1 g of `._(79)Au^(198)(t_(1//2)=65h)` gives stable mercury by `beta-` emission. What amount of mercury will left 260 h?

Options

A. 0.9375 g
B. 0.3758 g
C. 0.7586 g
D. 0.9000 g

Answer

To solve the problem, we need to determine how much mercury will be left after 260 hours when 1 gram of gold-198 undergoes beta decay.

### Step-by-Step Solution:

1. **Identify the Half-Life**:
The half-life of gold-198 (Au-198) is given as 65 hours.

2. **Determine the Total Time**:
We need to find out how many half-lives fit into the total time of 260 hours.
\[ \text{Number of half-lives} (n) = \frac{\text{Total time}}{\text{Half-life}} = \frac{260 \text{ hours}}{65 \text{ hours}} = 4 \]

3. **Calculate Remaining Mass of Gold-198**:
The remaining mass of the radioactive substance after \( n \) half-lives can be calculated using the formula:
\[ \text{Remaining mass} = \text{Initial mass} \times \left( \frac{1}{2} \right)^n \]
Here, the initial mass of Au is 1 gram.
\[ \text{Remaining mass of Au} = 1 \text{ g} \times \left( \frac{1}{2} \right)^4 = 1 \text{ g} \times \frac{1}{16} = \frac{1}{16} \text{ g} \]

4. **Calculate the Mass of Mercury Produced**:
Since 1 gram of Au-198 decays completely into mercury (Hg), the mass of mercury produced can be calculated as follows:
\[ \text{Mass of Hg produced} = \text{Initial mass of Au} - \text{Remaining mass of Au} \]
\[ \text{Mass of Hg produced} = 1 \text{ g} - \frac{1}{16} \text{ g} = \frac{16}{16} \text{ g} - \frac{1}{16} \text{ g} = \frac{15}{16} \text{ g} \]

5. **Convert to Decimal**:
To express the mass of mercury in decimal form:
\[ \frac{15}{16} \text{ g} = 0.9375 \text{ g} \]

### Final Answer:
The amount of mercury produced after 260 hours is **0.9375 grams**.