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Radiochemistry Question... - Shidoshi - 2011-03-06
I'm studying radiochemistry here and I have a test this week about it.
I'm having a bit of trouble in this question here (translated from French by me):
"A nuclei of mass A=100 desexcitates (goes to the fundamental state) by emitting a gamma ray of 100keV. By writting the laws of conservation of energy and impulsion, show that the recoil energy Er of the nuclei is negligable and of ~0,05eV."
I know that E² = m²c²² + p²c² but I don't really know how to get the recoil energy from this.
Radiochemistry Question... - OB3LISK - 2011-03-06
Your equation is confusing me too.
First of all, I think that (m^2)(c^22) should really be (m^2)(c^4). 22nd is kind of drastic exponent.
And uhh. Can you go over what the letters stand for F3?
E = Energy
m = mass
c = speed of light constant (3 x 10^8)
p = momentum = speed x mass
?
Radiochemistry Question... - Shidoshi - 2011-03-06
Yeah, it's to the 4th. It's the same as squared squared. Yes, that's what they mean.
Radiochemistry Question... - HellenzSin - 2011-03-06
I get E = 100000,05eV :l
Relevant http://en.wikipedia.org/wiki/M%C3%B6ssbauer_effect
Er= (Ey^2)/(2mc^2)
Radiochemistry Question... - Shidoshi - 2011-03-07
Thanks a lot found the answer. Er = 0,0533 eV
Radiochemistry Question... - OB3LISK - 2011-03-07
Good thing you had a genius like me to help. *pushes this Hellzen guy out a window*
Good luck on your test.
Radiochemistry Question... - Shidoshi - 2011-03-07
For anyone interested I just did momentum balace:
Pgama² = Patom²
Egama²/c² = Erecoil * 2M
For the Erecoil it came from:
E=p²/2M (from the classical kinetic energy formula)
For the Egama it was a bit harder:
p = h/2pi * k (from quantum mechanics, duality wave-particle)
k = 2pi/lambda (definition)
E = hf = hc/lambda (energy of a wave)
Then adding all 3 together:
p=E/c
So basically Erecoil = Egama²/2Mc²
Egama = 100 keV
M = 100 * 938MeV/c² (100 nucleons on this atom)