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I have two questions.

First, I understand that in a nuclear reaction $$Q:=K_{after}-K_{before}\equiv E_{0,before}-E_{0,after} \qquad (1)$$ where $K$ is the total kinetic energy, and $E_0$ is the total rest energy. My question is, in the reaction $^{9}Be\left(\gamma,n\right)^{8}Be$, where should I put the energy of the $\gamma$-photon? I mean, conservation of energy says $$Mc^{2}+h\nu=mc^{2}+M'c^{2}+K_{n}+K' \qquad(2)$$ where $M$=mass of $^{9}Be$, $m$=mass of the neutron, $M'$=mass of $^{8}Be$, and $K_{n}$, $K'$ are the kinetic energies of the neutron and of the $^{8}Be$, respectively.

Then, using (1) (with the rest energies version) I wrote $$Q=(M-m-M')c^2 \qquad (3)$$ but this is equivalent to: (using (2))

$$Q=K_{n}+K'-h\nu \qquad (4)$$ So, my 1st question is: is (4) the correct expression for $Q$?

Second, I want to calculate the kinetic energies $K_n$ and $K'$. In order to do this, I'm thinking to use (4) (if it's correct) with $K_{n}=\frac{1}{2}mv^{2}$ and $K'=\frac{1}{2}M'V^{2}$, and the conservation of linear momentum

$$\frac{h}{\lambda}=mv+M'V \qquad (5)$$

Assuming that the neutron and the $^{8}Be$ continue moving in the direction of the $\gamma$-photon.

So, my 2nd question is: can I make this last assumption? Or in other words, is (5) a correct expression for the linear momentum conservation?

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up vote 1 down vote accepted

Your equations (2), (3) and (4) are all correct. I don't like the way (1) is written, because it relies on an unstated understanding that the $E$'s represent only mass energy, but with that understanding it is also correct.

This is just the conservation of energy in it's relativistic form where mass is just another kind of energy (previously you had to worry about kinetic energy and various kinds of potential and internal degrees of freedom expressed as heat and so on; with the advent of relativity you have to add mass to that list). $Q$ is the difference of the masses expressed in terms of energy.

Your idea for getting the distribution of energy between the product nucleus on the neutron (use the conservation of momentum) is correct, but you must work in two dimensions not one (you can choose to work in the $x$--$y$ plane however). Choosing the direction of the photon as $\hat{x}$ () which you can do without loss of generality) it should be $$ h \nu \hat{x} = \vec{p}_n + \vec{P} $$ where I have not written the momenta in the Newtonian approximation, because you do not yet know that this is correct. You have a couple of choices here:

  1. Do in the full relativistic expression
  2. Do it in the Newtonian regime and then check to see if that was reasonable
  3. Assume that the nucleus will be Newtonian but treat the neutron relativistically, and then check to make sure the result was reasonable.

The important thing is that if you make the Newtonian approximation, you must go back and check to see that it was reasonable after.

Questions for the student:

  • What condition or conditions make the Newtonian approximation "reasonable"?
  • Can you deduce if the approximation will likely be reasonable before you begin calculating?
  • Why did I suggest treating only the recoiling nucleus in the Newtonian approximation as an option above?
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No because beta has to be a factorial product of the reaction in relation to the mass-rest energy.

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Can you be more specific please? – Ana May 27 '13 at 14:02
By the way, what do you mean with "beta"? – Ana May 27 '13 at 15:36

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