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A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is____(Nearest integer) (Given : antilog 0 . 125 = 1 . 333 , antilog 0 . 693 = 4 . 93 )

Correct answer

0

Step-by-step solution

For first order reaction, K = 2 . 303 t log N o N t ... 1 Also, K = 0 . 693 t 1 2 ... 2 Equating 1 and 2 Hence, 2 . 303 t log N o N t = 0 . 693 t 1 2 2 . 303 83 log N o N t = 0 . 693 200 N o N t = 4 3 % N t N 0 = 0 . 75 × 100 = 75 %

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