Question

The isotope Sr-85 is used in bone scans. It has a half-life of 64.9 days. If you start with a10-mg Sample, how much would be remaining after 50 days? Round to the nearest hundredth.

Answer 1

The formula for the half life is as follows:

`N(t)=N_0\mleft((1)/(2)\mright)^{\frac{t}{(t_{_{_(1))}}}}`

where N(t) is the final amount, N₀ is the initial amount, t is the time that passed, and t2 is the half-life.

The following are the given values in the problem:

`\begin{gathered} N_0=10 \\ t=50 \\ t2=64.9_{} \end{gathered}`

Substitute the values into the equation.

`N(50)=10\mleft((1)/(2)\mright)^{(50)/(64.9)}`

Simplify the right side of the equation. Divide 50 by 64.9 and then raise 1/2 by the obtained quotient. And finally, multiply the obtained value by 10.

`\begin{gathered} N(50)\approx10\mleft((1)/(2)\mright)^(0.7704160247) \\ \approx10(0.5862483959) \\ \approx5.862483959 \end{gathered}`

Therefore, after 50 days, it will become approximately 5.86 mg.