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e=mc^2 question |
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| Nov21-12, 03:39 PM | #1 |
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e=mc^2 question
1. Energy is proportional to mass.
2. Mass (of nucleus) is proportional to volume. 3. Volume can be determined from cross-sectional area. If this is the case, then is E=mc^2 equivalent to kE=(pi)r^2 where r is the radius of the nucleus and k is some constant? |
| Nov21-12, 03:43 PM | #2 |
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Is number 2 even correct? If I add one neutron to hydrogen to make deuterium, is the nuclear now bigger in volume?
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| Nov21-12, 03:49 PM | #3 |
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The relationship between volume and mass in a nucleus is not as simple as just a proportionality.
http://en.wikipedia.org/wiki/Semi-em...l_mass_formula |
| Nov21-12, 04:06 PM | #4 |
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e=mc^2 question
Thanks, that helps.
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| Nov21-12, 04:29 PM | #5 |
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Even if (2) were true (and DaleSpam has pointed out it isn't), you would expect an r3 term, not an r2 term. (Volume of a sphere isn't proportional to area).
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| Nov24-12, 07:39 AM | #6 |
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2. Mass = Volume x Density
What's the idea to measure volume of as a cross section anyway? Volume of n-sphere can be easily calculated without any cross sectional area using formulas. Cross section of 4D objects is 3D object. I have no idea how to calculate mass-radius relationship for nucleus, it's difficult because it's a compound, but for example mass of electron = coupling * Planck's mass * Planck's length / classical electron radius |
| Nov24-12, 09:35 AM | #7 |
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I think Andy Lee meant the two-dimensional cross-section which can be observed in particle physics - this can be used to estimate the volume of the nucleus.
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| Nov26-12, 07:03 AM | #8 |
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If 2. were correct I dont think atomic bombs would exist (at least not in the way they do in this universe!) since the energy as I understand it comes from the difference in mass of the nucleus and the constituents of the nucleus, so if they were proportional there wouldn't be any release of excess energy.
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