A Little Bit of Nuclear..

It’s been a while since I posted any cute physics problems, so here’s a little one to amuse you this rainy Thursday morning.

In the following the notation A(a,b)B means the reaction a+A→b+B. The atomic number of Oxygen is 8 and that of Fluorine is 9.

The Q-value (i.e. energy release) of the reaction 19O(p,n)19F is 4.036 MeV, but the minimum energy of a neutron which, incident on a carbon tetrafluoride target, can induce the reaction 19F(n,p)19O is 4.248 MeV. Account for the difference between these two values.

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9 Responses to “A Little Bit of Nuclear..”

  1. Does the energy released include the kinetic energy of the neutron coming in?

    • telescoper Says:

      The energy released is the difference between the rest-mass energies of the incoming and outgoing particles.

  2. I think Phillip’s on the right track. The key is energy and momentum conservation. For the second reaction, the incoming energy (4.248 MeV) is the kinetic energy of the neutron, which of course corresponds to a certain amount of momentum. That momentum is absorbed by the nucleus, which recoils. The kinetic energy associated with this recoil (of both 19O and p) accounts for the “missing” 0.2 MeV.

    To be a bit quantitative, the momentum is P = sqrt(2 m_n E) where E=4.248 MeV and m_n is the neutron mass. The kinetic energy of the recoiling nucleus is P^2 / (2m) = (m_n/m) E, where m is the mass of the recoiling bodies. When the incoming particle has the minimum possible energy, both 19O and p recoil together, so m is about 20 amu. In round numbers, therefore, the recoil kinetic energy is about E/20, which is about 0.2 MeV.

  3. telescoper Says:

    I’d put it slightly differently. Only a fraction A/(A+1) of the energy carried by the incoming neutron is available to activate the reaction, where A is the mass number of the target nucleus. This is because momentum has to be conserved, so the target recoils thus absorbing some KE. The incoming neutron energy therefore has to be 20/19 bigger than the Q value for the first reaction, which turns out to be spot on at 4.248 MeV.

    • Your description makes sense, but I’m confused about how the question is phrased. If the Q value is the difference in the rest (i.e. invariant, i.e. not including kinetic energy) of the participants before and after the reaction, why (apart from momentum conservation) does the incoming neutron need any kinetic energy at all? The Q value presumably goes into the kinetic energy of the proton and fluorine nucleus.

      • telescoper Says:

        The first reaction liberates energy. You would expect the second reaction to require an input of the same amount of energy as is liberated in the first one, i.e. you would think you’d have to put 4.036 MeV worth of energy in to get the reaction to work. In other words, the first reaction is exothermic while the second is endothermic. The required energy has to be supplied by the incoming neutron.

        However, as I tried to explain above, only a fraction of the energy of the incoming neutron is available in the centre of mass frame to feed the reaction because some of it must make the target recoil. Hence the energy of the neutron must be a bit higher than you would have first thought.

      • Yes, perfectly clear, I misread to which reaction the Q-value applies. 😦

  4. John Peacock Says:

    Peter: I think it’s a little inelegant to talk about the “fraction of the energy that’s available”. The whole process can be reversed, eating energy if run in one direction, but liberating the same amount of energy if reversed. This is with the “obvious” proviso that you use the same reference frame in each case, since kinetic energy depends on the observer’s motion. So the question is really a trick that turns on the fact that in case one you are in the centre of mass frame but in the other you aren’t. But the particular frames don’t matter: it would still be the identical energy change in both cases if you had the centre of mass flying past near v=c.

    • telescoper Says:

      It’s worth mentioning that this question came from the Cambridge Scholarship paper I sat in 1981 and which I posted on here a while ago.

      I thought about changing the wording before posting it, but decided to keep it as it was. You have to read between the lines a bit to realise the difference is due to centre-of-mass versus laboratory frame. I remember at the time thinking that something to do with the bonds in Carbon Tetrafluoride was responsible for the difference, but dismissed that because the energy would be too small…

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