problem43_77

problem43_77 - 1 1 min 50 Bq 10 5 7 min 25 Bq 10 6 3 min 10...

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43.77: (a) There are two processes occurring: the creation of I 128 by the neutron irradiation, and they decay of the newly produced I 128 . So K N K dt dN where λ - = is the rate of production by the neutron irradiation. Then = - N t dt N K N d 0 0 . λ ( 29 [ ] ( 29 ( 29 ( 29 . λ 1 λ ln λ ln λ λ ln λ 0 t N e K t N t K N K t N K - - = - = - - = - b) The activity of the sample is ( 29 ( 29 ( 29 × × = - = - s decays 10 5 . 1 1 λ 6 λ t e K t N - - t e min 25 693 . 0 1 . So the activity is ( 29 ( 29 t e 02772 . 0 6 1 s decays 10 5 . 1 - - × , with t in minutes. So the activity - dt N d at various times is: Bq; 10 5 . 1 min) 180 ( Bq; 10 3 . 1 min) 75 ( Bq; 10
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Unformatted text preview: 1 . 1 min) 50 ( Bq; 10 5 . 7 min) 25 ( Bq; 10 6 . 3 min) 10 ( Bq; 10 1 . 4 min) 1 ( 6 6 6 5 5 4 × = = ′-× = = ′-× = = ′-× = = ′-× = = ′-× = = ′-dt t N d t dt N d dt t N d t dt N d dt t N d t dt N d c) ( 29 atoms 10 2 . 3 02772 . ) 60 ( ) 10 5 . 1 ( λ 9 6 max × = × = = K N . d) The maximum activity is at saturation, when the rate being produced equals that decaying and so it equals . s decays 10 5 . 1 6 ×...
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This note was uploaded on 05/12/2008 for the course PHYS 262 taught by Professor Dougherty during the Spring '08 term at Lafayette.

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