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The activity of the radioactive sample d...

The activity of the radioactive sample drops to `(1/64)^(th)` of its original value in 2 hr find the decay constant `(lambda)`.

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The correct Answer is:
`lambda=2.079hr^(-1)`
`lambda=(1)/(2) ln. (1)/((1//64))=2.079 hr^(-1)`
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