Strontium 90 is one of the harmful nuclides resulting from nuclear fission explosions.
Strontium 90 decays by beta particle emission with a half life of 28 years. How long (in years) would it take for 99.0% of a sample of strontium-90 released in an atmospheric test of an atomic bomb to decay? (log 2 = 0.30)

Explanation

To solve the problem of how long it would take for 99.0% of a sample of strontium-90 to decay, we can follow these steps:

### Step 1: Understand the decay process
Strontium-90 decays by beta emission and has a half-life of 28 years. We need to find out how long it takes for 99.0% of the sample to decay. If 99.0% has decayed, then 1.0% remains.

### Step 2: Use the decay formula
The decay of a radioactive substance can be described by the equation:
\[ A = A_0 e^{-\lambda t} \]
Where:
- \( A \) is the remaining quantity after time \( t \)
- \( A_0 \) is the initial quantity
- \( \lambda \) is the decay constant
- \( t \) is the time elapsed

### Step 3: Set up the equation
Since we want to find the time when 1% remains, we can set:
\[ A = 0.01 A_0 \]
Thus, the equation becomes:
\[ 0.01 A_0 = A_0 e^{-\lambda t} \]

### Step 4: Simplify the equation
We can divide both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ 0.01 = e^{-\lambda t} \]

### Step 5: Take the natural logarithm of both sides
Taking the natural logarithm gives:
\[ \ln(0.01) = -\lambda t \]

### Step 6: Calculate \( \lambda \)
The decay constant \( \lambda \) can be calculated using the half-life:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Given that the half-life \( t_{1/2} = 28 \) years:
\[ \lambda = \frac{\ln(2)}{28} \]

### Step 7: Substitute \( \lambda \) back into the equation
Now we substitute \( \lambda \) back into the equation:
\[ \ln(0.01) = -\left(\frac{\ln(2)}{28}\right) t \]

### Step 8: Solve for \( t \)
Rearranging gives:
\[ t = -\frac{28 \ln(0.01)}{\ln(2)} \]

### Step 9: Calculate \( \ln(0.01) \)
We know that:
\[ \ln(0.01) = \ln(10^{-2}) = -2 \ln(10) \]
Using \( \ln(10) \approx 2.303 \):
\[ \ln(0.01) \approx -2 \times 2.303 = -4.606 \]

### Step 10: Substitute \( \ln(0.01) \) into the equation
Now substituting this value into the equation for \( t \):
\[ t = -\frac{28 \times (-4.606)}{\ln(2)} \]

### Step 11: Calculate \( \ln(2) \)
Using the value given in the problem:
\[ \ln(2) \approx 0.693 \]

### Step 12: Final calculation
Now substituting \( \ln(2) \):
\[ t = \frac{28 \times 4.606}{0.693} \]
Calculating this gives:
\[ t \approx \frac{128.968}{0.693} \approx 185.85 \text{ years} \]

### Conclusion
It would take approximately **185.85 years** for 99.0% of a sample of strontium-90 to decay.

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