## Radioactivity of U-238

This is just an example of how to calculate the radioactivity of an isotope from its halflife.

Uranium 238 has a halflife of 4.468×10^{9} years (4,468,000,000years). This is 4.468×10^{9 }x 365.25 x 24 x 60 x 60 = 1.41×10^{17} seconds.

We can now calculate the decay constant (i.e. the number of decays per second) using the formula derived in Rates of Decay:

Rearranging we get

So we get a decay constand of ln(2) / 1.41×10^{17 }=4.916×10^{-18} per second.

We now want to work out how many atoms there are in 1 tonne (1,000Kg = 1×10^{6}g) of Uranium 238. We know that 238g of U-238 contains 6.022×1023 atoms from the definition of atomic mass.

Therefore 1 tonne contains 1×10^{6}/238 x 6.022×10^{23} = 2.53 x 10^{27} atoms. Its activity would therefore be 2.53 x 10^{27 }x 4.916×10^{-18 }= 1.24×10^{10} becquerel.

However the decay product – Thorium – 234 is not stable (halflife 24 days) and decays via Î² decay to Pa-234 which undergoes Î² decay (halflife 1.17 seconds) to U-234 (halflife 240,00years). So you get many more emissions from on U-238 atom decaying than just the one Î± particle you might expect.

I would assume that that there is some U-234 in fresh nuclear fuel. However it is likely that its decay product Th-230 (halflife 77,000 years) was removed during the chemical separation of the uranium from the ore. It will take some time for the Th-230 to build up to such an extent that its decay products add considerably to the activity of the U-238.

We would expect the U-238,Â Th-234, Pa-234, U-234 and Th-230 to be in secular equilibrium in the fresh fuel i.e. their activities to be the same.

Therefore for every U-238 decay we get

U-238 | Î±,Î³ |

Th-234 | Î²,Î³ |

Pa-234 | Î²,Î³ |

U-234 | Î±,Î³ |

a total of eight emissions.

Hi Peter, is the decay constant correct? Should it not be 4.9e-18 instead of 4.9e-16?

You are correct. Not sure how I made this error but I shall correct it in the post. Thanks very much for pointing it out. Just had a look at it and it seems to be a typo since the figure of 1.24e10 Bq is correct.