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Relativistic energy and momentum questions.

by MathematicalPhysicist
Tags: energy, momentum, relativistic
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MathematicalPhysicist
#1
Sep24-07, 02:59 AM
P: 3,242
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.
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Hootenanny
#2
Sep24-07, 04:11 AM
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For question one, what is the 'rest energy' of the nucleus?
MathematicalPhysicist
#3
Sep24-07, 04:43 AM
P: 3,242
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.

Hootenanny
#4
Sep24-07, 04:50 AM
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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]
MathematicalPhysicist
#5
Sep24-07, 04:57 AM
P: 3,242
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?
MathematicalPhysicist
#6
Sep24-07, 05:00 AM
P: 3,242
i think that p=E_ph/c, am i wrong?
MathematicalPhysicist
#7
Sep24-07, 05:38 AM
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ok, i solved question number 2.
MathematicalPhysicist
#8
Sep25-07, 12:59 PM
P: 3,242
any news on question number 1?


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