What kind of energy is released in a nuclear fusion reaction?

[freddie_mclair]

Hi, I have a fundamental (and maybe silly question) but I couldn't find a proper answer anywhere yet:

For example, for a nuclear fusion reaction of Tritium (T) and Deuterium (D), we get an alpha particle ($\alpha$) a neutron (n) and energy release due to the mass difference $\Delta m=m_D+m_T-m_n-m_{\alpha}$, which means that: $D + T \rightarrow \alpha + n + \mbox{ 17.6 MeV}$ where $\mbox{ 17.6 MeV}= \Delta m c^2$. These 17.6MeV get split by the neutron (14.1MeV) and the alpha particle (3.5MeV).

Now, my question is: what does it mean, to release energy? What kind of energy is this?
Thanks!

 

[Hill]

AFAIK, kinetic energy of the products and gamma rays.

 

[freddie_mclair]

I agree with the kinetic energy of the products, but where are the gamma rays? What I understand is that the 17.6MeV are just split into the kinetic energy of the neutron by $KE_n=\Delta m c^2 \frac{m_{\alpha}}{m_{\alpha}+m_n}$ and the rest to the alpha particle $KE_{\alpha} = \Delta m c^2 - T_n$.

 

[Vanadium 50]

but where are the gamma rays

You posited a reaction without them.

 

[freddie_mclair]

You posited a reaction without them.

what would be the correct formulation then? and what amount of radiation would that be in terms of energy?

 

[Vanadium 50]

I can't tell you what reaction you are thinking of. Just that A+B → C+D and A+B → C+D+γ are not the same process.

 

[freddie_mclair]

For this specific reaction I mentioned it is just Deuterium + Tritium, there are no gamma rays, just an alpha particle and a neutron. But in several places it is indicated that, apart from the reaction products, there is also an energy release, like for example here.

 

[Vanadium 50]

here are no gamma rays

but where are the gamma rays?

Do you see why people are confused?

 

[freddie_mclair]

No, why? I asked Hill where are the gamma rays in the reaction I described.

 

[mfb]

The reaction without gamma rays is the most common outcome, all the released energy becomes kinetic energy of the reaction products:
$D + T \rightarrow \alpha + n$

This is possible, too:
$D + T \rightarrow \alpha + n + \gamma$
Here the photon energy is variable and the rest will be kinetic energy of the alpha and n.

 

[PeroK]

For this specific reaction I mentioned it is just Deuterium + Tritium, there are no gamma rays, just an alpha particle and a neutron. But in several places it is indicated that, apart from the reaction products, there is also an energy release, like for example here.

That's not a sufficiently detailed source for discussion. In this reaction, most of the energy is kinetic energy of the neutron. Hyperphysics has a little more detail:

http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/fusion.html

 

[freddie_mclair]

Thanks mfb, and PeroK.
To conclude: energy release in this specific fusion reaction can be totally kinetic (shared the n and ) or kinetic + EM radiation.

 

[sharmast]

The deuterium tritium reaction produces an alpha particle and a neutron. The energy of the reaction becomes the kinetic energy of the products. Gamma rays do not seem to be produced.

 

[mfb]

Gamma rays do not seem to be produced.

It's possible, as discussed, it's just rare. It even has a 16.75 MeV photon line corresponding to He-5 decaying to its ground state before emitting a neutron.

 

[nightvidcole]

When neutrons are emitted, nuclei of the surrounding material could capture a neutron and emit gamma rays. So even though the reaction itself may not emit gammas, you will get gamma rays "in real life" from any reaction with neutrons as a product.