The half-life of the isotope `._11 Na^24` is `15` hrs. How much time does it take for `(7)/(8) th` of a sample of this isotope to decay ?

Options

A. 75 hrs
B. 65 hrs
C. 55 hrs
D. 45 hrs

Explanation

To solve the problem, we need to determine how much time it takes for \( \frac{7}{8} \) of a sample of the isotope \( \text{Na}^{24} \) to decay, given that its half-life is 15 hours.

### Step-by-Step Solution:

1. **Understanding the Decay**:
- If \( \frac{7}{8} \) of the sample has decayed, then the remaining undecayed portion is:
\[ 1 - \frac{7}{8} = \frac{1}{8} \]

2. **Using the Decay Formula**:
- The relationship between the remaining quantity \( N \) and the initial quantity \( N_0 \) can be expressed using the decay formula:
\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
- Here, \( T_{1/2} \) is the half-life of the isotope, which is 15 hours.

3. **Setting Up the Equation**:
- We know that:
\[ \frac{N}{N_0} = \frac{1}{8} \]
- Substituting this into the decay formula gives:
\[ \frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{t}{15}} \]

4. **Expressing \( \frac{1}{8} \) as a Power of \( \frac{1}{2} \)**:
- We can express \( \frac{1}{8} \) as:
\[ \frac{1}{8} = \left( \frac{1}{2} \right)^3 \]
- Thus, we can rewrite the equation as:
\[ \left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{\frac{t}{15}} \]

5. **Equating the Exponents**:
- Since the bases are the same, we can equate the exponents:
\[ 3 = \frac{t}{15} \]

6. **Solving for \( t \)**:
- To find \( t \), multiply both sides by 15:
\[ t = 3 \times 15 = 45 \text{ hours} \]

### Final Answer:
It takes **45 hours** for \( \frac{7}{8} \) of the sample of \( \text{Na}^{24} \) to decay.

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