24.112 CP - Chemistry: Molecular Nature Of Matter And Change

Question

In the 1950s, radioactive material was spread over the land from aboveground nuclear tests. A woman drinks some contaminated milk and ingests 0.500 g of ⁹⁰Sr, which is taken up by bones and teeth and not eliminated. 

(a) How much ${}^{90}Sr(t_{1/2}=29yr)$ is present in her body after 10yr ? 

(b) How long will it take for 99.9% of the ⁹⁰Sr  to decay?

Step-by-Step Solution

Answer

(a) The amount ⁹⁰Sr of that is present in her body after ten years is$N_t=0.0394 g{}^{90}Sr$ .

(b) The time required for 99.9% of the ⁹⁰Sr to decay is t=125.52 yr.

Step 1: Half-life

The half-life of a radioactive substance is defined as the amount of time required for the decay of half the initial amount of a substance. Half-life is denoted by t₁₂ .

Radioactive decay is a first-order reaction. The expression for half-life is given below.

        $t_{1/2}=\frac{0.693}{k}$           

Where, k is the decay constant.

Step 2: Find the decay constant from the half-life

Identify the constant (k) using the formula for half-life.

$$\begin{gathered} t_{1/2}=\frac{\ln 2}{k}\to k=\frac{\ln 2}{t_{1/2}} \\ k=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{29yr}= 0.023902y{r}^{-} \end{gathered}$$

Step 3: Amount of 90 Sr present after 10 years

Radioactive decay us a first order reaction. Therefore, we will use the following equation to estimate the amount of ⁹⁰Sr present in the woman’s body after 10 years.

$$\begin{gathered} N_t=N_0{e}^{-kt} \\ {N}_t=0.0500\times \left(e^{-0.023902 y{r}^{-}\times 10yr}\right) g \\ {N}_t=0.0394 g \end{gathered}$$