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The half-life of radioisotopic bromine - 82 is 36 hours. The fraction which remains after one day is _________ × 10 - 2 . (Given antilog 0 . 2006 = 1 . 587 ) Round off to the nearest integer

Correct answer

0

Step-by-step solution

Half life of bromine - 82 = 36 hours t 1 / 2 = 0 . 693 K K = 0 . 693 36 = 0 . 01925 hr - 1 K = 2 . 303 t log N 0 N t t = 1 day = 24 hr log N 0 N t = 24 hr × 0 . 01925 hr - 1 2 . 303 = 0 . 2006 O r N 0 N t = anti log 0 . 2006 N 0 N t = 1 . 587 Or N t N 0 = 1 1 . 587 = 0 . 630 Hence, the fraction remain after day = 63 × 10 - 2

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