Question

The half-life of a radioactive substance is 11 days. After 12 days a sample of the substance has been reduced
to a mass of 4 mg.

What was the initial mass (in mg) of the sample?


Using your answer from above, after how many days will there only be 0.6 mg of substance left?


(Round your answer to 2 decimal places if needed)

Answer 1

The initial mass of the sample is 8.51 mg. The time taken for 0.6 mg of the sample to remain is 42 days.

How to the determination of the initial mass of the sample can be calculated?

We'll begin by calculating the number of half-lives that have elapsed

$n = t / t^{1/2}$

where Half-life (t½)  and Time (t)

Half-life (t½) = 11 days

Time (t) = 12days

A number of half-lives (n)

$n = t / t^{1/2}\\n = 12 / 11$

Number of half-lives (n) = 12/11

Amount remaining (N) = 4 mg

Initial amount (N₀)

$N_0 = N\times 2^n\\\\N_0 = 4\times 2^{12/11}\\\\N_0 = 8.51$

Therefore, the initial mass of the sample is 8.51 mg.

Determination of the time

Amount remaining (N) = 0.6 mg

Initial amount (N₀) = 8.51 mg

Number of half-lives (n)

$2^n= \frac{N_0}{N} \\\\2^n= \frac{8.51}{0.6} \\\\2^n= 14.19$

Take log both sides

$log 2^n= log14.19\\\\n log2= log14.19\\\\n= log14.19/log2\\\\n = 3.82$

Finally, we shall determine the time

Number of half-lives (n) = 3.8

Half-life (t½) = 11 days

Time(t) t = n × t½

t = 3.8 × 11

t = 42 days

Therefore, it will take 42 days for 0.6 mg of the sample to remain.

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