In a radioactive sample, 40_19 K nuclei either decay into stable 40_20 Ca nuclei with decay constant

Q: Why do we add the decay constants $\lambda_1$ and $\lambda_2$?

In parallel decay, the parent nucleus has independent probabilities of decaying through different pathways. The total probability of decay per unit time (which is the effective decay constant) is the sum of the probabilities of each individual pathway.

Q: Why is the number of stable nuclei equal to $N_0 - N$?

This is due to the conservation of nucleons. Every parent nucleus that decays must become one of the stable daughter nuclei. Therefore, the total number of daughter nuclei produced is exactly equal to the number of parent nuclei that have disappeared ($N_0 - N$).

Q: Does the individual ratio of Calcium to Argon matter for this calculation?

No, it doesn't. The question asks for the ratio of the *sum* of all stable nuclei to the remaining radioactive nuclei. Because we are looking at the total daughter population, we only need the effective decay constant, not the branching ratio.

Q: How did $\ln(100)$ become $2 \ln(10)$?

This is a standard property of logarithms. Since $100 = 10^2$, we can write $\ln(100)$ as $\ln(10^2)$. The power rule for logarithms states that $\ln(x^y) = y \ln(x)$, so $\ln(10^2)$ becomes $2 \ln(10)$.

Q: What if the problem gave half-lives instead of decay constants?

You would first need to convert the half-lives into decay constants using the relation $\lambda = \frac{\ln(2)}{T_{1/2}}$. Once you have $\lambda_1$ and $\lambda_2$, you can add them to find $\lambda_{\text{eff}}$ and proceed with the same steps.

Question

In a radioactive sample, $^{40}_{19}	ext{K}$ nuclei either decay into stable $^{40}_{20}	ext{Ca}$ nuclei with decay constant $4.5 	imes 10^{-10}$ per year or into stable $^{40}_{18}	ext{Ar}$ nuclei with decay constant $0.5 	imes 10^{-10}$ per year. Given that in this sample all the stable $^{40}_{20}	ext{Ca}$ and $^{40}_{18}	ext{Ar}$ nuclei are produced by the $^{40}_{19}	ext{K}$ nuclei only. In time $t 	imes 10^9$ years, if the ratio of the sum of stable $^{40}_{20}	ext{Ca}$ and $^{40}_{18}	ext{Ar}$ nuclei to the radioactive $^{40}_{19}	ext{K}$ nuclei is 99, the value of t will be : [Given: $\ln 10 = 2.3$]

Options:

(A) 9.2

(B) 1.15

(C) 4.6

(D) 2.3

Sigma Percentile · Accepted Answer

### The Parallel Decay Phenomenon

Imagine you are observing a sample of Potassium-40 ($^{40}_{19}	ext{K}$) nuclei. Unlike a simple decay process where a parent nucleus transforms into a single type of daughter nucleus, Potassium-40 has a choice. It undergoes what we call **parallel radioactive decay**.

Some of the Potassium-40 nuclei decay into Calcium-40 ($^{40}_{20}	ext{Ca}$) with a decay constant $\lambda_1 = 4.5 	imes 10^{-10} 	ext{ yr}^{-1}$. Simultaneously, other Potassium-40 nuclei decay into Argon-40 ($^{40}_{18}	ext{Ar}$) with a decay constant $\lambda_2 = 0.5 	imes 10^{-10} 	ext{ yr}^{-1}$.

### The Effective Decay Constant

When a radioactive nucleus can decay through multiple independent pathways, the overall rate at which the parent nuclei disappear is governed by the sum of the individual probabilities. Therefore, the **effective decay constant** ($\lambda_{	ext{eff}}$) is simply the sum of the individual decay constants.

$$\lambda_{	ext{eff}} = \lambda_1 + \lambda_2$$

Let's plug in the given values to find the effective decay constant for our sample:

$$\lambda_{	ext{eff}} = (4.5 	imes 10^{-10}) + (0.5 	imes 10^{-10}) = 5.0 	imes 10^{-10} 	ext{ yr}^{-1}$$

### Decoding the Nuclei Population

Now, let's analyze the populations of the nuclei over time. Let $N_0$ be the initial number of Potassium-40 nuclei at $t = 0$. At any later time $t$, let $N$ be the number of radioactive Potassium-40 nuclei still remaining.

Because every Potassium nucleus that decays turns into either a stable Calcium or a stable Argon nucleus, the total number of stable daughter nuclei produced must be exactly equal to the number of Potassium nuclei that have decayed. This is a direct consequence of the conservation of nucleons.

$$N_{	ext{stable}} = N_0 - N$$

The problem provides a crucial piece of information: the ratio of the sum of stable nuclei to the remaining radioactive nuclei is 99.

$$\frac{N_{	ext{stable}}}{N_{	ext{radioactive}}} = \frac{N_0 - N}{N} = 99$$

We can easily simplify this algebraic expression by splitting the fraction:

$$\frac{N_0}{N} - 1 = 99 \implies \frac{N_0}{N} = 100$$

This tells us that the initial number of nuclei was exactly 100 times the number of nuclei currently remaining.

### The Final Countdown

We know the fundamental law of radioactive decay, which states that the number of undecayed nuclei decreases exponentially over time:

$$N = N_0 e^{-\lambda_{	ext{eff}} t}$$

Rearranging this equation to match our ratio, we get:

$$\frac{N_0}{N} = e^{\lambda_{	ext{eff}} t}$$

Equating our two expressions for $\frac{N_0}{N}$, we have:

$$e^{\lambda_{	ext{eff}} t} = 100$$

To solve for time $t$, we take the natural logarithm ($\ln$) on both sides. Remember the logarithmic property $\ln(x^y) = y \ln(x)$:

$$\lambda_{	ext{eff}} t = \ln(100) = \ln(10^2) = 2 \ln(10)$$

Now, we substitute the value of $\lambda_{	ext{eff}}$ we calculated earlier and the given value of $\ln(10) = 2.3$:

$$(5.0 	imes 10^{-10}) t = 2(2.3)$$

$$(5.0 	imes 10^{-10}) t = 4.6$$

Finally, isolating $t$ gives us the answer:

$$t = \frac{4.6}{5.0 	imes 10^{-10}} = 0.92 	imes 10^{10} = 9.2 	imes 10^9 	ext{ years}$$

Comparing this result with the format given in the question ($t 	imes 10^9$ years), we find that the value of $t$ is **9.2**.

Select Answer:

Visualized Solution

$_{19}^{40} K Parallel Decay_{20}^{40} Ca +_{18}^{40} Ar$

$λ_{eff} = λ_{1} + λ_{2}$

$λ_{eff} = 5.0 \times 1 0^{- 10} yr^{- 1}$

$N_{stable} = N_{0} - N$

$\frac{N _{stable}}{N _{radioactive}} = 99$

$\frac{N _{0}}{N} = 100$

$N = N_{0} e^{- λ_{eff} t}$

$λ_{eff} t = ln (100)$

$t = 9.2 \times 1 0^{9} years$

The Sigma Insight: Radioactivity

The Parallel Decay Phenomenon

The Effective Decay Constant

Decoding the Nuclei Population

The Final Countdown

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1940​KParallel Decay​2040​Ca+1840​Ar

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λeff​=λ1​+λ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λeff​=5.0×10−10 yr−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nstable​=N0​−N

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nradioactive​Nstable​​=99

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } NN0​​=100

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=N0​e−λeff​t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λeff​t=ln(100)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=9.2×109 years