Question

An ancient discovery found a sample, where \(75\%\) of the original carbon \((C^{14})\) remains. Then the age of the sample is \((T_{1/2}(C^{14})=5730\ \text{years},\ \ln0.5=-0.7,\ \ln(0.75)=-0.3)\)

• A. \(2300\ \text{years}\)
• B. \(2456\ \text{years}\)
• C. \(2546\ \text{years}\)
• D. \(3456\ \text{years}\)

Hint:

For radioactive decay, \[ N=N_0e^{-\lambda t} \] and \[ \lambda=\frac{\ln2}{T_{1/2}} \] Always convert percentage remaining into the ratio \(\dfrac{N}{N_0}\).

Answer 1

The Correct Option is B

Step 1: Write the radioactive decay law.
The number of radioactive nuclei remaining after time $t$ follows: \[ N = N_0 e^{-\lambda t} \] where $\lambda$ is the decay constant and $N_0$ is the initial number of nuclei.
Step 2: Set up the equation with the given information.
Since $75\%$ of the original C-14 remains: \[ \frac{N}{N_0} = 0.75 \] Substituting into the decay law: \[ 0.75 = e^{-\lambda t} \] Taking natural logarithm of both sides: \[ \ln(0.75) = -\lambda t \] Given $\ln(0.75) = -0.3$, so: \[ -0.3 = -\lambda t \implies \lambda t = 0.3 \]
Step 3: Relate the decay constant to the half-life.
The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by: \[ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{T_{1/2}} \] Using the given hint $\ln 0.5 = -0.7$, so $\ln 2 = 0.7$: \[ \lambda = \frac{0.7}{5730}\,\text{year}^{-1} \]
Step 4: Solve for the age $t$.
From $\lambda t = 0.3$: \[ t = \frac{0.3}{\lambda} = \frac{0.3 \times 5730}{0.7} = \frac{1719}{0.7} \approx 2455.7\,\text{years} \]
Step 5: Round to the nearest given option.
$t \approx 2456\,\text{years}$, which matches option (2).
Step 6: State the final answer.
The age of the ancient sample based on Carbon-14 dating is: \[ \boxed{t \approx 2456\,\text{years}} \]