Question

Half-life of \(^{14}C\) is 5730 years. Suppose we start with 1 billion \(^{14}C\) atoms, and after a certain interval of time only 125 million \(^{14}C\) atoms remain, the number of half-lives that has elapsed is ________ (Answer in integer).

Hint:

For half-life problems, use the decay formula and logarithms to calculate the number of half-lives.

Answer 1

Step 1: Understanding the half-life decay formula.
The decay of a substance follows the exponential decay formula: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{t/T} \] where:
\(N(t)\) is the remaining number of atoms after time \(t\),
\(N_0\) is the initial number of atoms,
\(T\) is the half-life of the substance.
Step 2: Given values and calculation.
Initial number of atoms, \(N_0 = 10^9\),
Remaining number of atoms, \(N(t) = 125 \times 10^6\),
Half-life of \(^{14}C\), \(T = 5730\) years.
We need to find the number of half-lives \(n\), such that: \[ \frac{N(t)}{N_0} = \left(\frac{1}{2}\right)^n \] Substitute the values: \[ \frac{125 \times 10^6}{10^9} = \left(\frac{1}{2}\right)^n \quad \Rightarrow \quad 0.125 = \left(\frac{1}{2}\right)^n \] Step 3: Solving for \(n\). Taking the logarithm on both sides: \[ \log(0.125) = n \log\left(\frac{1}{2}\right) \] \[ n = \frac{\log(0.125)}{\log(0.5)} = 3 \] Thus, the number of half-lives that have elapsed is \(3\).