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adoni [48]
3 years ago
8

In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones o

f people alive at the time. If the half-life of strontium-90 is 29 years, what fraction of the strontium-90 absorbed in 1964 remained in people's bones in 1991?
Chemistry
1 answer:
Advocard [28]3 years ago
5 0

Answer:

52.54 %

Explanation:

Half life = 29 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{29.0}\ {years}^{-1}

The rate constant, k = 0.023902 hour⁻¹

From 1964 to 1991:

Time = 27 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,  

\frac {[A_t]}{[A_0]}=e^{-0.023902\times 27}

\frac {[A_t]}{[A_0]}=0.5245

<u>The strontium-90 remains in the bone = 52.54 %</u>

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