High School

Radioactive copper-64 versenate is often injected intravenously to scan the brain. It can often detect brain tumours, since the concentration of copper remaining is much greater in a tumour than in normal brain tissue. The half-life of copper-64 versenate is 12.7 hours. A patient is injected with 100 units of copper-64 versenate. Determine, to the nearest unit, how many units would be expected to remain after two days.

Answer :

To solve this problem, we need to determine how much of the initial 100 units of copper-64 versenate will remain after two days (48 hours).

  1. Understanding Half-Life: The half-life of a substance is the time it takes for half of it to decay. The half-life of copper-64 versenate is 12.7 hours.

  2. Number of Half-Life Periods: First, we calculate how many half-life periods fit into 48 hours.

    [tex]\text{Number of half-life periods} = \frac{48 \text{ hours}}{12.7 \text{ hours/half-life}} \approx 3.78[/tex]

  3. Decay Calculation: Each half-life reduces the remaining substance to half of its current amount. If you start with 100 units, it decreases by the factor of [tex]\frac{1}{2}[/tex] for each half-life period.

    The formula for the remaining quantity after a certain number of half-lives is:

    [tex]N = N_0 \times \left(\frac{1}{2}\right)^n[/tex]

    where:

    • [tex]N_0[/tex] is the initial quantity (100 units),
    • [tex]n[/tex] is the number of half-life periods.

    Substituting the values gives:

    [tex]N = 100 \times \left(\frac{1}{2}\right)^{3.78}[/tex]

  4. Calculate the Remaining Units: Calculating the above expression:

    [tex]N \approx 100 \times 0.08268 \approx 8.268[/tex]

    Rounding to the nearest unit, approximately 8 units of copper-64 versenate are expected to remain after two days.

Hence, the answer is that 8 units of copper-64 versenate would be expected to remain in the patient's body after 48 hours.

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