Step-by-Step Solution

1. Understanding the Decay Process The radioactive nucleus ${}_{19}^{40}K$ decays through two parallel paths into stable nuclei ${}_{20}^{40}Ca$ and ${}_{18}^{40}Ar$. This is a case of parallel radioactive decay.

   • Decay 1: ${}_{19}^{40}K \rightarrow {}_{20}^{40}Ca$ with decay constant $\lambda_{Ca} = 4.5 \times 10^{-10}$ per year.
   • Decay 2: ${}_{19}^{40}K \rightarrow {}_{18}^{40}Ar$ with decay constant $\lambda_{Ar} = 0.5 \times 10^{-10}$ per year.

2. Calculating the Total Decay Constant ($\\\lambda_{total}$) For parallel decay, the total decay constant is the sum of the individual decay constants. $\lambda_{total} = \lambda_{Ca} + \lambda_{Ar}$ $\lambda_{total} = (4.5 \times 10^{-10}) + (0.5 \times 10^{-10}) \text{ per year}$ $\lambda_{total} = 5.0 \times 10^{-10} \text{ per year}$

3. Radioactive Decay Law Let $N_0$ be the initial number of ${}_{19}^{40}K$ nuclei at time $T=0$. The number of ${}_{19}^{40}K$ nuclei remaining at time $T$, denoted by $N_K$, is given by the radioactive decay law: $N_K(T) = N_0 e^{-\lambda_{total} T}$

4. Relating Parent and Daughter Nuclei The problem states that all stable ${}_{20}^{40}Ca$ and ${}_{18}^{40}Ar$ nuclei are produced from the decay of ${}_{19}^{40}K$. Therefore, the total number of daughter nuclei is equal to the number of parent nuclei that have decayed. Number of decayed ${}_{19}^{40}K$ nuclei = $N_0 - N_K(T)$. Sum of stable daughter nuclei = $N_{Ca}(T) + N_{Ar}(T)$. So, $N_{Ca}(T) + N_{Ar}(T) = N_0 - N_K(T)$.

5. Using the Given Ratio We are given that at a specific time $T$, the ratio of the sum of stable nuclei to the radioactive nuclei is 99. $\frac{N_{Ca}(T) + N_{Ar}(T)}{N_K(T)} = 99$ Substituting the expression from Step 4: $\frac{N_0 - N_K(T)}{N_K(T)} = 99$

6. Solving for Time (T) Let's solve the equation from Step 5. $\frac{N_0}{N_K(T)} - 1 = 99$ $\frac{N_0}{N_K(T)} = 100$ Now, substitute the decay law from Step 3, $N_K(T) = N_0 e^{-\lambda_{total} T}$. $\frac{N_0}{N_0 e^{-\lambda_{total} T}} = 100$ $e^{\lambda_{total} T} = 100$ Taking the natural logarithm (ln) of both sides: $\ln(e^{\lambda_{total} T}) = \ln(100)$ $\lambda_{total} T = \ln(10^2) = 2 \ln(10)$ $T = \frac{2 \ln(10)}{\lambda_{total}}$

7. Numerical Calculation We are given $\ln(10) = 2.3$ and we calculated $\lambda_{total} = 5.0 \times 10^{-10}$ per year. $T = \frac{2 \times 2.3}{5.0 \times 10^{-10}} \text{ years}$ $T = \frac{4.6}{5.0 \times 10^{-10}} \text{ years}$ $T = 0.92 \times 10^{10} \text{ years}$ $T = 9.2 \times 10^9 \text{ years}$

8. Finding the value of 't' The problem states that the time is $t \times 10^9$ years. Comparing this with our calculated time $T$: $t \times 10^9 \text{ years} = 9.2 \times 10^9 \text{ years}$ $t = 9.2$

Therefore, the value of t is 9.2, which corresponds to option A.