Question

A rock formed at time \( t_0 \) with number of \( ^{14}C \) atoms = \( 10^5 \). The number of \( ^{14}C \) atoms (in log10) after a time of \( 8 \times 10^3 \) years is \(\underline{\hspace{1cm}}\) (round off to 3 decimal places). (Use a decay constant of \( 1.25 \times 10^{-4} \, \text{yr}^{-1} \))

Hint:

For calculating the remaining amount of a radioactive substance, use the exponential decay formula and its logarithmic form.

Answer 1

Correct Answer: 4.565 - 4.567

The number of \( ^{14}C \) atoms after a given time can be calculated using the formula: \[ N(t) = N_0 e^{-\lambda t} \] where: - \( N_0 \) is the initial number of atoms, - \( \lambda \) is the decay constant, and - \( t \) is the time. First, we calculate the number of \( ^{14}C \) atoms at \( t = 8 \times 10^3 \) years using the logarithmic form: \[ \log_{10} N(t) = \log_{10} N_0 - \frac{\lambda t}{2.303} \] Given:
- \( N_0 = 10^5 \),
- \( \lambda = 1.25 \times 10^{-4} \, \text{yr}^{-1} \),
- \( t = 8 \times 10^3 \, \text{years} \).
Substituting the values into the formula: \[ \log_{10} N(t) = \log_{10} (10^5) - \frac{1.25 \times 10^{-4} \times 8 \times 10^3}{2.303} \] \[ \log_{10} N(t) = 5 - \frac{1}{2.303} \times 1.0 = 5 - 0.434 = 4.566 \] Thus, the number of \( ^{14}C \) atoms is approximately \( 4.566 \) in log10.