The power irradiated by the Earth through its surface is about 40 TW. A large fraction of

Question:

The power irradiated by the Earth through its surface is about 40 TW. A large fraction of this energy is due the decay of radioactive isotopes. The most important are 238U, 232Th and 40K. In the ‘standard’ model built by the geologists, the Bulk Silicate Earth Model, radioactivity is assumed to contribute 50% of the total power,i.e. with 20 TW. The model contains important uncertainties and can be tested by detecting the electron antineutrinos produced in the decays. Their flux can be measured by observing the process ν̅e + p → e+ + n in a medium containing free protons such as a liquid scintillator. The energy threshold is Eν ≥ 1.8 MeV (see Problem 10.8). The maximum antineutrino energy is Eν, max = 3.26 MeV for 238 U,Eν, max = 2.25 MeV for 232Th and Eν, max = 1.31 MeV for 40K, so only U and Th produce antineutrinos above threshold. The total antineutrino flux at the surface dueto 238U and 232Th is expected to be Φν = 3.5 x 1010 m-2 s-1. It is evaluated that the fraction Pee  ≈  0.6 will reach our detector as ν̅e due to the oscillations. Moreover,only the fraction f ≃ 0.05 of the flux is above detection threshold. Assume an average cross-section value σ ≃ 10-47 m2. We take as scintillator a blend of 20% PXE (C16H18) and 80% dodecane (C16H26) as proposed by the LENA proposal. Calculate the sensitive scintillator mass necessary to observe 1000 events.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  answer-question
Question Posted: