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# Q: Calculating Ke in nuclear fusion reaction

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1. I've had a look at the mark scheme and examiner report but i still don't understand how they calculated kinetic energy.

Q: Calculate the kinetic energy, in MeV, of the neutron released by the fusion ofdeuterium and tritium nuclei. Assume that the net momentum of the nucleibefore fusion is zero.

This is the relevant data given to you in the question:
17.5 MeV of energy isreleased when this fusion reaction takes place.

Mass / MeV/c2
Neutron 939.6
Deuterium 1875.6
Tritium 2808.9
Helium 3727.4

Source: Question 19 (iv)
https://a4942901ab27cf2817f7a4f7497d...%20Physics.pdf

Mark Scheme:
https://a4942901ab27cf2817f7a4f7497d...%20Physics.pdf
2. Okay, I don't know how *they* figured it out, but I have a method that works for these questions (although I'm not sure how).
It's all about ratios, so what I do is create a substitute momentum based on the idea that 17.5 MeV is the kinetic energy of the combined masses of the Helium nucleus and the neutron:

3727.4 + 939.6 = m, Ek = 17.5 = 0.5 * m * v^2,

So to get what I'm going to call the "virtual velocity" of the combined mass, you divide 17.5 by (0.5 * m) and then square root that.

This gives you v = 0.087 (rounded for the sake of explanation)

Use this to find the "virtual momentum" by multiplying it by m

This gives you roughly 404.16

Divide this by 2 since the momentum of both particles should be the same.

From this calculate the "virtual" kinetic energies of the individual particles using their individual masses.

You should get around 43 for the neutron and 11 for the helium. These aren't the real kinetic energies of the particles, but they are of the correct ratio.

Add them together and divide the "virtual" kinetic energy of the neutron (43ish) by that to get the ratio of the kinetic energy of the neutron.

Times this ratio by 17.5 MeV to get the correct answer of 14 MeV (if you've rounded you'll be slightly off).

I'm sure there's a simpler way, but this method works 100% of the time and as of writing this no-one else has helped, so in the case that no-one else gives an answer, use this method.

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