Half life of 90Sr is 6.93 years. In a child body 1 mg of 90Sr dopped in place of calcium, how many years will it take to reduce its concentration by 90% (Assume no involvement of Sr in metabolism)

Answer

To solve the problem of how long it will take for the concentration of 90Sr to reduce by 90%, we can follow these steps:

### Step 1: Understand the Problem
We know that the half-life of 90Sr (Strontium-90) is 6.93 years. We need to find out how long it will take for 1 mg of 90Sr to reduce to 0.1 mg (which is 10% of the original amount, meaning a reduction of 90%).

### Step 2: Calculate the Decay Constant (k)
The decay constant \( k \) can be calculated using the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
where \( t_{1/2} \) is the half-life. Plugging in the values:
\[ k = \frac{0.693}{6.93} \approx 0.1 \text{ years}^{-1} \]

### Step 3: Use the First-Order Kinetics Formula
For first-order kinetics, the relationship between the initial amount \( A_0 \), the final amount \( A \), and time \( t \) is given by:
\[ t = \frac{1}{k} \ln\left(\frac{A_0}{A}\right) \]
Here, \( A_0 = 1 \text{ mg} \) and \( A = 0.1 \text{ mg} \).

### Step 4: Substitute Values into the Formula
Substituting the known values into the equation:
\[ t = \frac{1}{0.1} \ln\left(\frac{1}{0.1}\right) \]
\[ t = 10 \ln(10) \]

### Step 5: Calculate \( \ln(10) \)
Using the value of \( \ln(10) \approx 2.303 \):
\[ t = 10 \times 2.303 = 23.03 \text{ years} \]

### Conclusion
Thus, it will take approximately **23.03 years** for the concentration of 90Sr in the child's body to reduce by 90%.

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