The half-life of radioactive material is `3 h`. If the initial amount is `300 g`, then after `18 h`, it will remain:
(1.) 4.68g
(2.) 46.8g
(3.) 9.375g
(4.) 93.75g

Options

A. `4.68 g`
B. `46.8 g`
C. `9.375 g`
D. `93.75 g`

Answer

To solve the problem, we will use the formula for radioactive decay based on half-life. The formula to calculate the remaining amount of a radioactive substance after a certain time is given by:

\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]

Where:
- \( N \) is the remaining amount of the substance.
- \( N_0 \) is the initial amount of the substance.
- \( t \) is the total time elapsed.
- \( t_{1/2} \) is the half-life of the substance.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial amount, \( N_0 = 300 \, \text{g} \)
- Half-life, \( t_{1/2} = 3 \, \text{h} \)
- Total time, \( t = 18 \, \text{h} \)

2. **Calculate the number of half-lives that have passed:**
\[ \text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{18 \, \text{h}}{3 \, \text{h}} = 6 \]

3. **Substitute the values into the decay formula:**
\[ N = 300 \left( \frac{1}{2} \right)^{6} \]

4. **Calculate \( \left( \frac{1}{2} \right)^{6} \):**
\[ \left( \frac{1}{2} \right)^{6} = \frac{1}{64} \]

5. **Calculate the remaining amount \( N \):**
\[ N = 300 \times \frac{1}{64} = \frac{300}{64} = 4.6875 \, \text{g} \]

6. **Round the result to two decimal places:**
\[ N \approx 4.69 \, \text{g} \]

### Final Answer:
After 18 hours, the remaining amount of the radioactive material is approximately \( 4.68 \, \text{g} \).