Nuclei of a radioactive element A are being produced at a constant rate alpha. The element has a dec

Q: Why is the decay rate $\lambda N$ and not just $\lambda$?

The decay constant $\lambda$ represents the probability of decay per unit time for a *single* nucleus. If you have $N$ nuclei, the total number decaying per unit time is $\lambda \times N$.

Q: What does the negative sign in the integration step mean?

When integrating $\frac{1}{\alpha - \lambda N}$, we use substitution. Let $u = \alpha - \lambda N$, then $du = -\lambda dN$. This introduces the $-\frac{1}{\lambda}$ factor in the result.

Q: Why does the number of nuclei reach a steady state instead of growing infinitely?

As the number of nuclei $N$ increases, the decay rate $\lambda N$ also increases. Eventually, the decay rate becomes exactly equal to the constant production rate $\alpha$. At this point, for every nucleus produced, one decays, keeping the total number constant.

Q: How did we get $e^{-\lambda t} = 1/2$ at one half-life?

The half-life $t_{1/2}$ is defined as $\frac{\ln 2}{\lambda}$. Substituting this into $e^{-\lambda t}$ gives $e^{-\lambda (\ln 2 / \lambda)} = e^{-\ln 2} = e^{\ln(2^{-1})} = 1/2$.

Q: Is it possible for the number of nuclei to decrease over time?

Yes! If the initial decay rate $\lambda N_0$ is greater than the production rate $\alpha$, the net rate of change $dN/dt$ will be negative, and the number of nuclei will decrease until it reaches the equilibrium state.

Question

Nuclei of a radioactive element $A$ are being produced at a constant rate $\alpha$. The element has a decay constant $\lambda$. At time $t = 0$, there are $N_0$ nuclei of the element.
(a) Calculate the number $N$ of nuclei of $A$ at time $t$.
(b) If $\alpha = 2N_0\lambda$, calculate the number of nuclei of $A$ after one half-life of $A$ and also the limiting value of $N$ as $t \rightarrow \infty$.

Sigma Percentile · Accepted Answer

### The Tug of War: Production vs. Decay

Imagine a container where new radioactive nuclei are constantly being added, while some of the existing ones are decaying away. This creates a fascinating dynamic—a tug of war between creation and destruction. To understand how the total number of nuclei changes over time, we must look at the net rate of change.

The net rate of change is simply the rate at which nuclei are produced minus the rate at which they decay. Mathematically, we can express this as:

$$\frac{dN}{dt} = 	ext{Rate of production} - 	ext{Rate of decay}$$

### Setting Up the Mathematics

We are given that the nuclei are produced at a constant rate $\alpha$. We also know from the fundamental law of radioactivity that the decay rate is proportional to the number of nuclei present, which is $\lambda N$.

Substituting these into our rate equation, we get a first-order linear differential equation:

$$\frac{dN}{dt} = \alpha - \lambda N$$

To solve this, we need to separate the variables. Let's bring all the $N$ terms to one side and the time terms to the other:

$$\frac{dN}{\alpha - \lambda N} = dt$$

### Solving the Differential Equation

Now, let's integrate both sides. We know that at time $t = 0$, the initial number of nuclei is $N_0$. At any arbitrary time $t$, let the number be $N$. These will serve as our limits of integration:

$$\int_{N_0}^{N} \frac{dN}{\alpha - \lambda N} = \int_{0}^{t} dt$$

The integral of $\frac{1}{\alpha - \lambda N}$ is $\frac{\ln(\alpha - \lambda N)}{-\lambda}$. Applying the limits and simplifying, we get:

$$\ln\left( \frac{\alpha - \lambda N}{\alpha - \lambda N_0} ight) = -\lambda t$$

To isolate $N$, we take the exponential of both sides:

$$\frac{\alpha - \lambda N}{\alpha - \lambda N_0} = e^{-\lambda t}$$

After a bit of algebraic rearrangement, we find the expression for $N$ as a function of time:

$$N = \frac{1}{\lambda} [\alpha - (\alpha - \lambda N_0)e^{-\lambda t}]$$

This answers the first part of our question, giving us a general formula for any production rate $\alpha$.

### A Specific Scenario: The Approach to Equilibrium

Moving on to part (b), we are given a specific production rate: $\alpha = 2N_0\lambda$. Let's substitute this into our general equation for $N$:

$$N = \frac{1}{\lambda} [2N_0\lambda - (2N_0\lambda - \lambda N_0)e^{-\lambda t}]$$

Simplifying this, we get a much neater expression:

$$N = 2N_0 - N_0 e^{-\lambda t}$$

### The Half-Life Milestone and the Infinite Future

We need to find the number of nuclei after one half-life. Remember, the half-life $t_{1/2}$ is defined as $\frac{\ln 2}{\lambda}$. At one half-life, the term $e^{-\lambda t}$ simply becomes $\frac{1}{2}$. Substituting this, we find:

$$N = 2N_0 - N_0 \left(\frac{1}{2}ight) = \frac{3}{2}N_0$$

Finally, what happens after a very long time? As $t ightarrow \infty$, the exponential term $e^{-\lambda t}$ decays to zero. This leaves us with:

$$N = 2N_0$$

This is the steady-state or equilibrium number of nuclei. At this point, the rate of production exactly balances the rate of decay, and the number of nuclei remains constant forever.

Visualized Solution

$Rate of Change of Nuclei$

$Differential Equation$

$Rearranging for Integration$

$Integration with Limits$

$Solving the Integral$

$Expression for N (t)$

$Part (b): Substituting Given Values$

$Number of Nuclei at Half-Life$

$Limiting Value as t \to \infty$

The Sigma Insight: Radioactivity

The Tug of War

Setting Up the Mathematics

Solving the Differential Equation

A Specific Scenario

The Half-Life Milestone and the Infinite Future

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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 of Change of Nuclei

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expression for N(t)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part (b): Substituting Given Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Nuclei at Half-Life

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Value as t→∞