Explanation Write the expression for the number of nuclei of 94 238 Pu . N 0 = m N A M (I) Here, N 0 is the number of nuclei, m is the mass of the source, N A is the Avogadro number , M is the molar mass. Write the expression for the initial activity of the radioactive nuclei. R 0 = λ N 0 (II) Here, R 0 is the initial activity of the radioactive nuclei, λ is the decay constant. Write the expression for λ . λ = ln 2 T 1 / 2 (III) Here, T 1 / 2 is the half-life of the radioactive nuclei. Use equation (III) in (II) to solve for R 0 . R 0 = ( ln 2 T 1 / 2 ) N 0 (IV) The nuclear reaction is 94 238 Pu → 92 234 U + 2 4 He . Write the expression for the energy released in each reaction. Q = [ M 94 238 Pu − M 92 234 U − M 2 4 He ] ( 931.5 MeV / u ) (V) Here, Q is the energy released in the reaction, M 94 238 Pu is the mass of 94 238 Pu , M 92 234 U is the mass of 92 234 U , M 2 4 He is the mass of 2 4 He . Write the expression for the initial power output of the nuclear power source. P = η R 0 Q (VI) Here, P is the initial power output of the nuclear power source, η is the electrical transmission. Conclusion: Substitute 3.80 kg for m , 6.02 × 10 23 nuclei / mol for N A , 238.049560 g / mol for M in equation (I) to find N 0 . N 0 = ( 3

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Chapter 30, Problem 57P

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