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Q.

If a radioactive substance reduces to $\frac{1}{16}$ of its original mass in 40 days, what is its halt-life?

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a

20 days

b

None of these

c

40 days

d

10 days

answer is A.

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Detailed Solution

$\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow \frac{1}{16} = {(\frac{1}{2})}^{4} = {(\frac{1}{2})}^{n} \Rightarrow n = 4 Also n = \frac{t}{T_{1 / 2}} \Rightarrow T_{1 / 2} = \frac{40}{4} = 10 days$

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