What is the heat released from the decay of 1 kg of 40K atoms?

[Hypatio]

The radioisotope 40K decays either by electron capture (~10% of the time), releasing about 1.31 MeV, or by beta decay (90% of the time), releasing about 1.51 MeV. Thus the mean heat released is about 1.33 MeV per decay.

There are 1.5x10^25 40K atoms in 1 kilogram. Therefore the energy which could be released from complete decay of 1 kg is $$\phi$$=3.2x10^12 Joules.

The half-life of 40K is T=1.277x10^9 years, or 4.027x10^16 sec.

Therefore, we can calculate the heat produced (W/kg) from the decay of 1 kg of 40K atoms as:

$$=\phi\exp(-tln(2)/T)\frac{ln(2)}{T}=\phi\frac{ln(2)}{T}$$

I calculate 5.53x10^-5 W/kg, but numerous sources which I would otherwise consider authoritative give values which are closer to 3x10^-5 W/kg. I don't understand why our values are so different. I have calculated the heat released from many other radioisotopes (e.g. 235U, 232Th, etc.) without problems.

Am I correct, or where have I gone wrong?

 

[mfb]

Thus the mean heat released is about 1.33 MeV per decay.

This looks like the mean energy released. A significant fraction is released as neutrinos, which do not heat the material unless you happen to be in a neutron star or similar environments.

 

[Hypatio]

ok well the equations for 40K decay are

40K ---> 40Ca + e- + n + 1.31 MeV
40K+ e- ---> 40Ar + v + 1.51 MeV

where e- is an electron, n is an electron anti-neutrino, and v is an electron neutrino.

So if the energy indicated doesn't give you the heat, how do you calculate it?

 

[mfb]

You need the fraction of the energy which is released as neutrinos. For electron capture, this is nearly the whole energy. In the beta decay, electron and neutrino (not a neutron) share the released energy. The kinetic energy of the nucleus should be negligible here.
The positron annihilates afterwards, producing two gamma rays. If they get absorbed, they give additional 1.022 MeV.