During nuclear explosion , one of the products is `""^(90)Sr` with half -life of 28.1 years. If 1 `mu g` of `""^(90)Sr` was absorbed in the bones of a newly born baby instead of calcium , how much of it will remain after 10 years.

Answer

To solve the problem of how much \(^{90}Sr\) will remain after 10 years, we will use the principles of first-order kinetics and the half-life of the isotope. Here’s a step-by-step solution:

### Step 1: Understand the Half-Life
The half-life (\(t_{1/2}\)) of \(^{90}Sr\) is given as 28.1 years. This means that every 28.1 years, half of the remaining quantity of \(^{90}Sr\) will decay.

### Step 2: Calculate the Rate Constant
The rate constant (\(k\)) for a first-order reaction can be calculated using the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the half-life:
\[ k = \frac{0.693}{28.1} \approx 0.0247 \text{ year}^{-1} \]

### Step 3: Use the Integrated Rate Equation
The integrated rate equation for a first-order reaction is given by:
\[ \ln\left(\frac{[A_0]}{[A_t]}\right) = kt \]
Where:
- \([A_0]\) is the initial amount (1 µg)
- \([A_t]\) is the amount remaining after time \(t\) (10 years)
- \(k\) is the rate constant
- \(t\) is the time in years

### Step 4: Substitute the Known Values
We can rearrange the equation to solve for \([A_t]\):
\[ \ln\left(\frac{1 \, \mu g}{[A_t]}\right) = 0.0247 \times 10 \]
Calculating the right side:
\[ \ln\left(\frac{1 \, \mu g}{[A_t]}\right) = 0.247 \]

### Step 5: Exponentiate to Solve for \([A_t]\)
To find \([A_t]\), we exponentiate both sides:
\[ \frac{1 \, \mu g}{[A_t]} = e^{0.247} \]
Calculating \(e^{0.247}\):
\[ e^{0.247} \approx 1.28 \]
Thus,
\[ [A_t] = \frac{1 \, \mu g}{1.28} \approx 0.7816 \, \mu g \]

### Conclusion
After 10 years, approximately \(0.7816 \, \mu g\) of \(^{90}Sr\) will remain in the baby’s bones.