Horace Heffner writes > I have done plenty of tritium counting using liquid scintillation counting. > I think it is more difficult to count water borne tritium by other means. > Scintillation couters can reliably and automatically discriminate between > tritium and say carbon 14. There is almost no penetrating power for 20 keV > beta particles, so counting 201 Tl without interference from tritium is > easy.
Despite your expertise, your conclusion is debatable, depending on the sophistication of the detector... and perhaps depending on an operator with less extenisive background ;-) . See below. > BTW, my handbook shows 201 Tl decaying by electron capture (1.36 MeV) with > Hg and K shell x-rays of 135.28 keV and 167.40 keV. This stuff should > stand out like the sun on a clear day. Let me direct your attention to "Thallium online" http://www.rxlist.com/cgi/generic/thallium.htm You will see that over 95% of the gammas in this situation would have a mean energy between 68-80 KeV but are coming from the transitory mercury isotope as the Tl life is so short. After an extended run, and with such a small amount used, and with a starting half-life of only 70+ hours, there is almost no Tl left to measure at the end of the run. As you say, the end point for tritium betas is around 20 KeV and nearly all would be absorbed in the water. The Radiation Yield (Y) from bremsstrahlung can be calculated using the following Y=(6x10^-4(ZT))/(1+6x10^-4(ZT)) Where Z is the atomic #; T is the Kinetic E. of the beta in MeV. for an average energy of 6keV you get: Y=(6x10^-4(4*.006)/(1+6x10^-4(4*.006)) =1.44x10^-5 Which is the fraction of the 6 keV converted to photons as the Beta particle slows down. ...or, the standard approximation is ZE/3000 where E is the maximum beta energy i.e. 0.0186 MeV. From Evan's "The Atomic Nucleus" ... This gives (for Be) 4 x 0.0186/3000 or 2.5 E-5, roughly twice the value above. Anyway if lots of tritium was being produced, a fair amount of the bremsstrahlung gamma photons of about 3-6 keV would be seen. This "should be" easily discriminated from the Tl emission, but not necessarily so - depending on the detector used and how the results were interpreted. That is why I asked the question. Jones
