p43_043

# p43_043 - 1 / 2 = (ln 2) / (2 . 58 h) = 0 . 269 h 1 = 7 ....

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43. (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 41, R = λN after a long time has passed. Now, λN =8 . 88 × 10 10 s 1 , so R =8 . 88 × 10 10 s 1 . (b) They decay at the same rate as they are produced, 8 . 88 × 10 10 s 1 . (c) We use N = R/λ .I f T 1 / 2 is the half-life, then the disintegration constant is λ =(
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Unformatted text preview: 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 . (d) The mass of a 56 Mn nucleus is (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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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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