First slide

Radio activity

Question

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

Easy

A

10 days
B

20 days
C

40 days
D

None of these

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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