# Gamow peak and nuclear reaction rate

It's known that the nuclear reaction rate (inside a Star) can be determined with

$$R_{ab}=n_a n_b\left<\sigma v\right> \, \approx \, n_a n_b \Big(\frac{8}{\pi m_e}\Big)^{1/2} \frac{S(E_0)}{(k_BT)^{3/2}} \Delta \frac{\sqrt{\pi}}{2} e^{-3E_0/k_BT},$$

where $k_B$ is Boltzmann constant, $T$ the temperature, $v$ the velocity, $\sigma$ the cross section, $E$ the energy, and $$\Delta=\frac{4}{\sqrt{3}}\Big(\frac{b}{2}\Big)^{1/3}(k_B T)^{5/6}.$$

The above formula is found by using Maxwell velocity distribution and tunneling probability, since $$\left<\sigma v\right>=\int_0^{\infty} \sigma(E)v(E)f(E)dE.$$

The maximum of the reaction rate is called Gamow peak and I guess that is achieved with a proper trade-off of the Maxwell velocity distribution and the tunneling probability.

As the temperature tends to infinity, the reaction rate approaches to 0. Is it due to the fact that according to Maxwell velocity distribution there are less particles with higher temperature and, thus, there will be less probability that two atoms collide?

I believe the first formula you give is obtained by approximating the integrand $S(E) e^{-(E/k_BT-\sqrt{E_G/E})}$, where $E_G$ is the Gamow energy, by a Gaussian centered about the maximum value $E_0$. However, if you take $T\rightarrow \infty$, then the integrand becomes $$e^{-\sqrt{E_G/E}}\rightarrow 1,$$ and the integrated cross section blows up, as expected. So issue is that your approximate formula is only valid for $k_BT \ll E_G$.