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relativistic energy and momentum questions. |
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| Sep24-07, 02:59 AM | #1 |
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relativistic energy and momentum questions.
problems statement:
1. a nucleus of mass m initially at rest absorbs a gamma ray (photon) and is excited to a higher energy state such that its mass now is 1.01m, find the energy of the incoming photon needed to carry out this excitation. 2. A moving radioactive nucleus of known mass M emits a gamma ray in the forward direction and drops to its stable nonradiactive state of known mass m. Find the energy E_A of the incoming nucleus such that the resulting mass m nucleus is at rest. The unknown energy E_c of the outgoing gamma ray should not appear in the answer. attempt at solution 1.well, for the first question i think this is fairly simple: from conservation of 4-momentum we have before 4-momentum is:(mc,0) after (E/c+E_ph/c,P) so we have : (mc)^2=(E/c+E_ph/c)^2-P^2=(E/c)^2-P^2+2EE_ph/c^2+(E_ph/c)^2=(1.01mc)^2+2EE_ph/c^2+(E_ph/c)^2 where (E/c)^2-P^2=(1.01mc)^2, here im kind of stuck with E which is not given, any hints? 2.for the second the answer in the book is E_A=((M^2+m^2)/2m)c^2 but i dont get it, here's my attempt to solve it: the before 4 momentum is (E_A/c,P) after: (E_c/c,0)+(mc,0)=(E_c/c+mc,0) which by the square of the momentums we get that: (E_A/c)^2-P^2=(E_c/c+mc)^2=(Mc)^2 but im not given P so im kind of stuck here again, i thought perhaps calculate it in the rest frame of M which means that the before is: (E_A/c,0) the after is (E_c/c,0)+(E/c,-P) but still don't get far with it, any help is appreciated, thanks in advance. |
| Sep24-07, 04:11 AM | #2 |
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For question one, what is the 'rest energy' of the nucleus?
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| Sep24-07, 04:43 AM | #3 |
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well, if it wans't clear in my post, obviously it's mc^2, and i wrote in 4 momentum notation (mc,0) for the before the absorption of the photon.
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| Sep24-07, 04:50 AM | #4 |
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relativistic energy and momentum questions.
Perhaps I'm missing something here, but couldn't you write;
[tex]p^2 = (mc)^2 - (1.01mc)^2[/tex] |
| Sep24-07, 04:57 AM | #5 |
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well first, it should be minus that ofocurse cause this way we get a negatrive value where everything there is positive.
and im not sure, what's wrong with what i wrote, first we have (mc,0) after that we have the absorption: (E/c+E_ph/c,p) now (E/c)^2-p^2=(1.01mc)^2 and E_ph=mc-E/c=mc-sqrt((p)^2+(1.01mc)^2) but how do you find p? |
| Sep24-07, 05:00 AM | #6 |
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i think that p=E_ph/c, am i wrong?
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| Sep24-07, 05:38 AM | #7 |
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ok, i solved question number 2.
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| Sep25-07, 12:59 PM | #8 |
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any news on question number 1?
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