Coulomb correction for beta decay spectrum

1. Jul 18, 2010

statphys

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

2. Jul 19, 2010

hamster143

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. Jul 19, 2010

statphys

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 text books 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. Jul 19, 2010

the_house

Check to see if the book is using http://en.wikipedia.org/wiki/Gaussian_units" [Broken], 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: May 4, 2017
5. Jul 19, 2010

hamster143

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. Jul 19, 2010

statphys

Thanks house and hamster that's cleared up that little mystery for me.