JEE Challenger
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PhyMedAdv 2019 • Paper 1

Time Elapsed for Radioactive Decay Ratio of Potassium

In a radioactive sample, 1940K^{40}_{19}\text{K} nuclei can transform either into stable 2040Ca^{40}_{20}\text{Ca} nuclei with decay constant 4.5×10−10 year−14.5 \times 10^{-10}\text{ year}^{-1} or into stable 1840Ar^{40}_{18}\text{Ar} nuclei with decay constant 0.5×10−10 year−10.5 \times 10^{-10}\text{ year}^{-1}. Assume that all stable 2040Ca^{40}_{20}\text{Ca} and 1840Ar^{40}_{18}\text{Ar} nuclei present in the sample originate exclusively from 1940K^{40}_{19}\text{K} nuclei. After a time duration of t×109 yearst \times 10^9\text{ years}, if the ratio of the combined count of stable 2040Ca^{40}_{20}\text{Ca} and 1840Ar^{40}_{18}\text{Ar} nuclei to the remaining radioactive 1940K^{40}_{19}\text{K} nuclei equals 9999, determine the value of tt. [Given: ln⁡10=2.3\ln 10 = 2.3]

Options

A

1.151.15

B

9.29.2

Correct
C

2.32.3

D

4.64.6

Topics & Concepts

NucleiRadioactivity

Step-by-Step Solution

The total decay constant λ\lambda for the parallel radioactive decay of 1940K^{40}_{19}\text{K} nuclei into stable 2040Ca^{40}_{20}\text{Ca} and 1840Ar^{40}_{18}\text{Ar} nuclei is the sum of the individual decay constants:

λ=λ1+λ2\lambda = \lambda_1 + \lambda_2

Given: λ1=4.5×10−10 year−1\lambda_1 = 4.5 \times 10^{-10}\text{ year}^{-1} λ2=0.5×10−10 year−1\lambda_2 = 0.5 \times 10^{-10}\text{ year}^{-1}

Therefore, the total decay constant is: λ=4.5×10−10+0.5×10−10=5.0×10−10 year−1\lambda = 4.5 \times 10^{-10} + 0.5 \times 10^{-10} = 5.0 \times 10^{-10}\text{ year}^{-1}

Let N0N_0 be the initial number of 1940K^{40}_{19}\text{K} nuclei present at time T=0T = 0. At time TT, the number of remaining 1940K^{40}_{19}\text{K} nuclei is given by:

NK(T)=N0e−λTN_{\text{K}}(T) = N_0 e^{-\lambda T}

Since all stable 2040Ca^{40}_{20}\text{Ca} and 1840Ar^{40}_{18}\text{Ar} nuclei originate exclusively from the decayed 1940K^{40}_{19}\text{K} nuclei, the total number of stable product nuclei formed by time TT is equal to the number of decayed 1940K^{40}_{19}\text{K} nuclei:

NCa+NAr=N0−NK(T)N_{\text{Ca}} + N_{\text{Ar}} = N_0 - N_{\text{K}}(T)

We are given that the ratio of the combined count of stable nuclei to the remaining 1940K^{40}_{19}\text{K} nuclei at time TT is equal to 9999:

NCa+NArNK(T)=99\frac{N_{\text{Ca}} + N_{\text{Ar}}}{N_{\text{K}}(T)} = 99

Substituting NCa+NAr=N0−NK(T)N_{\text{Ca}} + N_{\text{Ar}} = N_0 - N_{\text{K}}(T) into the equation:

N0−NK(T)NK(T)=99\frac{N_0 - N_{\text{K}}(T)}{N_{\text{K}}(T)} = 99

N0NK(T)−1=99\frac{N_0}{N_{\text{K}}(T)} - 1 = 99

N0NK(T)=100\frac{N_0}{N_{\text{K}}(T)} = 100

Substitute NK(T)=N0e−λTN_{\text{K}}(T) = N_0 e^{-\lambda T}:

N0N0e−λT=100  ⟹  eλT=100\frac{N_0}{N_0 e^{-\lambda T}} = 100 \implies e^{\lambda T} = 100

Taking the natural logarithm on both sides:

λT=ln⁡(100)=2ln⁡(10)\lambda T = \ln(100) = 2 \ln(10)

Given ln⁡(10)=2.3\ln(10) = 2.3:

λT=2×2.3=4.6\lambda T = 2 \times 2.3 = 4.6

The time duration is specified as T=t×109 yearsT = t \times 10^9\text{ years}. Substituting λ\lambda and TT:

(5.0×10−10 year−1)×(t×109 years)=4.6(5.0 \times 10^{-10}\text{ year}^{-1}) \times (t \times 10^9\text{ years}) = 4.6

5.0×10−1×t=4.65.0 \times 10^{-1} \times t = 4.6

0.5t=4.60.5 t = 4.6

t=4.60.5=9.2t = \frac{4.6}{0.5} = 9.2

Thus, the value of tt is 9.29.2, which corresponds to Option B.

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