The activity of the radioactive sample drops to 1 / 64 of its original value in 2 hr find the decay constant ?(λ).
A
(λ)=1.079hr−1
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B
(λ)=1.579hr−1
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C
(λ)=2.079hr−1
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D
(λ)=2.579hr−1
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Solution
The correct option is D(λ)=2.079hr−1 Now, we know λ=0.693halftime here activity drops to 164 in 2 hours. so half life is 20 mins then using above formula and putting half life time into it we get Ans-λ=2.579hr−1