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Gamow peak and nuclear reaction rate

  1. May 4, 2015 #1
    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?
  2. jcsd
  3. May 4, 2015 #2


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    If the temperature is approaching infinity, then the number of particles with high temperatures should increase, not decrease, right?
  4. May 5, 2015 #3

    Ken G

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    There must be something wrong with that formula, because fusion rates generally rise, and quite drastically, with temperature. That formula would give a peak in the fusion rate as a function of temperature, but there is not such a peak. The "Gamow peak" is a peak in what you said, the product (at a given temperature) of the number of particles with a given energy, times the fusion rate for that energy per particle. It's analogous to the Planck function, which is like the excitation number of the states at a given energy, times the number of states at that energy per photon. The Planck function also peaks with energy for a given T, but it is a monotonically increasing function of T at any energy.
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