A sample of a radioactive isotope used in medical imaging has an initial activity of 320\t… — JAMB UTME

A sample of a radioactive isotope used in medical imaging has an initial activity of $320\text{ MBq}$ . If its activity decreases to $20\text{ MBq}$ after an elapsed time of $15\text{ hours}$ , what is the half-life of the isotope in hours?

Answer: 3.75 hours

Answer

The half-life of the radioactive isotope is

3.75\text{ hours}

The initial activity of

320\text{ MBq}

drops to

20\text{ MBq}

, which is a reduction to

\frac{20}{320} = \frac{1}{16}

of its original value. Since

\frac{1}{16} = \left(\frac{1}{2}\right)^4

, exactly 4 half-lives have elapsed over the period of

15\text{ hours}

. Therefore, one half-life is

\frac{15\text{ hours}}{4} = 3.75\text{ hours}

Step-by-Step Solution

Calculate the fraction of initial activity remaining

Fraction remaining =

\frac{20\text{ MBq}}{320\text{ MBq}} = \frac{1}{16}

The ratio of current activity to initial activity determines the fraction of un-decayed nuclei remaining.

Determine the number of half-lives (

n

) that have elapsed

n = 4

half-lives, since

\left(\frac{1}{2}\right)^4 = \frac{1}{16}

Each half-life reduces the remaining sample activity by half.

Calculate the half-life duration (

T_{1/2}

)

T_{1/2} = \frac{t}{n} = \frac{15\text{ hours}}{4} = 3.75\text{ hours}

Dividing total elapsed time by the number of half-lives gives the duration of one half-life.

Key Concept

Radioactive Decay Law and relationship between remaining activity fraction, number of half-lives, and total elapsed time.

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