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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.

Sagot :

The formula for the half life is as follows:

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

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:

[tex]\begin{gathered} N_0=10 \\ t=50 \\ t2=64.9_{} \end{gathered}[/tex]

Substitute the values into the equation.

[tex]N(50)=10\mleft(\frac{1}{2}\mright)^{\frac{50}{64.9}}[/tex]

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.

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

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