What is the correct solution for a nucleus with a diameter of 8x10^(-15)m?

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The discussion centers on calculating the atomic mass of a nucleus with a diameter of 8x10^(-15)m, with initial calculations yielding 512, which does not match the provided multiple-choice answers. The diameter of nucleons, particularly neutrons and protons, is noted to vary, leading to different mass estimates based on the assumed size. It is emphasized that without a precise value for nucleon diameter, the question lacks a definitive answer. Participants suggest using a proton radius of 1.2 femtometers for calculations and highlight the sensitivity of results to small variations in diameter. Overall, the conversation underscores the importance of accurate measurements in nuclear physics calculations.
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i have the following for a multiple choice question
.." If a nucleus had a diameter of 8x10^(-15)m what would its atomic mass be?..."

I have calculated the answer to be 512, however this is not a solution. The solutions are 7, 37, 64, 296. Is the correct solution here? Thanks
 
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The neutron is "of the order" of 10^(-15) meters in diameter, and using that, I get the same answer as you (8^3 = 512).

The problem here is that order of estimates are not precise enough for this calculation. Particles ranging from 0.5*10^(-15) to 5*10^(-15) in diameter can legitimately be said to have an order of magnitude of 10^(-15) meters. Of course, using different actual values will give wildly different estimates.

So, I don't know. Unless you have an authoritative source for the diameter of a neutron, the question has no right answer. I would just go with the one that's closest to the calculation and hope for the best. I'd also complain about the question after the test.
 
prolong199 said:
i have the following for a multiple choice question
.." If a nucleus had a diameter of 8x10^(-15)m what would its atomic mass be?..."

I have calculated the answer to be 512, however this is not a solution. The solutions are 7, 37, 64, 296. Is the correct solution here? Thanks
Yes. Use proton radius = 1.2 fm.

AM
 
Andrew Mason said:
Yes. Use proton radius = 1.2 fm.

AM

Just for my own edification, where can you find reference values like these ?
 
Just as an illustration of how wildly other estimates can vary, http://hypertextbook.com/facts/1999/YelenaMeskina.shtml (various figures given, including a bad estimate of 2*10^(-14) m)

http://www.glafreniere.com/sa_protons.htm (10^(-15) m)

http://www.coral-lab.org/~marc/libro/insignificance_full.html (1.6*10^(-15) m)

Needless to say, each of these different values will give a different answer to the question. Which is why a question like this is bad unless one is given the diameter of a nucleon to be assumed when working out the question.
 
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Andrew Mason said:
http://en.wikipedia.org/wiki/Atomic_nucleus#Nucleus_size" or "[URL
[/URL] are pretty good for finding just about anything, I find.

AM

Thanks for the references, but my concern is that there are various other estimates out there. This sort of question is exquisitively sensitive to small variations in the assumed nucleon diameter.
 
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Curious3141 said:
Thanks for the references, but my concern is that there are various other estimates out there. This sort of question is exquisitively sensitive to small variations in the assumed nucleon diameter.
I think the diameter of a proton at 1.2 fm is determined by extrapolation from measurements of larger nucleii on the theory that the nucleus has uniform density. If you plot r against A2/3 where r is the radius and A is the number of nucleons, you should get a line of slope r0 = radius of proton/neutron.

AM
 
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Andrew Mason said:
I think the diameter of a proton at 1.2 fm is determined by extrapolation from measurements of larger nucleii on the theory that the nucleus has uniform density. If you plot r against A2/3 where r is the radius and A is the number of nucleons, you should get a line of slope r0 = radius of proton/neutron.

AM

Ah, thanks for the clarification. :smile:
 
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