`t_(1//2)` of `C^(14)` isotope is 5770 years. time after which 72% of isotope left is:

Options

A. 2740 years
B. 274 years
C. 2780 years
D. 278 years

Answer

To solve the problem of determining the time after which 72% of the C-14 isotope is left, we can follow these steps:

### Step 1: Understand the Half-Life
The half-life (t₁/₂) of C-14 is given as 5770 years. This means that every 5770 years, half of the original amount of C-14 will remain.

### Step 2: Determine the Remaining Percentage
We need to find the time after which 72% of the isotope is left. This implies that 72% remains, which means that 28% has decayed.

### Step 3: Use the First-Order Kinetics Formula
For a first-order reaction, the relationship between the initial amount (A₀), the remaining amount (Aₜ), and time (t) is given by the formula:
\[ t = \frac{2.303}{k} \log\left(\frac{A₀}{Aₜ}\right) \]

Where:
- \( k \) is the decay constant,
- \( A₀ \) is the initial amount (100%),
- \( Aₜ \) is the remaining amount (72%).

### Step 4: Calculate the Decay Constant (k)
The decay constant \( k \) can be calculated using the half-life:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the half-life:
\[ k = \frac{0.693}{5770} \approx 1.202 \times 10^{-4} \text{ year}^{-1} \]

### Step 5: Substitute Values into the Formula
Now we can substitute \( A₀ = 100 \) and \( Aₜ = 72 \) into the formula:
\[ t = \frac{2.303}{k} \log\left(\frac{100}{72}\right) \]
Calculating \( \frac{100}{72} \):
\[ \frac{100}{72} \approx 1.3889 \]

Now calculate the logarithm:
\[ \log(1.3889) \approx 0.143 \]

### Step 6: Calculate Time (t)
Substituting \( k \) and the logarithm value back into the equation:
\[ t = \frac{2.303}{1.202 \times 10^{-4}} \times 0.143 \]
\[ t \approx 19175.0 \times 0.143 \]
\[ t \approx 2742.0 \text{ years} \]

### Conclusion
The time after which 72% of the C-14 isotope is left is approximately **2742 years**.

---