Starting with a sample of pure `^66Cu`, `3/4` of it decays into Zn in 15 minutes. The corresponding half-life is

Options

A. (a) 5 minutes
B. (b) 7.5 minutes
C. (c) 10 minutes
D. (d) 3.5 minutes

Explanation

To find the half-life of the sample of pure \( ^{66}Cu \), we can follow these steps:

### Step 1: Understand the decay process
We start with a sample of \( ^{66}Cu \). According to the problem, \( \frac{3}{4} \) of the sample decays into \( Zn \) in 15 minutes. This means that only \( \frac{1}{4} \) of the original sample remains.

### Step 2: Set up the equation for decay
The relationship between the remaining quantity of a substance and its half-life can be expressed as:
\[ N = N_0 \left(\frac{1}{2}\right)^n \]
where:
- \( N \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( n \) is the number of half-lives that have passed.

### Step 3: Determine the remaining quantity
From the problem, we know that:
\[ N = \frac{1}{4} N_0 \]
This indicates that \( \frac{3}{4} \) of the sample has decayed, which means \( N_0 - N = \frac{3}{4} N_0 \).

### Step 4: Relate the remaining quantity to half-lives
Since \( N = N_0 \left(\frac{1}{2}\right)^n \) and we have \( N = \frac{1}{4} N_0 \), we can set up the equation:
\[ \frac{1}{4} N_0 = N_0 \left(\frac{1}{2}\right)^n \]
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{4} = \left(\frac{1}{2}\right)^n \]

### Step 5: Solve for \( n \)
Recognizing that \( \frac{1}{4} = \left(\frac{1}{2}\right)^2 \), we can equate the exponents:
\[ n = 2 \]
This means that 2 half-lives have passed in 15 minutes.

### Step 6: Calculate the half-life
Since 2 half-lives correspond to 15 minutes, we can find the duration of one half-life (\( t_{1/2} \)):
\[ 2 \cdot t_{1/2} = 15 \text{ minutes} \]
Thus,
\[ t_{1/2} = \frac{15}{2} = 7.5 \text{ minutes} \]

### Final Answer
The corresponding half-life of \( ^{66}Cu \) is \( 7.5 \) minutes.

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