Question

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

Answer 1

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,  

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

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

The strontium-90 remains in the bone = 52.54 %