Understanding the expression for Nuclear Reaction rate I don't get, How the expression for Nuclear reaction rate can be written as
$$r=n_1n_2\langle \sigma v\rangle $$
where $$\langle \sigma v\rangle =\int_0^\infty \sigma(E)v f(E)dE$$
Now, $\sigma(E)$ is the probability for the nuclear reaction to occure for energy $E$ and $f(E)$ is the probability density of the particle to have energy $E$. From here, it's clear that $\sigma(E) f(E)dE$ is the probability that nuclei have energy $E$ to $E+dE$ and nuclear reaction will occure.
The object $\langle \sigma v\rangle  $ is the expectation value of velocity. I don't find how does it is related to reaction rate? Please help me understand the expression.
 A: As a commenter writes: $\sigma$ is related to reaction probability, but it's actually a cross section, with units of area.  So you are incorrect when you write "$\langle\sigma v\rangle$ is the expectation value of the velocity." The quantity $\langle\sigma v\rangle$ has units of volume per unit time.
Let's take a step back from quantum mechanics and imagine a gas of hard spheres, which only interact when they touch each other.  Each sphere has radius $r$ and cross-sectional area $\sigma = \pi r^2$.  Imagine that each sphere leaves a trail behind it of the part of the space it has touched, like a three-dimensional version of a paintbrush.  If the sphere is moving with velocity $v$, then $\sigma v$ is the rate at which new volume is added to the paint trail.
For your reaction rate, imagine that species 1 with number density $n_1$ is stationary, like nuclei in a solid, and species 2 with number density $n_2$ is moving.  If there are $N_2$ total moving spheres, the volume they sweep out is increasing at a rate $N_2 \sigma v$, and the fraction of space they have swept out is increasing at a rate $n_2 \sigma v$.  You get an interaction when the swept-out volume grows to include a new member of species 1.  That gives
$$ r = n_1 n_2 \sigma v $$
as the rate of interactions per unit volume.
Your definition of the average $\left<\sigma v\right>$ takes into account that $\sigma$ varies with energy, and weights the average by the population with each energy, as you've said.
I have found that most confusing questions about cross-sections can be resolved by

*

*carefully squinting at the units, so I understand what I'm computing, then

*forgetting about quantum mechanics and thinking about hard spheres.

A: The answer about your reply can easily find on the following book:
Rolfs - Cauldrons in the Cosmos: Nuclear Astrophysics
At the page number 139 you can find the answer at your question (and more and more about this fantastic relation).

Generally, the reactions have cross sections σ(E) that depend strongly
on center-of-mass energy $E$.
This dependence is caused by a repulsive Coulomb barrier, so that:
$\sigma(E) = E-1 exp(-2\pi\eta)S(E)$,
for an ‘astrophysical S-factor’ $S(E)$ that is relatively less variable
with energy.
In the case of neutrons have a $\sigma(E)∝ 1/v$ behavior for
projectile-target relative velocity $v$ near zero.
Or it maybe because of resonances giving sharply peaked cross
sections, like:
$\Gamma/[(E - E_R)2 + \Gamma^2/4]$,
for a single resonance centered at ER with a full width at half
maximum of $Г$.
In a stellar plasma of a mixture of two or more nuclear species, there
will be a considerable range of relative energies $E$ (or, velocities $v$)
because of the statistical distribution of thermal energy among all
the particles in the plasma. The actual rate of nuclear reactions will
therefore require an averaging of the cross sections $\sigma(E)$ over the
thermal distribution of relative energies.
We will therefore define an average reaction rate as the number of
reactions per second per unit volume, and find an expression for this
in terms of $\sigma(E)$ and the distribution $\phi(v)$ of the relative velocities
of the interacting particles.

