One gram of a radioactive substance takes 50 s to lose 1 centigram. Find its half-life period

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Solution

Using law of radio activity , $N = N_{0} e^{- λ t}$
Here, $N_{0} = 1$ gram and rest substance after decay $N = (1 - 1 / 100) = 0.99$ gram ; $t = 50 s$
Now, $0.99 = e^{- 50 λ}$ or $e^{50 λ} = 1 / 0.99$
Taking ln, $50 λ = l n 1 - l n 0.99 = 0 - (- 0.01) = 0.01$
$λ = 0.0002 s^{- 1}$
Half life $t_{1 / 2} = 0.693 / λ = 0.693 / 0.0002 = 3465 s$

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Question 139

A half-life is the amount of time that it takes for a radioactive substance to decay one-half of its original quantity.
Suppose radioactive decay causes 300 grams of a substance to decrease $300 \times 2^{- 3}$ grams after 3 half-lives. Evaluate $300 \times 2^{- 3}$ to determine how many grams of the substance is left.
Explain why the expression $300 \times 2^{- n}$ can be used to find the amount of the substance that remains after n half-lives.