At time t=0, a material is composed of two radioactive atoms A and B, where N_A(0) = 2N_B(0). The de

Q: Why is the rate of formation of B equal to $\lambda N_A$?

Every time an atom of A decays, an atom of B is created. Therefore, the rate at which B is formed is exactly equal to the rate at which A decays, which is given by the radioactive decay law as $\lambda N_A$.

Q: Why do we use an integrating factor here?

The rate equation $\frac{dN_B}{dt} + \lambda N_B = \lambda N_A(t)$ is a standard first-order linear differential equation of the form $\frac{dy}{dx} + P y = Q$. The integrating factor $e^{\int P dt}$ allows us to compress the left side into the exact derivative of a product, making integration possible.

Q: How did we find the constant of integration $C$?

We used the initial conditions provided in the problem. At time $t=0$, the number of B atoms is $N_B(0)$. Substituting $t=0$ and $N_B(t) = N_B(0)$ into our integrated equation allows us to solve algebraically for $C$.

Q: Why does the graph start at 1 instead of 0?

The y-axis of the graph represents the ratio $\frac{N_B(t)}{N_B(0)}$. At $t=0$, this ratio is exactly $\frac{N_B(0)}{N_B(0)}$, which simplifies to 1. If the graph were just plotting $N_B(t)$, it would start at $N_B(0)$.

Q: Is it possible for the peak to be less than 1?

In this specific problem, because $N_A(0) = 2N_B(0)$ and the decay constants are equal, the formation rate initially exceeds the decay rate, causing the ratio to rise above 1. If $N_A(0)$ was very small, the decay rate of B might dominate from the start, and the graph would monotonically decrease.

Question

At time $t=0$, a material is composed of two radioactive atoms $A$ and $B$, where $N_A(0) = 2N_B(0)$. The decay constant of both kind of radioactive atoms is $\lambda$. However, $A$ disintegrates to $B$ and $B$ disintegrates to $C$. Which of the following figures represents the evolution of $N_B(t)/N_B(0)$ with respect to time $t$?
$\left[ \begin{array}{l} N_A(0) = 	ext{Number of } A 	ext{ atoms at } t=0 \ N_B(0) = 	ext{Number of } B 	ext{ atoms at } t=0 \end{array} ight]$

Sigma Percentile · Accepted Answer

The problem of successive radioactive decays is a beautiful intersection of nuclear physics and calculus. When a parent nucleus decays into a radioactive daughter nucleus, the population of the daughter nucleus is governed by a delicate balance between its rate of formation and its rate of decay. Let's dive into the mathematics of this relay race!

### Analyzing the Setup

We are given a decay chain where atom $A$ disintegrates into $B$, and $B$ further disintegrates into $C$. Both $A$ and $B$ share the exact same decay constant, $\lambda$.

Initially, at $t = 0$, the sample contains twice as many $A$ atoms as $B$ atoms. Mathematically, this is written as:
$$N_A(0) = 2N_B(0)$$

Our goal is to find how the ratio $\frac{N_B(t)}{N_B(0)}$ evolves over time. To do this, we must construct the rate equation for $N_B(t)$.

### The Master Equation

The number of $B$ atoms changes due to two simultaneous processes:
1. **Formation:** $B$ is created when $A$ decays. The rate of formation is $\lambda N_A(t)$.
2. **Decay:** $B$ is destroyed when it decays into $C$. The rate of decay is $\lambda N_B(t)$.

Combining these, the net rate of change of $B$ is:
$$\frac{dN_B(t)}{dt} = \lambda N_A(t) - \lambda N_B(t)$$

We already know that $A$ undergoes simple exponential decay, so $N_A(t) = N_A(0) e^{-\lambda t}$. Substituting this into our rate equation gives:
$$\frac{dN_B(t)}{dt} = \lambda N_A(0) e^{-\lambda t} - \lambda N_B(t)$$

### The Integrating Factor Magic

This is a classic first-order linear differential equation. Let's rearrange it into the standard form:
$$\frac{dN_B(t)}{dt} + \lambda N_B(t) = \lambda N_A(0) e^{-\lambda t}$$

To solve this, we multiply the entire equation by an integrating factor, which is $e^{\int \lambda dt} = e^{\lambda t}$. This brilliant mathematical trick collapses the left side into the exact derivative of a product:
$$\frac{d}{dt} \left( N_B(t) e^{\lambda t} \right) = \lambda N_A(0)$$

Now, we substitute our initial condition $N_A(0) = 2N_B(0)$:
$$\frac{d}{dt} \left( N_B(t) e^{\lambda t} \right) = 2\lambda N_B(0)$$

Integrating both sides with respect to time $t$:
$$N_B(t) e^{\lambda t} = 2\lambda N_B(0) t + C$$

To find the constant of integration $C$, we apply the initial condition at $t = 0$:
$$N_B(0) \cdot e^0 = 0 + C \implies C = N_B(0)$$

### Final Calculation and Graphing

Substituting $C$ back into our equation and dividing by $e^{\lambda t}$, we get the explicit function for $N_B(t)$:
$$N_B(t) = N_B(0) (1 + 2\lambda t) e^{-\lambda t}$$

Dividing by $N_B(0)$ gives the exact ratio we need to plot:
$$\frac{N_B(t)}{N_B(0)} = (1 + 2\lambda t) e^{-\lambda t}$$

To determine the shape of the graph, we need to find if and where this ratio reaches a maximum. We differentiate the expression with respect to $t$ and set it to zero:
$$\frac{d}{dt} \left[ (1 + 2\lambda t) e^{-\lambda t} \right] = 2\lambda e^{-\lambda t} - \lambda (1 + 2\lambda t) e^{-\lambda t} = 0$$

Factoring out $\lambda e^{-\lambda t}$:
$$\lambda e^{-\lambda t} (2 - 1 - 2\lambda t) = 0$$
$$1 - 2\lambda t = 0 \implies t = \frac{1}{2\lambda}$$

At this peak time, the value of the ratio is:
$$\frac{N_B}{N_B(0)} = \left(1 + 2\lambda \cdot \frac{1}{2\lambda}\right) e^{-1/2} = \frac{2}{\sqrt{e}} \approx 1.21$$

Since $1.21 > 1$, the graph starts at $1$, rises to a peak at $t = \frac{1}{2\lambda}$, and then decays exponentially towards zero. This perfectly matches the visual representation in **Option (c)**.

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Visualized Solution

$Initial Setup$

$Rate Equation$

$Substitution$

$Integrating Factor$

$Integration$

$Ratio Expression$

$Finding the Peak$

$Final Conclusion$

$Secular Equilibrium$

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation

The Integrating Factor Magic

Final Calculation and Graphing

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Visualized Solution

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

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

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

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

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

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

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

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

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