During the nuclear explosion, one of the products is `._(90)Sr` with half life of 6.93 years. If `mug" of "._(90)Sr` was absorbed in the bones of a newly born in place of Ca, how much time (in years) is required to reduce it by `90%`. If it is not lost metabolically? Report your answer by rounding it up to a nearest whole number.

Answer

To solve the problem, we need to determine how long it takes for the amount of Strontium-90 (90Sr) to reduce by 90%. We will use the half-life of 90Sr and the first-order kinetics of radioactive decay.

### Step-by-Step Solution:

1. **Identify the Half-Life**:
The half-life of 90Sr is given as 6.93 years.

2. **Calculate the Rate Constant (k)**:
The relationship between half-life (t₁/₂) and the rate constant (k) for a first-order reaction is given by the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the half-life:
\[ k = \frac{0.693}{6.93} \approx 0.1 \text{ years}^{-1} \]

3. **Determine Initial and Final Concentrations**:
Let's assume the initial concentration of 90Sr is 1 microgram. To reduce this by 90%, the final concentration (C) will be:
\[ C = 1 \text{ microgram} - 0.9 \text{ microgram} = 0.1 \text{ microgram} \]

4. **Use the First-Order Kinetics Formula**:
The formula to calculate the time (t) for first-order reactions is:
\[ t = \frac{2.303}{k} \log\left(\frac{[C_0]}{[C]}\right) \]
Where:
- \( [C_0] \) = initial concentration = 1 microgram
- \( [C] \) = final concentration = 0.1 microgram

5. **Substitute the Values**:
Now we can substitute the values into the formula:
\[ t = \frac{2.303}{0.1} \log\left(\frac{1}{0.1}\right) \]
\[ t = \frac{2.303}{0.1} \log(10) \]
Since \(\log(10) = 1\):
\[ t = \frac{2.303}{0.1} \times 1 = 23.03 \text{ years} \]

6. **Round Up the Answer**:
Rounding 23.03 to the nearest whole number gives us:
\[ t \approx 23 \text{ years} \]

### Final Answer:
The time required to reduce the amount of 90Sr by 90% is approximately **23 years**.