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 of how much mercury will be left after 260 hours from 1 g of gold (Au-198) that undergoes beta decay, we can follow these steps:

### Step 1: Understand the decay process
Gold-198 (Au-198) decays into stable mercury (Hg) through beta emission. The half-life of Au-198 is given as 65 hours.

### Step 2: Determine the number of half-lives in 260 hours
To find out how many half-lives fit into 260 hours, we divide the total time by the half-life:
\[ \text{Number of half-lives} = \frac{260 \text{ hours}}{65 \text{ hours}} = 4 \]

### Step 3: Calculate the remaining amount of gold after 4 half-lives
The formula for the remaining quantity of a radioactive substance after a certain number of half-lives is:
\[ N_t = N_0 \left(\frac{1}{2}\right)^n \]
where \(N_t\) is the remaining quantity, \(N_0\) is the initial quantity, and \(n\) is the number of half-lives.

Here, \(N_0 = 1 \text{ g}\) and \(n = 4\):
\[ N_t = 1 \text{ g} \left(\frac{1}{2}\right)^4 = 1 \text{ g} \times \frac{1}{16} = \frac{1}{16} \text{ g} \]

### Step 4: Calculate the amount of mercury produced
Since the initial amount of gold was 1 g, the amount of gold remaining after 260 hours is \(\frac{1}{16} \text{ g}\). The amount of gold that has decayed into mercury is:
\[ \text{Amount of mercury} = \text{Initial amount} - \text{Remaining gold} = 1 \text{ g} - \frac{1}{16} \text{ g} = \frac{16}{16} \text{ g} - \frac{1}{16} \text{ g} = \frac{15}{16} \text{ g} \]

### Step 5: Convert to decimal
To express \(\frac{15}{16}\) in decimal form:
\[ \frac{15}{16} = 0.9375 \text{ g} \]

### Final Answer
The amount of mercury left after 260 hours is **0.9375 g**.

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