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85sr is a short-lived (half-life 65 days) isotope used in bone scans. a typical patient receives a dose of 85sr with an activity of 0.10 mci.

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Final answer:

The question asks about the time it takes for the activity of strontium-85 (85Sr) to reduce to 0.100 mCi. Given the half-life of 85Sr and the initial activity, we can use a formula to calculate the time required.

Step-by-step explanation:

The question asks about the time it takes for the activity of strontium-85 (85Sr) to reduce to 0.100 mCi. Given that the half-life of 85Sr is 65 days and a typical patient receives a dose with an activity of 0.10 mCi, we can calculate the time it takes for the activity to decrease to 0.100 mCi.

To do this, we can use the formula:Final activity = Initial activity * (0.5)^(t/half-life)

Where:
Final activity = 0.100 mCi
Initial activity = 0.10 mCi
Half-life = 65 days
t = time in days

By substituting the given values into the formula, we can solve for t to find the time required for the activity to reduce to 0.100 mCi.

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User Phil Leh
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