# Reaction of tritium and deuterium

1. Apr 13, 2014

### skrat

1. The problem statement, all variables and given/known data
$_{1}^{2}\textrm{D}+_{1}^{3}\textrm{T}\rightarrow _{4}^{2}\textrm{He}+n$

How much energy releases?

2. Relevant equations

Semi empirical mass formula that we use:

$W=-w_0A+w_1A^{2/3}+w_2\frac{Z(Z-1)}{A^{1/3}}+w_3\frac{(a-2Z)^2}{A}+w_4\frac{\delta _{ZN}}{A^{3/4}}$

for
$w_0=15.6MeV$, $w_1=17.3MeV$, $w_2=0.07MeV$, $w_3=23.3MeV$ and $w_4=33.5MeV$.

3. The attempt at a solution

Now we did plenty of examples yet elements usually had much more neutrons than this reaction, meaning we could simply forget if there were any additional neutrons on the RHS of the reaction.

This is obviously not the case and I have no idea what to do with that extra neutron.

For everything else:

$W=W(He)+W(n)-W(D)-W(T)$

$W=-29.768MeV+W(n)-16.18MeV+3.047MeV$

I hope all the numbers are ok. The result should be 17.6 Mev.

2. Apr 14, 2014

### Naz93

In addition to the binding energy terms W, each of the deuterium, tritium, helium and neutron has mass energy. If you calculate the mass energy for each term, including that for the neutron. Mass energy is just [(number of neutrons) x (neutron mass)] + [(number of protons) x (proton mass)]. Put in these mass energies along with the binding energies. Yes, the numbers will be big compared to the binding energies, but the big numbers should cancel... I haven't worked it through myself to check, but this would be how I'd attempt the problem...

3. Apr 14, 2014

### skrat

Hmmm, your idea my work, yet I have found out what I was supposed to do.

The idea is too ignore semi empirical mass formula but to rather look in the table and find out that:
$m(n)=1.008665 au$
$m(D)=2.014102 au$
$m(T)=3.016050 au$ and
$m(He)=4.002603 au$.

This now gives me the right result $W=c^2[m(He)+m(n)-m(T)-m(D)]=17.6MeV$