problem43_37

# problem43_37 - = ⋅ ∆ =-γ E N E So the absorbed dose is...

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43.37: a) We need to know how many decays per second occur. . s 10 79 . 1 ) y s 10 156 . 3 ( ) y 3 . 12 ( 693 . 0 693 . 0 λ 1 9 7 2 1 - - × = × = = T The number of tritium atoms is 1 9 10 s 10 1.79 ) Ci Bq 10 (3.70 Ci) (0.35 λ 1 ) 0 ( - - × × = = dt dN N 18 10 2540 . 7 ) 0 ( × = N nuclei. The number of remaining nuclei after one week is just 18 ) s 3600 ( ) 24 ( ) 7 ( ) s 10 79 . 1 ( 18 λ 10 7.2462 week) 1 ( ) 10 25 . 7 ( ) 0 ( ) week 1 ( 1 9 × = × = = - - × - - N e e N N t nuclei decays. 10 7.8 week) 1 ( ) 0 ( 15 × = - = N N N So the energy absorbed is J. 6.24 ) eV J 10 (1.60 eV) 5000 ( ) 10 8 . 7 ( 19 15 total = × ×
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Unformatted text preview: = ⋅ ∆ =-γ E N E So the absorbed dose is rad. 5 . 12 kg J 125 . kg) (50 J) (6.24 = = Since RBE = 1, then the equivalent dose is 12.5 rem. b) In the decay, antinentrinos are also emitted. These are not absorbed by the body, and so some of the energy of the decay is lost (about 12 keV )....
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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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