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Problem

The nuclear equation for the decay of calcium-47 into scandium-47 is given by:

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The following data are available: Mass of calcium-47 nucleus = 46.95455 u Mass of scandium-47 nucleus = 46.95241 u

1) Using the data, determine the max KE, in MeV, of the products in the decay of calcium-47.

2) State why the KE will be less than your value in question 1

What I don't get

1.48 MeV is what I calculated for question 1 (which is correct according to my answer book). However, why does it say in question 2 that the actual KE will be less than 1.48 MeV? The way I got to 1.48 MeV is converting the mass defect to energy, and the antineutrino (which is X in the equation) has no mass so this process of calculating 1.48 MeV doesn't take into account the KE of the antineutrino. Therefore, shouldn't the actual KE value be greater than 1.48 MeV because we are adding the antineutrino's KE to this 1.48 MeV?

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Did you take the nuclear binding energy of the decay product into account? Then the 1.48MeV would be split between the kinetic energy and the binding energy.

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  • $\begingroup$ But binding energy is not actual energy it is like an energy debt. The 1.48 MeV goes purely to the kinetic energy but the binding energy increases because you are getting more stable nuclei in the product. In other words, more energy is needed to break apart these more stable nuclei but the binding energy isn't actual energy. Here is the answer to question 2 according to my textbook: "does not account for energy of antineutrino." I have trouble understanding this textbook answer. $\endgroup$ – user532874 Apr 21 at 1:15
  • $\begingroup$ @user532874 In the case that we ignore the antineutrino, the max kinetic energy goes straight into the daughter nucleus and the electron. Even when you take the antineutrino into account, the max kinetic energy in the system can't exceed 1.48MeV since that's just the maximum energy you can get out of the reaction. Thus, it must be split between the daughter nucleus, the electron and also the antineutrino which would reduce the kinetic energy of the daughter nucleus and electron. Maybe that's the solution? $\endgroup$ – Aditya Kondapuram Apr 21 at 1:45
  • $\begingroup$ Wait what is KE specifically referring to in problem 2? The total KE or the KE of the daughter nucleus and electron? I'm not even sure what "KE" is referring to. $\endgroup$ – user532874 Apr 21 at 5:44
  • $\begingroup$ @user532874 The total KE (Kinetic Energy) should be referring to the energy released during the reaction (which was 1.48MeV). In problem 2, I'm assuming that they're referring to the KE of the daughter nucleus and the electron, but the question isn't entirely clear. It doesn't specifically state which KE it is looking for. Based on my previous assumption, the KE would be less since it must share some of its energy with the antineutrino. On the other hand, if the KE is referring to the total KE, then that would not be the correct answer. $\endgroup$ – Aditya Kondapuram Apr 21 at 18:52

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