relativistic energy and momentum questions.by MathematicalPhysicist Tags: energy, momentum, relativistic 

#1
Sep2407, 02:59 AM

P: 3,175

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 4momentum we have before 4momentum is:(mc,0) after (E/c+E_ph/c,P) so we have : (mc)^2=(E/c+E_ph/c)^2P^2=(E/c)^2P^2+2EE_ph/c^2+(E_ph/c)^2=(1.01mc)^2+2EE_ph/c^2+(E_ph/c)^2 where (E/c)^2P^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)^2P^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. 



#2
Sep2407, 04:11 AM

Emeritus
Sci Advisor
PF Gold
P: 9,789

For question one, what is the 'rest energy' of the nucleus?




#3
Sep2407, 04:43 AM

P: 3,175

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.




#4
Sep2407, 04:50 AM

Emeritus
Sci Advisor
PF Gold
P: 9,789

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] 



#5
Sep2407, 04:57 AM

P: 3,175

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)^2p^2=(1.01mc)^2 and E_ph=mcE/c=mcsqrt((p)^2+(1.01mc)^2) but how do you find p? 



#6
Sep2407, 05:00 AM

P: 3,175

i think that p=E_ph/c, am i wrong?




#7
Sep2407, 05:38 AM

P: 3,175

ok, i solved question number 2.




#8
Sep2507, 12:59 PM

P: 3,175

any news on question number 1?



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