# Effective Coulomb barrier for deuteron

What is the effective Coulomb barrier for a Deuterium-deuterium fusion reaction?

I am seeing temperatures of about $40 \times 10^7 K$ online, but have no idea how they are getting this.

If we have

$^2H+^2H \rightarrow ^{3}He + ^1n$

and the coulomb barrier is: $U=\frac{ke^2}{r}$ which needs to be overcome for fusion and the strong force to dominate.

Isn't r just $1.3 (A_1^{1/3} + A_1^{1/3} )f= 1.3 (2^{1/3} + 2^{1/3})f=3.2758f \quad (f=10^{-15} m)$

Plugging this into the coulomb equation I get about 476 KeV which is about $552\times10^7$ K

This isn't for an assignment but for my own studies.

• You appear not to be familiar with tunneling. See duplicate physics.stackexchange.com/q/73970 – Rob Jeffries Mar 22 '15 at 19:35
• I am familiar with it, but how do they get a temperature of $40 \times 10^7 K$ online? Which converts to about 50 KeV. Does this mean if I put a deuterium atom in a 50kV electric field so it is in constant acceleration it will fuse together with another deuterium atom once they collide? – sci-guy Mar 22 '15 at 19:43
• Give your source. Do they say this is the kT equivalent to the Coulomb barrier? Or are they saying this is the temperature required to get D to fuse in some density condition or other? The two are completely different. Note also that fusion does not occur with high probability via tunneling, so your latter scenario would not work very often. – Rob Jeffries Mar 22 '15 at 19:49
• to overcome the coulomb barrier: hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c4 at the bottom – sci-guy Mar 22 '15 at 19:51
• The answer (or at least the reason why your perfectly respectable classical calculation doesn't yield the right tmperature) is at hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c2 – Rob Jeffries Mar 22 '15 at 20:17

According to your question and the discussion in the comments there seems to be some confusion. Let's first try to answer your original question (Coulomb barrier for a D-D reaction):

The Coulomb barrier is usually defined as the point where the strong force overcomes the repulsive Coulomb force of two positively charged nuclei. A proper estimation is that the nuclei need to barely touch, thus have a distance of $$R = R_1 + R_2,$$ where $R_1$ and $R_2$ are the radii of the nuclei. As you pointed out, a useful approximation for a nucleus' radius is $$R=r_0 A^{1/3},$$ with $A$ the mass number and $r_0\approx1.3\cdot10^{-15}\,\mathrm{m}$. The distance at which the strong force starts to set in can thus be written as $$R\approx r_0\left(A_1^{1/3} + A_2^{1/3}\right).$$ Let's now plug that into the Coulomb potential: \begin{eqnarray} V_{Coulomb}&\approx&\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_0\left(A_1^{1/3} + A_2^{1/3}\right)}\\ &=&\frac{e^2}{4\pi\epsilon_0}\frac{Z_1Z_2}{r_0\left(A_1^{1/3} + A_2^{1/3}\right)}, \end{eqnarray} with $q_1$, $q_2$ the charge of the nuclei and $Z_1$ and $Z_2$ their charge numbers.

Inserting the numbers for a D-D reaction, this results in $$V_{Coulomb}\approx440\,\mathrm{keV},$$ which corresponds to the value you got.

This corresponds to the potential wall we need to overcome to let the two Deuterium nuclei fuse. Luckily, though, there are two effects leading to a on average lower temperature at which fusion occurs:

1. Tunneling
2. Particles in the Maxwellian tail

The fusion rate coefficient or reactivity, $\left<\sigma_{fus}v\right>$, can therefore be calculated as the convolution of a Maxwellian distribution $f_M$ and the fusion cross section $\sigma_{fus}$ (which we know from extensive experiments). So to get the average temperature at which D-D fusion has the highest probability, we need to have a look at the fusion reactions:

1. D + D $\rightarrow$ T + p
2. D + D $\rightarrow$ $^3$He + n

They both occur with roughly the same probability. To calculate their actual reactivity, we look up data from experiments and might get something like what is shown in the following plot.

As you can see, the rate coefficient of the D-D reaction is basically increasing with energy (temperature), it will asymptotically approach a very broad maximum at temperatures which are larger by an order of magnitude (sorry for not showing but I couldn't find the necessary data). For comparison, the D+T reaction is also shown and you can see why this is favored in the lab: its reactivity is larger by two orders of magnitude almost over the full range shown here.

To summarize, you don't need to actually climb up all the Coulomb wall to let the particles fuse, finite probability for fusion does occur at much lower temperatures.