$$H_1^2 + H_1^2 \rightarrow He^4_2+ \text{energy}$$

How can energy be released, if the left hand side of this equation has less mass/energy than the right?

$\text{BE}_H \approx 1 \text{MeV}\\ \text{BE}_{He} \approx 11 \text{MeV}$

BE = binding energy per nucleon

The energy equation would therefore be

$$ 2\cdot 2 \cdot (1\text{MeV}) \rightarrow 4\cdot (11\text{MeV}) + n$$

Making $n$ negative.

I know this fusion is usually done with $H_1^3$ (Tritirium?), does this make a difference?

With this, a proton would be emitted, would that be the energy from equation 1?

  • $\begingroup$ actually the process is rather more complicated (at least in stars); see en.m.wikipedia.org/wiki/… or here en.m.wikipedia.org/wiki/… $\endgroup$ Commented Mar 7, 2017 at 1:54
  • $\begingroup$ 2.62 x 10^12 joules per mole when you multiply weight difference of those sides by square of speed of light. 1proton+1neutron + 1proton+1neutron is not equal in weight with 2 proton 2 neutron $\endgroup$ Commented Mar 7, 2017 at 2:10
  • 3
    $\begingroup$ We've done this before. Binding energy has a confusing sign convention. See physics.stackexchange.com/q/230933 physics.stackexchange.com/q/100352 and links therein as well as some others. Surprisingly none of our questions on the topic seem to be very highly voted, so there isn't a great choice for an authoritative link. I started with this search: physics.stackexchange.com/search?q=binding+energy+mass+energy $\endgroup$ Commented Mar 7, 2017 at 2:10
  • $\begingroup$ @huseyintugrulbuyukisik what are you talking about? The differences in mass have been compensated for. $\endgroup$
    – Tobi
    Commented Mar 7, 2017 at 2:38
  • 1
    $\begingroup$ I got it later within the night. Binding energy goes up, as the mass goes down, as the sum of these equal the individual masses of the nucleons. Students, along with my teacher (today), were similarly baffled by what was going on. I think it's misleading having minding energies in any of these equations as they don't simply add up, like masses. $\endgroup$
    – Tobi
    Commented Mar 7, 2017 at 12:45

1 Answer 1


I think you got the fusion reaction slightly wrong, what is happening in the end is

\begin{eqnarray} ^{2}\mbox{H} +\ ^{2}\mbox{H} &\rightarrow& ^{3}\mbox{H} +\ ^{1}\mbox{H}\\ ^{2}\mbox{H} +\ ^{2}\mbox{H} &\rightarrow& ^{3}\mbox{He} + \mbox{n}, \end{eqnarray}

which both have approximately the same probability to occur. Note that the released kinetic energy is not included in the reactions above.

Insert the atomic masses results in

\begin{eqnarray} 2.01410177812\cdot u + 2.01410177812\cdot u &=& 3.0160492779\cdot u + 1.00782503223\cdot u + x_1\\ 2.01410177812\cdot u + 2.01410177812\cdot u &=& 3.0160293201 \cdot u + 1.00866491588\cdot u + x_2 \end{eqnarray}

where $x_{1,2}$ corresponds to the released energy. The data for the atomic masses is from the NIST website and from wikipedia.

Solving the equatione yields

\begin{eqnarray} x_1 &\approx& 0.00432924611\cdot u\\ x_2 &\approx& 0.00350932026\cdot u \end{eqnarray}

corresponding to ($1\cdot u$ corresponds approximately to $931.5$ MeV)

\begin{eqnarray} x_1 &\approx& 4.03\,\mathrm{MeV}\\ x_2 &\approx& 3.27\,\mathrm{MeV}. \end{eqnarray}

So this is the released energy for both reactions, does the reaction ''equation'' now makes more sense to you?


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