`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: Understanding 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 remaining C-14 will decay.

### Step 2: Determine the Remaining Percentage
If 72% of the isotope is left, then the percentage that has decayed is:
\[ 100\% - 72\% = 28\% \]
This means that 28% of the original C-14 has decayed.

### Step 3: Calculate the Rate Constant (k)
For a first-order reaction, the relationship between the half-life and the rate constant (k) is given by:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the half-life of C-14:
\[ k = \frac{0.693}{5770 \text{ years}} \]
Calculating this gives:
\[ k \approx 1.202 \times 10^{-4} \text{ years}^{-1} \]

### Step 4: Use the First-Order Kinetics Equation
The first-order kinetics equation is:
\[ t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]_t}\right) \]
Where:
- \([A]_0\) is the initial concentration (100%),
- \([A]_t\) is the concentration at time t (72%).

### Step 5: Substitute Values into the Equation
Substituting the values into the equation:
\[ t = \frac{2.303}{1.202 \times 10^{-4}} \log\left(\frac{100}{72}\right) \]

### Step 6: Calculate the Logarithm
First, calculate the logarithm:
\[ \log\left(\frac{100}{72}\right) \approx \log(1.3889) \approx 0.143 \]

### Step 7: Calculate Time (t)
Now substitute the log value back into the equation:
\[ t = \frac{2.303}{1.202 \times 10^{-4}} \times 0.143 \]
Calculating this gives:
\[ t \approx 2742 \text{ years} \]

### Final Answer
Thus, the time after which 72% of the C-14 isotope is left is approximately **2740 years**.

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