Question

The half-life of a radioactive substance is 15 days. After 10 days a sample of the substance has been reduced to a mass of 4 mg. A. What was the inital mass (in mg) of the sample? B. Using your answer from above, after how many days will there only be 0. 3 mg of substance left?.

Answer 1

A. The initial mass of the sample is 6.35

B. The time taken for 0.3 mg of the sample to remain is 66 days

A. Determination of the initial mass of the sample

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

Half-life (t½) = 15 days

Time (t) = 10 days

Number of half-lives (n) =?

n = t / t½

n = 10 / 15

n = ⅔

• Finally, we shall determine the initial mass of the sample.

Number of half-lives (n) = ⅔

Amount remaining (N) = 4 mg

Initial amount (N₀) =?

$N_0 = N \times {2}^{n} \\ N_0 = 4 \times {2}^{2 \div 3} \\ N_0 = 6.35 \: mg$

Therefore, the initial mass of the sample is 6.35 mg

B. Determination of the time

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

Amount remaining (N) = 0.3 mg

Initial amount (N₀) = 6.35 mg

Number of half-lives (n) =?

${2}^{n} = \frac{N_0}{N} \\ \\ {2}^{n} = \frac{6.35}{0.3} \\ \\ {2}^{n} = 21.167 \\ take \: the \: log \: of \: both \: side \\ log \: {2}^{n} = log \: 21.167 \\ nlog2 = log \: 21.167 \\ divide \: both \: side \: by \: log2 \\ \\ n = \frac{log \: 21.167}{log2} \\ \\ n = 4.4$

• Finally, we shall determine the time

Number of half-lives (n) = 4.4

Half-life (t½) = 15 days

Time(t) =?

t = n × t½

t = 4.4 × 15

t = 66 days

Therefore, it will take 66 days for 0.3 mg of the sample to remain.

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