# Thermonuclear fusion

1. Jun 10, 2006

### Soilwork

I just need a little help in interpreting a question.

Question
Identify the terms in the above two expression and give their physical significance

What do you think they mean by physical significance?

These are the two expressions by the way.

$$\epsilon = n_a n_x \frac{< \sigma V >} {\rho}$$

$$< \sigma V > =( \frac {8} {m \pi})^{1/2} (kT)^{-3/2} \int_{0}^{100000} S(E) e^{(\frac{-E} {kT} - \frac {b} {E^{1/2}})} dE$$

What I've done is explain all the terms in these expressions

na = number of particles per unit volume of nuclide a
nx = number of particles per unit volume of nuclide x
Q = Energy per reaction
$$\rho$$ = density
$$\epsilon$$ = power generated per unit mass

$$<\sigma V >$$= the velocity integrated cross-section
m = reduced mass of the target-projectile system.
k = Boltzmann’s constant.
T = temperature of the system
S(E) = cross-section factor
E = Energy
b = this is just a term that simplifies the equation and here all I did was I gave the full expression.

I was thinking that maybe the physical significance would be to say that $$e^{\frac {-b} {E^{1/2}}$$ is the probability of penetration of the energy barrier and $$e^{\frac{-E} {kT}}$$ is the maxwellian distribution etc.??

P.S. The integral is meant to be infinity but I didn't know the syntax

Last edited: Jun 10, 2006
2. Jun 10, 2006

### Staff: Mentor

$$\infty$$ "[tag]\infty[/tag]" gives infinity. Check the LaTeX thread.

I believe b is the impact parameter, or is related. The 'b' must have units of $\sqrt{energy}$, so is related to velocity or momentum.

From what text is the second expression for $<\sigma V >$?

Here is some useful discussion on distributions, particularly Maxwell-Boltzmann. See plate 3.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/disfcn.html

http://hyperphysics.phy-astr.gsu.edu/hbase/math/disfcn.html#c1

Last edited: Jun 11, 2006
3. Jun 10, 2006

### bugon

The exp(-E/kT) comes from the Boltzmann distribution, the exp(-b/sqrt(E)) comes from the penetration factor. The first function decreases with energy (there are less fast particles), the second increases (it is easier for fast particles to tunnel through the potential barrier. When you multiply the two functions, you get the Gamow Peak. Check it out in google.

b comes from the tunneling calculation. It is: $b \equiv {\sqrt{2\mu} \pi Z_a Z_b e^2 \over \hbar}$. The full tunnelling + reaction includes also the nuclear cross section which enters through S(E).

In the sun, the Gamow peak comes out to be at about 5 times kT while other reactions have it much higher (e.g., 20 times kT for typical CNO reactions)

NJS

4. Jun 11, 2006

### Soilwork

Thanks a lot for you help guys. That's cleared the question up.