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: Use the radioactive decay law.
Radioactive decay is given by \[ N=N_0e^{-\lambda t} \] where, \[ N_0=\text{initial amount} \] \[ N=\text{amount remaining after time }t \] \[ \lambda=\text{decay constant} \] Given, \[ \frac{N}{N_0}=0.75 \] Thus, \[ 0.75=e^{-\lambda t} \] Taking natural logarithm, \[ \ln(0.75)=-\lambda t \] Given, \[ \ln(0.75)=-0.3 \] Hence, \[ \lambda t=0.3 \]

Step 2: Use relation between decay constant and half-life.
The decay constant is related to half-life by \[ \lambda=\frac{\ln2}{T_{1/2}} \] Given, \[ T_{1/2}=5730\ \text{years} \] and \[ \ln0.5=-0.7 \] Since, \[ \ln2=0.7 \] Therefore, \[ \lambda=\frac{0.7}{5730} \]

Step 3: Calculate the age of the sample.
Using \[ \lambda t=0.3 \] \[ \frac{0.7}{5730}t=0.3 \] \[ t=\frac{0.3\times5730}{0.7} \] \[ t=2455.7 \] \[ t\approx2456\ \text{years} \]

Step 4: Final conclusion.
Hence, the age of the sample is \[ \boxed{2456\ \text{years}} \]