The ratio of Pu238 and O2 in Plutonium dioxide??? Hello, guys. I was wondering if I can find a ratio of Pu238 and O2 in PuO2. In my problem set, these are all I have. - 238/94 Pu -----86yrs----> 4/2 He + 234/92 U + 5.59 Mev - 238PuO2 : 6000 gram in a sphere (10 cm diameter) - Density of PuO2 : 11.46 g/cc Q1) Calculate the number of curies of 238 Pu in this sphere Q2) Calculate the power generated by this sphere (watt) Since I couldn't find how much Pu or O2 is in PuO2 respectively, I couldn't simple figure the curies of Pu. Thus I've tried to find the ration for a couple of hours on the web but nothing came out. Does anyone help me out? Thanks, Ryan.
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? Assuming that PuO_{2} is stoichiometric, the ratio O/Pu = 2. That's a simple chemical balance. There might be a surplus or deficiency of O depending on how the material is fabricated, e.g., reducing or oxidizing environment. But O/M is usually well controlled and probably on the order of 2 +/- 0.002. Note that 94-Pu-234 decays by alpha emission to 92-U-234 (also formed from Th-234 following two successive beta decays), but U likes to form UO_{2} as well, and ThO_{2} is also likely. http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radser.html#c3 Now if the measured density is greater than 100% theoretical density (of stoichiometric PuO_{2}), then there is likely a deficiency of O. If the density is less than 100% theoretical, then there is a possibility of an excess of O, or there is some porosity, or the metal is not entirely Pu.
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? The molar mass of 238-PuO2 is 276.06 g/mol by Wiki. So 6000g / (276.06g/mol) = 21.73 mol of PuO2. As the stoichiometric of PuO2, 21.73 mol / 3 = 7.24 mol 238Pu: 7.24 mol O2: 14.49 mol, right? Molar mass of 238 Pu is 244.06g/mol and O2 is 16g/mol 244.06 g/mol x 7.24mol =1766.56g 16 g/mol x 14.49 mol = 231.84g The sum is only 1998.4g. It is a lot less than 6000g. What did I wrong?
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? Relying on Wiki (Wikipedia) for starters. Please use a chemistry or physics textbook! Pu-238 should have a molar mass of ~238 gm/g-mole! Then O_{2} - with two O^{16} atoms would have a molar mass of 32 g/g-mole So the molar mass of Pu^{238}O_{2} should be about 238 + 32 = 270. If it's heavier, by 6 g/g-mol, then it's probably got some heavier isotopes of Pu or other TUs. Also, one mole of PuO_{2} has one (1) mole of Pu and 2 moles of O, or one mole of O_{2}. For atomic masses see http://www.iupac.org/publications/pac/pdf/2003/pdf/7506x0683.pdf http://iupac.org/publications/pac/pdf/2006/pdf/7811x2051.pdf or NIST - http://www.nist.gov/pml/data/comp.cfm
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? Actually I quote the atomic weigh of 238 Pu in here. http://www.webqc.org/molecular-weight-of-Pu238.html They said Symbol Element Atomic weight Number of atoms Mass percent Pu Plutonium 244.0642 238 100.0000 %
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? That's the atomic mass of Pu-244, which has a half-life of 8E7 yrs (the longest living isotope of Pu). Normally an element's atomic mass/weight is given based on naturally occurring isotopic vector, which is how one would find it in nature. But for some TU elements, it may be the atomic mass of the most stable isotope. See - http://www.nndc.bnl.gov/chart/reCenter.jsp?z=94&n=146 (select Zoom 1). Normally, if one is going to make Pu-238, one tries to ensure that most of it is Pu-238, and not heavier isotopes, 239 - 246.
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? PuO2 is a molecule, so there is no cause to redivide by 3 to segregate the Pu and O2 moles. That is the problem with your factor of 3 difference. The rest is rounding and possibly isotope composition.
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? Could you explain about why it doesn't need to redivide by 3? Then how much Pu 238 is in PuO2? 244.06 g/mol x 7.24mol =1766.56g x 3(factor) = 5299.68g 16 g/mol x 14.49 mol = 231.84g x 3 = 695.52 g
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? If you have 21.73 mols of PuO2, that is also equal to 21.73 mols each of O2 and of Pu. Matter is neither created nor destroyed in chemistry and decomposing a mol of a molecule yields same number of mols of the constituent atoms. So there is no call for a factor of 3 as you had included.
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? I appreciate to clear that up. I didn't thought about the basic rule. The last question. How to get the energy? 238/94 Pu -----86yrs----> 4/2 He + 234/92 U + 5.59 Mev Does this mean that 1 mole of Pu generates 5.5Mev?
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? I think that merely means that the decay of 1 atom of Pu 238 produces 1 atom of Helium 4 and one of U 234, along with an electron of 5.59 Mev energy. You could then use this times the number of Pu atoms in the sphere and the number of seconds in 86 years to get the total emission power.
Re: The ratio of Pu238 and O2 in Plutonium dioxide??? You are being asked to find activity from half life. λ=^{ln(2)}/_{(half life)}. A = λN = ln(2)*(Avagadro's number)*(moles)/(half life in seconds) This is the number of events (Bacquerel), each of which is a 5.5 MeV alpha. Divide by 3.7e10 to get curies. Multiply by W/eV (1.609e-19) to get Watts. Remember your freshman chemistry stoichiometric rules. One mole of PuO2 is one mole of Pu and two moles of O2. Pu238 has an exact mass of 238.0495534 AMU, Oxygen would go by the 15.9994 standard mass, since its not isotopicly separated. So your molar mass of PuO2 is 270.04835. Every 270.04835 grams will have one mole of Plutonium.