
#1
Mar1810, 03:49 PM

P: 113

1. The problem statement, all variables and given/known data
A proton (1.6726×1027 kg) and a neutron (1.6749×1027 kg) at rest combine to form a deuteron, the nucleus of deuterium or "heavy hydrogen". In this process, a gamma ray (highenergy photon) is emitted, and its energy is measured to be 2.39 MeV (2.39×106 eV). Keeping all five significant figures, what is the mass of the deuteron? (Assume you can neglect the small kinetic energy of the recoiling deuteron.) 2. Relevant equations Rest Energy = m c^2 E_{before}=E_{after} 3. The attempt at a solution I found the energy before the combining by c^2 (1.6726×1027+1.6749×1027) = 3.0086*10^10 J this should equal the energy of the deuteron (mc^2) + the energy of the photon energy of the photon in joules: 3.8292*10^13 J therefore: 3.0086*10^10 = m_{d}c^2 + 3.8292*10^13 m_{d}=3.3433*10^27 kg However, this isn't right 



#2
Mar1810, 06:33 PM

P: 1,877

Do you know the correct answer? You should be right, unless there is a rounding error, but I know that the mass of deuteron is just about what you've stated. Make sure you're giving the answer in the correct units (usually we talk about particle masses in terms of rest mass and give the units in MeV).




#3
Mar1910, 11:45 PM

P: 4

3.0086*10^10 = m_{d}c^2  3.8292*10^13, so m_{d}=3.34318*10^27 kg only a fraction out. What gets me is that, where Ed is the deuteron binding in joules, Up and Un are the proton and neutron magnetions respectivey and Uo is the magnetic permeability of free space: (UpUnUo/Ed)^1/3 = 0.746 fm in other words a coupling distance between proton and neutron forming a deuteron of about half a nuclear radius from simple magnetic dipole binding, without having to invoke an extranucleonic strong force at all. Protons will also bind nicely once the magnetic attraction overcomes electrostatic repulsion at about 1.8 fm, at an energy of about 532Kev. Of course there are no quantum numbers involved here so the digits are subject to change, but not the scale, which is one of nuclear interactions. 



#4
Mar2010, 04:50 AM

Emeritus
Sci Advisor
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Thanks
PF Gold
P: 11,530

Energy of a deuteron 



#5
Mar2010, 06:00 PM

P: 4

http://hyperphysics.phyastr.gsu.edu...ne/nucbin.html..
Ep=En (roughly) so (2x931 Mev  2.39 Mev)/(Nqc^2). About 0.3% Less than the sum of the individual nucleons was the point :) 



#6
Mar2911, 04:06 PM

P: 1

I know what you happened. You got all the steps right, but you probably need to use the speed of light as 3*10^9 in the program. I did this same problem for my class and i figured it out by your post since it didn't except my answer wither until i did that.




#7
Jul111, 12:38 PM

P: 4




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