A radioactive nucleus X decays to a nucleus Y with a decay constant lambda_x = 0.1 s^-1, Y further d

Q: Why is the rate of change of $Y$ given by $\lambda_X N_X - \lambda_Y N_Y$?

Because $Y$ is being formed by the decay of $X$ at a rate $\lambda_X N_X$, and it is simultaneously decaying into $Z$ at a rate $\lambda_Y N_Y$.

Q: Why do we set $\frac{dN_Y}{dt} = 0$ to find the maximum population?

In calculus, the maximum or minimum of a function occurs where its first derivative is zero. Here, it represents the equilibrium point where the formation rate equals the decay rate.

Q: Can we find $N_Y$ without using the long formula?

Yes! At the maximum point, $\lambda_X N_X = \lambda_Y N_Y$. Once you find $N_X$, you can easily calculate $N_Y$ using this relation.

Q: Why is the total number of nuclei conserved?

In this specific decay chain $X \to Y \to Z$, every $X$ nucleus eventually becomes a $Y$ nucleus, which then becomes a $Z$ nucleus. No nuclei are lost from the system.

Q: What happens to the populations as $t \to \infty$?

As time goes to infinity, both $X$ and $Y$ will completely decay, and all the initial nuclei will end up as the stable nucleus $Z$. So, $N_Z \to N_0$.

Question

A radioactive nucleus $X$ decays to a nucleus $Y$ with a decay constant $\lambda_x = 0.1	ext{ s}^{-1}$, $Y$ further decays to a stable nucleus $Z$ with a decay constant $\lambda_y = 1/30	ext{ s}^{-1}$. Initially, there are only $X$ nuclei and their number is $N_0 = 10^{20}$. Set-up the rate equations for the populations of $X, Y$ and $Z$. The population of $Y$ nucleus as a function of time is given by $N_y(t) = \{N_0\lambda_x/(\lambda_x - \lambda_y)\} [\exp(-\lambda_y t) - \exp(-\lambda_x t)]$. Find the time at which $N_Y$ is maximum and determine the populations $X$ and $Z$ at that instant.

Sigma Percentile · Accepted Answer

### Introduction to Successive Radioactive Decay

In this problem, we are dealing with a successive radioactive decay chain where a parent nucleus $X$ decays into an intermediate nucleus $Y$, which in turn decays into a stable nucleus $Z$. This is a classic scenario in nuclear physics that requires setting up and solving a system of differential equations.

### Setting up the Rate Equations

The rate of change of the population of each nucleus depends on its formation rate and its decay rate.

For nucleus $X$, it only decays, so its rate equation is simply:

$$\frac{dN_X}{dt} = -\lambda_X N_X$$

For nucleus $Y$, it is formed by the decay of $X$ and it decays into $Z$. Therefore, its rate equation is the difference between its formation rate and its decay rate:

$$\frac{dN_Y}{dt} = \lambda_X N_X - \lambda_Y N_Y$$

For the stable nucleus $Z$, it is only formed by the decay of $Y$, so its rate equation is:

$$\frac{dN_Z}{dt} = \lambda_Y N_Y$$

### Finding the Time for Maximum Intermediate Population

The population of the intermediate nucleus $Y$ will initially increase, reach a maximum, and then eventually decrease to zero. To find the time at which $N_Y$ is maximum, we set its derivative with respect to time to zero:

$$\frac{dN_Y}{dt} = 0 \implies \lambda_X N_X = \lambda_Y N_Y$$

We are given the explicit function for $N_Y(t)$ and we know that $N_X(t) = N_0 e^{-\lambda_X t}$. Substituting these into our condition yields:

$$\lambda_X (N_0 e^{-\lambda_X t}) = \lambda_Y \frac{N_0 \lambda_X}{\lambda_X - \lambda_Y} [e^{-\lambda_Y t} - e^{-\lambda_X t}]$$

By simplifying this equation, we can isolate $t$ to find the standard formula for the time of maximum activity:

$$t = \frac{1}{\lambda_X - \lambda_Y} \ln\left(\frac{\lambda_X}{\lambda_Y}ight)$$

Substituting the given values $\lambda_X = 0.1	ext{ s}^{-1}$ and $\lambda_Y = 1/30	ext{ s}^{-1}$:

$$t = \frac{1}{0.1 - 1/30} \ln\left(\frac{0.1}{1/30}ight) = 15 \ln(3) \approx 16.48 	ext{ s}$$

### Calculating the Populations at the Maximum Instant

Now that we have the time $t = 16.48	ext{ s}$, we can calculate the populations of all three nuclei.

First, the population of $X$:

$$N_X = N_0 e^{-\lambda_X t} = 10^{20} e^{-0.1 	imes 16.48} \approx 1.92 	imes 10^{19}$$

To find the population of $Y$, we can cleverly use the condition we derived earlier, $\lambda_X N_X = \lambda_Y N_Y$, which is much simpler than evaluating the full exponential expression:

$$N_Y = \frac{\lambda_X}{\lambda_Y} N_X = \frac{0.1}{1/30} (1.92 	imes 10^{19}) = 5.76 	imes 10^{19}$$

Finally, because the total number of nuclei in the system is conserved, the population of $Z$ is simply the initial total minus the remaining $X$ and $Y$ nuclei:

$$N_Z = N_0 - N_X - N_Y = 10^{20} - 1.92 	imes 10^{19} - 5.76 	imes 10^{19} = 2.32 	imes 10^{19}$$

### Conclusion

By carefully setting up the rate equations and utilizing the condition for a maximum, we were able to elegantly solve for the time and the populations of all nuclei in the decay chain without getting bogged down in overly complex algebra.

Visualized Solution

$Rate Equations$

$Condition for Maximum N_{Y}$

$Substituting Expressions$

$Solving for Time t$

$Calculating Time t$

$Population of X at t = 16.48 s$

$Population of Y at t = 16.48 s$

$Population of Z at t = 16.48 s$

The Sigma Insight: Radioactivity

Introduction to Successive Radioactive Decay

Setting up the Rate Equations

Finding the Time for Maximum Intermediate Population

Calculating the Populations at the Maximum Instant

Conclusion

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Maximum NY​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Expressions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Time t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Time t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Population of X at t=16.48 s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Population of Y at t=16.48 s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Population of Z at t=16.48 s