8 kg of `Cu^(66)` undergones radioactive decay and after 15 minutes only 1 kg remains . The half - life , in minutes , is then

Options

A. (a)5
B. (b)15/8
C. (c)`8//15`
D. (d)15 In (2)

Answer

To solve the problem of determining the half-life of the radioactive decay of copper, we can follow these steps:

### Step 1: Understand the Problem
We start with 8 kg of copper and after 15 minutes, only 1 kg remains. We need to find the half-life of this radioactive decay.

### Step 2: Identify Initial and Remaining Amounts
Let:
- \( N_0 \) = initial amount of copper = 8 kg
- \( N_t \) = remaining amount after time \( t \) = 1 kg
- \( t \) = time elapsed = 15 minutes

### Step 3: Use the Decay Formula
The relationship between the initial amount, remaining amount, and time can be expressed using the formula:
\[ N_t = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Where \( t_{1/2} \) is the half-life we want to find.

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 1 = 8 \left( \frac{1}{2} \right)^{\frac{15}{t_{1/2}}} \]

### Step 5: Simplify the Equation
Dividing both sides by 8 gives:
\[ \frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{15}{t_{1/2}}} \]

### Step 6: Rewrite \( \frac{1}{8} \)
Recognizing that \( \frac{1}{8} = \left( \frac{1}{2} \right)^3 \), we can rewrite the equation as:
\[ \left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{\frac{15}{t_{1/2}}} \]

### Step 7: Set Exponents Equal
Since the bases are the same, we can set the exponents equal to each other:
\[ 3 = \frac{15}{t_{1/2}} \]

### Step 8: Solve for Half-Life
Rearranging gives:
\[ t_{1/2} = \frac{15}{3} = 5 \text{ minutes} \]

### Conclusion
The half-life of the radioactive decay of copper is **5 minutes**.

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