The mean lives of a radioactive substance are 1620 years& and 405 years for α and β emission respectively. If it is decaying by both α and β & emission simultaneously, then& the time during which three-fourths of the

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Correct answer: B. 449 & years

λ = λ _ α + λ _ β cases = 1 T _ α + 1 T _ β & [∵ λ = (1)/(T)] cases = (1)/(1620) + (1)/(405) [given, T _ α = 1620yr and T _ β = 405yr ] = (5)/(1620)yr^ - 1 (3)/(4)th sample will decay, i.e. remaining (1)/(4)th, N = N _ 0 ((1)/(2)) ^ n N _ 0 4 = N _ 0 ((1)/(2)) ^ n ⇒ n = 2 ∴ t = n T _ (1)/(2) = n( 2)/(λ) = 2 × (0.693)/((5)/(1620)) = 449yr

The mean lives of a radioactive substance are 1620 years& and 405...

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