10^8)^2

=1.478643489x10^17J ?
 
  • #10
amazingphysics2255 said:
so 1.64293721x10^-28x(3.00x10^8)^2

=1.478643489x10^17J ?
Re-check your calculation. The mantissa looks right, but the exponent is off.

Also, if you're going to use that much precision with your mass (you originally had 6 digits), you might as well use a more precise value for the speed of light, [itex]c = 2.99792 \times 10^8\ \rm{\frac{m}{s}}[/itex].
 
  • #11
so 1.642938473x10^-28 kg is the changed conversion of 0.09894u

So 1.642938473 can be rounded to approx 1.642938 (6 s.f) so 1.642938x10^-28(3.00x10^8)^2 = 1.4786442x10^-11J Is this closer ?
 
  • #12
amazingphysics2255 said:
so 1.642938473x10^-28 kg is the changed conversion of 0.09894u

So 1.642938473 can be rounded to approx 1.642938 (6 s.f) so 1.642938x10^-28(3.00x10^8)^2 = 1.4786442x10^-11J Is this closer ?
That looks mostly good. Probably close enough. Again, there might be an issue with precision (due to the lower precision with the speed of light value used) but it's probably not a big deal.
 
  • #13
Actually, because of the subtraction of two large but close numbers, the precision left over is only 4 digits as already shown in the OP !
Nevertheless, 3e8 for ##c## is too crude.
 

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