Coulomb correction for beta decay spectrum

In summary, the conversation discusses an expression, F(Z,E), used to correct the beta decay spectrum for the coulomb effect of the nucleus on the ejected beta particle. The source of the expression is mentioned and the values for the constants used are given. The conversation then delves into a discrepancy with the values obtained and the possibility of a change in units since the textbook was written. It is eventually concluded that the discrepancy is due to a missing factor in the expression when using SI units. The correct equation in SI units is provided.
  • #1
statphys
18
0
Hello

I am re-reading some of my old textbooks and have come across a simplified (non-relativistic) term used to correct the beta decay spectrum for the coulomb effect of the nucleus on the ejected beta particle. The expression is;
F(Z,E) = (2*Pi*n) / (1 - exp(-2*Pi*n))
where n = (Z*e*e)/(hbar*v)
Z= atomic no of daughter nucleus
e= charge of the electron
hbar = Plancks constant (divided by 2 Pi)
v= speed of the electron

Using SI values for each of the constants and assuming Z is 50 and that the speed of the beta particle is 10% of the speed of light I get a really small number for n and hence non-sensical values for F(Z,E).

The book I am using is "Theoretical Nuclear Physics" by Blatt and Weisskopf which is circa 1955 so perhaps there is an issue with units? Any help sorting out this issue would be greatly appreciated.

Thanks
 
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  • #2
I have a source that gives n = Z*alpha/beta, where alpha is 1/137 and beta is v/c. For Z=50 and the particle at 10% the speed of light, that gives n=3.6 and F~23.

If n were to be really small, F=2*pi*n/(1-exp(-2*pi*n)) ~ 2*pi*n/(2*pi*n-(2*pi*n)^2/2) ~ 1 + pi*n.
 
  • #3
hamster143 said:
I have a source that gives n = Z*alpha/beta, where alpha is 1/137 and beta is v/c. For Z=50 and the particle at 10% the speed of light, that gives n=3.6 and F~23.

If n were to be really small, F=2*pi*n/(1-exp(-2*pi*n)) ~ 2*pi*n/(2*pi*n-(2*pi*n)^2/2) ~ 1 + pi*n.

Hi Hamster

Thanks for the reply. I think what you have is more or less consistent with the expression I stated. The value you quote for alpha looks to be the fine structure constant, which is defined in textbooks as
e^2/(hbar*c)
where c = speed of light
trouble is with SI units for everything the equation needs to be divided by 4*Pi*epsilon
where epsilon = 8.8542E-12 and is known as the permittivity of free space.

What is really confusing me is that my old textbook doesn't mention this extra factor and I was wondering whether there has been some subtle change of units involved since this textbook was written - any old physicists out there got any clues?
 
  • #4
statphys said:
trouble is with SI units for everything the equation needs to be divided by 4*Pi*epsilon
where epsilon = 8.8542E-12 and is known as the permittivity of free space.

Check to see if the book is using http://en.wikipedia.org/wiki/Gaussian_units" , or some other variant of cgs units. That's often the cause of missing factors of 4*Pi*epsilon_0 compared to SI units, although I don't know off the top of my head what the correct formula should be in your case.
 
Last edited by a moderator:
  • #5
Right, if you work in SI, you need an extra term of epsilon_0 to get things right. In CGS/Gaussian, charge has dimensionality that's expressed through other units and the expression (Z*e*e)/(hbar*v) can be dimensionless. In SI, the same equation is not dimensionless because you have two powers of coulomb (or ampere) and nothing to balance them. You need to divide by epsilon_0 to get a dimensionless quantity.
 
  • #6
Thanks house and hamster that's cleared up that little mystery for me.
 

What is the Coulomb correction for beta decay spectrum?

The Coulomb correction for beta decay spectrum is a mathematical adjustment that takes into account the repulsive Coulomb force between the beta particle and the daughter nucleus in beta decay. This correction is necessary for accurately predicting the shape and energy of the beta decay spectrum.

What is the purpose of the Coulomb correction?

The Coulomb correction is important because it accounts for the influence of the strong electromagnetic force on the beta decay process. Without this correction, the predicted spectrum of beta particles would not accurately reflect the actual distribution of energies observed in experiments.

How is the Coulomb correction calculated?

The Coulomb correction is calculated using a mathematical formula that takes into account the energy of the beta particle, the mass difference between the parent and daughter nuclei, and the distance between the two nuclei. This calculation can be quite complex and is typically done using computer programs.

What is the significance of the Coulomb correction in nuclear physics?

The Coulomb correction plays a crucial role in nuclear physics, as it allows for more accurate calculations of beta decay rates and energies. This is important for understanding the stability and decay of radioactive nuclei, as well as for applications such as nuclear power and medical imaging.

How does the Coulomb correction affect the shape of the beta decay spectrum?

The Coulomb correction typically results in a distortion of the beta decay spectrum, causing it to shift towards lower energies. This is due to the repulsive Coulomb force, which reduces the maximum energy that the beta particle can have after being emitted from the nucleus. This results in a peak at lower energies and a longer tail in the spectrum.

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