Explanation Given data: 99.28% of 238 U and 0.72% of 235 U is present. Half-life of 235 U is t 1 2 = 7.04 × 10 8 year and 238 U is t 1 2 = 4.47 × 10 9 year . Formula used: Consider the expression for decaying of nuclei. N = N 0 ( 1 2 ) t t 1 2 Here, N 0 is the initial value of nuclei, t is the time taken for decay, and t 1 2 is the half-life. Explanation: Ratio of 238 U is to 235 U at the present time is, N 238 N 235 = ( N 0 ) 238 ( 1 2 ) t t 1 2 ( N 0 ) 235 ( 1 2 ) t t 1 2 0.9928 0.0072 = 1 × ( 1 2 ) t ( t 1 2 ) 238 ( 1 2 ) t ( t 1 2 ) 235 137.89 = ( 1 2 ) t ( 1 ( t 1 2 ) 238 − 1 ( t 1 2 ) 235 ) Take natural logarithm on both sides

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College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)

Ch. 30 - Prob. 1CQ Ch. 30 - Prob. 2CQ Ch. 30 - Prob. 3CQ Ch. 30 - Prob. 4CQ Ch. 30 - Prob. 5CQ Ch. 30 - Prob. 6CQ Ch. 30 - Figure Q30.7 shows how the number of nuclei of one...Ch. 30 - Prob. 8CQ Ch. 30 - Prob. 9CQ Ch. 30 - Prob. 10CQ

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Suppose you have a pure radioactive material with a half-life of T1/2. You begin with N0 undecayed nuclei of the material at t = 0. At t=12T1/2, how many of the nuclei have decayed? (a) 14N0 (b) 12N0(C) 34N0 (d) 0.707N0 (e) 0.293N0
(a) Calculate the energy released in the a decay of 238U . (b) What fraction of the mass of a single 238U is destroyed in the decay? The mass of 234Th is 234.043593 u. (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?
(a) How many 239Pu nuclei must fission to produce a 20.0kT yield, assuming 200 MeV per fission? (b) What is the mass of this much 239Pu?

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