`1.0 g` of a radioactive isotope left `125 mg` after 24 hr. The half-life period of the isotope is
a. 8 hr b. 24 hr c. 6 hr d. 4 hr

Options

A. a. 8 hr
B. b. 24 hr
C. c. 6 hr
D. d. 4 hr

Explanation

To solve the problem of determining the half-life of a radioactive isotope that decayed from 1.0 g to 125 mg in 24 hours, we can follow these steps:

### Step 1: Convert units
First, we need to express the initial and final amounts in the same units.
- Initial amount, \( A_0 = 1.0 \, \text{g} = 1000 \, \text{mg} \)
- Final amount, \( A = 125 \, \text{mg} \)

### Step 2: Use the radioactive decay formula
The formula for radioactive decay is given by:
\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Where:
- \( N_0 \) is the initial quantity (1000 mg)
- \( N \) is the remaining quantity (125 mg)
- \( t \) is the time elapsed (24 hours)
- \( t_{1/2} \) is the half-life we want to find.

### Step 3: Set up the equation
We can rearrange the formula to solve for \( t_{1/2} \):
\[ 125 = 1000 \left( \frac{1}{2} \right)^{\frac{24}{t_{1/2}}} \]

### Step 4: Simplify the equation
Dividing both sides by 1000 gives:
\[ 0.125 = \left( \frac{1}{2} \right)^{\frac{24}{t_{1/2}}} \]
Recognizing that \( 0.125 = \frac{1}{8} = \left( \frac{1}{2} \right)^3 \), we can rewrite the equation as:
\[ \left( \frac{1}{2} \right)^{\frac{24}{t_{1/2}}} = \left( \frac{1}{2} \right)^3 \]

### Step 5: Equate the exponents
Since the bases are the same, we can equate the exponents:
\[ \frac{24}{t_{1/2}} = 3 \]

### Step 6: Solve for \( t_{1/2} \)
Now, we can solve for \( t_{1/2} \):
\[ t_{1/2} = \frac{24}{3} = 8 \, \text{hours} \]

### Conclusion
The half-life of the radioactive isotope is **8 hours**. Therefore, the answer is option **a**.

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