The Gamow factor, Sommerfeld factor or Gamow–Sommerfeld factor,[1] named after its discoverer George Gamow or after Arnold Sommerfeld, is a probability factor for two nuclear particles' chance of overcoming the Coulomb barrier in order to undergo nuclear reactions, for example in nuclear fusion.
By classical physics, there is almost no possibility for protons to fuse by crossing each other's Coulomb barrier at temperatures commonly observed to cause fusion, such as those found in the Sun.
When George Gamow instead applied quantum mechanics to the problem, he found that there was a significant chance for the fusion due to tunneling.
The probability of two nuclear particles overcoming their electrostatic barriers is given by the following equation:[2] where
While the probability of overcoming the Coulomb barrier increases rapidly with increasing particle energy, for a given temperature, the probability of a particle having such an energy falls off very fast, as described by the Maxwell–Boltzmann distribution.
Gamow found that, taken together, these effects mean that for any given temperature, the particles that fuse are mostly in a temperature-dependent narrow range of energies known as the Gamow window.
Gamow[3] first solved the one-dimensional case of quantum tunneling using the WKB approximation.
Considering a wave function of a particle of mass m, we take area 1 to be where a wave is emitted, area 2 the potential barrier which has height V and width l (at
This is solved for given A and α by taking the boundary conditions at the both barrier edges, at
, this is easily solved by ignoring the time exponential and considering the real part alone (the imaginary part has the same behavior).
(assumed not very large, since V is greater than E not marginally): Next Gamow modeled the alpha decay as a symmetric one-dimensional problem, with a standing wave between two symmetric potential barriers at
, and emitting waves at both outer sides of the barriers.
Solving this can in principle be done by taking the solution of the first problem, translating it by
Due to the symmetry of the problem, the emitting waves on both sides must have equal amplitudes (A), but their phases (α) may be different.
This gives a single extra parameter; however, gluing the two solutions at
requires two boundary conditions (for both the wave function and its derivative), so in general there is no solution.
, each having a different factor that depends on k and α, the factor of the sine must vanish, so that the solution can be glued symmetrically to its reflection.
Since the factor is in general complex (hence its vanishing imposes two constraints, representing the two boundary conditions), this can in general be solved by adding an imaginary part of k, which gives the extra parameter needed.
The physical meaning of this is that the standing wave in the middle decays; the emitted waves newly emitted have therefore smaller amplitudes, so that their amplitude decays in time but grows with distance.
The decay constant, denoted λ, is assumed small compared to
λ can be estimated without solving explicitly, by noting its effect on the probability current conservation law.
Since the probability flows from the middle to the sides, we have: Note the factor of 2 is due to having two emitted waves.
is negligible relative to its exponential dependence, we may write: Remembering the imaginary part added to k is much smaller than the real part, we may now neglect it and get: Note that
Finally, moving to the three-dimensional problem, the spherically symmetric Schrödinger equation reads (expanding the wave function
amounts to enlarging the potential, and therefore substantially reducing the decay rate (given its exponential dependence on
The main effect of this on the amplitudes is that we must replace the argument in the exponent, taking an integral of
, where we assume the nuclear potential energy is still relatively small, and
, which is where the nuclear negative potential energy is large enough so that the overall potential is smaller than E. Thus, the argument of the exponent in λ is: This can be solved by substituting
For a radium alpha decay, Z = 88, z = 2 and m = 4mp, EG is approximately 50 GeV.
with respect to E at an energy of 5 MeV to be ~ 1014 J−1, compared to the experimental value of 0.7×1014 J−1.