The half-life of a certain radioactive element is such that 7/8 of a given quantity decays in 12 days. What fraction remains undecayed after 24 days?

Options

A. 0
B. `1/128`
C. `1/64`
D. `1/32`

Explanation

To solve the problem, we need to determine the fraction of the radioactive element that remains undecayed after 24 days, given that 7/8 of the quantity decays in 12 days.

### Step-by-Step Solution:

1. **Understanding the Decay**:
- We start with an initial quantity \( A \).
- After 12 days, 7/8 of this quantity has decayed. This means the remaining quantity after 12 days is:
\[ A - \frac{7}{8}A = \frac{1}{8}A \]

2. **Finding the Decay Constant (\( \lambda \))**:
- We use the radioactive decay formula:
\[ N(t) = N_0 e^{-\lambda t} \]
- Here, \( N_0 \) is the initial quantity \( A \), and \( N(t) \) is the quantity remaining after time \( t \).
- After 12 days, we have:
\[ \frac{1}{8}A = A e^{-\lambda \cdot 12} \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{8} = e^{-\lambda \cdot 12} \]
- Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{8}\right) = -\lambda \cdot 12 \]
- We can rewrite \( \frac{1}{8} \) as \( 2^{-3} \):
\[ \ln(2^{-3}) = -3 \ln(2) = -\lambda \cdot 12 \]
- Therefore, we can solve for \( \lambda \):
\[ \lambda = \frac{3 \ln(2)}{12} = \frac{\ln(2)}{4} \]

3. **Finding the Remaining Quantity After 24 Days**:
- Now we need to find the quantity remaining after 24 days:
\[ N(24) = A e^{-\lambda \cdot 24} \]
- Substituting \( \lambda \):
\[ N(24) = A e^{-\frac{\ln(2)}{4} \cdot 24} = A e^{-6 \ln(2)} = A \left(e^{\ln(2)}\right)^{-6} = A \left(2^{-6}\right) \]
- Thus, the remaining quantity is:
\[ N(24) = \frac{A}{64} \]

4. **Finding the Fraction Remaining**:
- The fraction of the original quantity that remains undecayed after 24 days is:
\[ \text{Fraction remaining} = \frac{N(24)}{A} = \frac{A/64}{A} = \frac{1}{64} \]

### Final Answer:
The fraction of the radioactive element that remains undecayed after 24 days is \( \frac{1}{64} \).

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