Question

The radioactive nuclei \({}^{40}\text{K}\) decay to \({}^{40}\text{Ar}\) with a half-life of \(1.25 \times 10^9\) years. The \({}^{40}\text{K}/{}^{40}\text{Ar}\) isotopic ratio for a particular rock is found to be 50. The age of the rock is \(m \times 10^7\) years. The value of \(m\) is .............. (Round off to 2 decimal places)

Hint:

In radioactive dating, isotope ratios help estimate geological ages using \(N/N_D = e^{-\lambda t}/(1 - e^{-\lambda t})\). Always use natural logarithms.

Answer 1

Correct Answer: 3.55

Step 1: Radioactive decay relation.
The number of parent nuclei remaining after time \(t\) is given by: \[ N = N_0 e^{-\lambda t} \] and the number of daughter nuclei formed is: \[ N_D = N_0 - N = N_0 (1 - e^{-\lambda t}) \] Given isotopic ratio: \[ \frac{N}{N_D} = 50 $\Rightarrow$ \frac{e^{-\lambda t}}{1 - e^{-\lambda t}} = 50 \]

Step 2: Solve for decay constant term.
\[ e^{-\lambda t} = 50 - 50 e^{-\lambda t} $\Rightarrow$ 51 e^{-\lambda t} = 50 $\Rightarrow$ e^{-\lambda t} = \frac{50}{51} \]

Step 3: Decay constant.
\[ \lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{1.25 \times 10^9} = 5.544 \times 10^{-10}\, \text{yr}^{-1} \]

Step 4: Substitute to find \(t\).
\[ -\lambda t = \ln{\frac{50}{51}} = -0.0198 $\Rightarrow$ t = \frac{0.0198}{5.544 \times 10^{-10}} = 3.57 \times 10^7\, \text{years} \] \[ m = 3.57 \]

Step 5: Conclusion.
Hence, the age of the rock is \(3.57 \times 10^7\, \text{years}\), i.e., \(m = 3.57.\)