# [Special Relativity]Slow Neutron Capture

## Homework Statement

A helium-4 nucleus (rest mass 4.00151 u) is formed when a helium-3 nucleus (rest mass 3.01493 u), stationary in the laboratory frame, captures a slow neutron (rest mass 1.00867 u), in the following reaction.
3He+n→4 He+γ
Determine the energy (in MeV) of the γ-ray in the laboratory frame. (1 u ≡ 1.66 × 10−27 kg.)

Given Above

## The Attempt at a Solution

As far as I can see classically this would be a doddle. With Conservation of Energy and momentum to be taken into consideration before and after. However without the initial velocities of the neutron or the recoil velocity of the He-4 Nucleus im not sure how to calculate the energy of the gamma ray. I'm also pretty unsure where the relativity comes into this.

The neutron is slow, so not travelling relativistically, the initial nucleus is at rest. The gamme ray will travel at c, which makes it a pain all in all that i can tell since working out the Lorentz factor is impossible when V=C.

Thanks for any input.
Adam

## Answers and Replies

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vela
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The slow neutron is essentially at rest. In other words, its kinetic energy is negligible compared to its rest energy. You can assume the He-4 nucleus moves non-relativistically and solve for the energy of the photon. Then go back and check if the He-4 nucleus does indeed move non-relativistically.

If that dubious sort of logic doesn't sit well with you, you can solve the relativistic conservation of energy and momentum equations simultaneously. If you go with this approach and it's not working out for you, post your work here so we can see where you're getting stuck.

Thanks for your help. This way my attempt:

Considering Consv. of E

E$$_{n}$$+E$$_{He3}$$=E$$_{He4}$$+$$_{\gamma}$$

Since the neutron is a slow neutron its Kinetic Energy is negligable compared with its rest energy. Hence:

E$$_{n}$$=m$$_{n}$$*c$$^{2}$$
E$$_{He3}$$=m$$_{He3}$$*c$$^{2}$$
E$$_{He4}$$=m$$_{He4}$$*c$$^{2}$$
E$$_{\gamma}$$=Unknown

Thus:

E$$_{\gamma}$$=m$$_{n}$$*c$$^{2}$$+m$$_{He3}$$*c$$^{2}$$-m$$_{He4}$$*c$$^{2}$$

=c$$^{2}$$*u(1.00867+3.01493-4.00151)
=c$$^{2}$$*u(0.02209)

Therefore:

E$$_{\gamma}$$=3e8*1.66e-27*0.02209 = 1.1e-20 J

My issue with this is, there is no relativity involved. I have treated it purely classically, but since nothing but the gamma-ray is travelling relativistically I don't understand my mistakes.

Thanks again for any input.

P.S The powers are meant to be subscripts, for some reasons the latex _{blah-de-blah} command isn't working. Apologies

vela
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When you used E=mc2, you used relativity. What you found was that some rest mass disappeared in creating He-4, and the missing mass was converted to the energy of the photon. The concept that mass and energy are equivalent is not part of Newtonian physics.

You forgot to square c when calculating the energy of the photon, and you should verify that your answer is consistent with the assumption that the kinetic energy of He4 is negligible. (Find its momentum and use that to calculate its kinetic energy.)

Oh why thank you, I had not noticed that I had only used c.
And thanks for the input. Being 3rd Year now they expect our answer to have the rigour of textbook apparently, so ill make sure to add some explanation of what you said about mass and energy being equivalent just being a relativistic principle.

Ill also make sure to confirm the negligability of the KE before filling it in.

Thanks again =]]