ch42-p045 - T 1/2 = (ln 2)/(2.58 h) = 0.269 h 1 = 7.46 10 5...

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45. (a) The sample is in secular equilibrium with the source and the decay rate equals the production rate. Let R be the rate of production of 56 Mn and let λ be the disintegration constant. According to the result of Problem 42-43, R = λ N after a long time has passed. Now, λ N = 8.88 × 10 10 s – 1 , so R = 8.88 × 10 10 s – 1 . (b) We use N = R / λ . If T 1/2 is the half-life, then the disintegration constant is λ = (ln 2)/
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Unformatted text preview: T 1/2 = (ln 2)/(2.58 h) = 0.269 h 1 = 7.46 10 5 s 1 , so N = (8.88 10 10 s 1 )/(7.46 10 5 s 1 ) = 1.19 10 15 . (c) The mass of a 56 Mn nucleus is m = (56 u) (1.661 10 24 g/u) = 9.30 10 23 g and the total mass of 56 Mn in the sample at the end of the bombardment is Nm = (1.19 10 15 )(9.30 10 23 g) = 1.11 10 7 g....
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