Question

Today, a part of the Earth's radiogenic heat is produced along with the radioactive
decays in the chains of 2332 Th (T12 14,0 x 10 year), 238U (T1/2 = 4.47 x 109 year), 235U
(T1/2=0.70 x 109 year), and those of the 40K isotope (T12 = 1.28 x 109 year).
The overall decay schemes and the heat released in each of these decays are summarized
in the following equations:
238 U 2136 Pb + 8 + Be + 6v. + 51.7 MeV,
235U
207 Pb + 7a+ 4e +4, +46.4 Mev,
208 Pb + 6 + 4e + 40, +427 MeV.
40K 40 Ca + e +7,+ 1.31 MeV (89.3%),
40K+ e - "Ar + v +1.505 MeV (10.7%).
2321h
1. Compute for the heat energy contributions of the decay of each isotope above in terms of
TWh. Given that 1 MeV is approximately equal to 4.45E-29 TWh.​

Answer 1

Answer:

Today, a part of the Earth's radiogenic heat is produced along with the radioactive

decays in the chains of 2332 Th (T12 14,0 x 10 year), 238U (T1/2 = 4.47 x 109 year), 235U