IDontKnowReally
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Geothermal energy originates as energy released in the radioactive decay of ht uranium isotope U-238. Each nucleus that decays releases 4.2MeV.
Calculate the mass of U-238 that would release energy at a rate of 500MW.
Half life = 4.5 x 10^9 years
Molar mass of U-238 = 0.238kg/mol

Here is what I attempted:
Energy released per second = 500MJ = 3.125 x 10^27 eV
If each decay releases 4.2MeV of energy, then per second there must be around 7.44 x 10^20 decays.
The no. of decays a second = Activity = Decay Constant * No of nuclei
Decay constant = ln(2)/half life in seconds = 4.88 x 10^-18
Subbing that back into the Activity equation gave N = 1.52 x 10^38
Since N = number of moles * Avogadros constant, and mass = molar mass * moles, I got mass to be 6.02 x 10^13 kg.

However the answer is 5.90 x 10^13 kg.
Where have I gone wrong? Is the method incorrect?
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Eimmanuel
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#2
Report 4 years ago
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(Original post by IDontKnowReally)
Geothermal energy originates as energy released in the radioactive decay of ht uranium isotope U-238. Each nucleus that decays releases 4.2MeV.
Calculate the mass of U-238 that would release energy at a rate of 500MW.
Half life = 4.5 x 10^9 years
Molar mass of U-238 = 0.238kg/mol

Here is what I attempted:
Energy released per second = 500MJ = 3.125 x 10^27 eV
If each decay releases 4.2MeV of energy, then per second there must be around 7.44 x 10^20 decays.
The no. of decays a second = Activity = Decay Constant * No of nuclei
Decay constant = ln(2)/half life in seconds = 4.88 x 10^-18
Subbing that back into the Activity equation gave N = 1.52 x 10^38
Since N = number of moles * Avogadros constant, and mass = molar mass * moles, I got mass to be 6.02 x 10^13 kg.

However the answer is 5.90 x 10^13 kg.
Where have I gone wrong? Is the method incorrect?
Nothing is wrong. I think the given answer uses 3.1E7 s in one year while you use 3600*24*365 s.
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sc4rface
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Report 2 years ago
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Could you please explain how you got the number of nuclei? I dont get how the molar mass is used in these questions.

Thanks
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