Starting with a sample of pure `.^66 Cu, 7//8` of it decays into `Zn` in `15 min`. The corresponding half-life is.

Options

A. 5 min
B. `7(1)/(2)min`
C. 10 min
D. 15 min

Explanation

To find the half-life of the decay of copper-66 into zinc, we can follow these steps:

### Step 1: Understand the decay process
We start with a sample of pure \( ^{66}Cu \). According to the problem, \( \frac{7}{8} \) of the sample decays into zinc in 15 minutes. This means that only \( \frac{1}{8} \) of the original sample remains.

### Step 2: Set up the decay equation
The decay of a radioactive substance can be described by the equation:
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
where:
- \( N \) is the remaining quantity of the substance,
- \( N_0 \) is the initial quantity of the substance,
- \( t \) is the time elapsed,
- \( t_{1/2} \) is the half-life of the substance.

### Step 3: Substitute known values
From the problem, we know:
- \( \frac{N}{N_0} = \frac{1}{8} \) (since \( \frac{7}{8} \) has decayed),
- \( t = 15 \) minutes.

Substituting these values into the equation gives:
\[ \frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{15}{t_{1/2}}} \]

### Step 4: Express \( \frac{1}{8} \) in terms of powers of \( \frac{1}{2} \)
We can express \( \frac{1}{8} \) as:
\[ \frac{1}{8} = \left(\frac{1}{2}\right)^{3} \]
Thus, we can rewrite our equation as:
\[ \left(\frac{1}{2}\right)^{3} = \left(\frac{1}{2}\right)^{\frac{15}{t_{1/2}}} \]

### Step 5: Equate the exponents
Since the bases are the same, we can equate the exponents:
\[ 3 = \frac{15}{t_{1/2}} \]

### Step 6: Solve for \( t_{1/2} \)
To find \( t_{1/2} \), we rearrange the equation:
\[ t_{1/2} = \frac{15}{3} = 5 \text{ minutes} \]

### Conclusion
The half-life of \( ^{66}Cu \) is 5 minutes.

---