A sample of a radioactive nucleus A disintegrates to another radioactive nucleus B, which in turn di

Q: Why doesn't the number of atoms of B keep increasing forever?

Because B is also radioactive! As its quantity increases, its rate of decay also increases. Eventually, it decays faster than it is produced from A.

Q: What is the exact condition for the maximum number of B atoms?

The maximum occurs when the rate of production of B perfectly balances its rate of decay. Mathematically, this is when $\lambda_A N_A = \lambda_B N_B$.

Q: Why does the graph end up looking like a simple exponential decay curve?

After a long time, almost all of A has decayed, so no new B is being formed. B then simply undergoes standard exponential decay into C without any new replenishment.

Q: Can the number of B atoms ever exceed the initial number of A atoms?

No. By the conservation of nucleons, the total number of atoms $N_A + N_B + N_C$ is constant and equals the initial number of A atoms. Therefore, $N_B$ can never exceed $N_{A0}$.

Q: What if B was a stable nucleus instead of radioactive?

If B were stable, its decay rate would be zero ($\lambda_B = 0$). The number of B atoms would continuously increase and eventually plateau at the initial number of A atoms, never decreasing.

Question

A sample of a radioactive nucleus $A$ disintegrates to another radioactive nucleus $B$, which in turn disintegrates to some other stable nucleus $C$. Plot of a graph showing the variation of number of atoms of nucleus $B$ versus time is
(Assume that at $t = 0$, there are no $B$ atoms in the sample)

Sigma Percentile · Accepted Answer

### The Phenomenon of Sequential Decay

Imagine a cascading waterfall. The water from the top pool flows into a middle pool, which in turn flows into a bottom pool. This is exactly what happens in **sequential radioactive decay**.

We start with a parent nucleus $A$. It is unstable and decays into a daughter nucleus $B$ with a decay constant $\lambda_A$. But the story doesn't end there. Nucleus $B$ is also radioactive! It decays into a stable nucleus $C$ with its own decay constant $\lambda_B$.

This creates a dynamic tug-of-war for the population of nucleus $B$. It is constantly being created by the decay of $A$, and simultaneously being destroyed by its own decay into $C$.

### The Mathematics of Growth and Decay

To understand how the number of atoms of $B$ ($N_B$) changes over time, we need to look at its rate equation. The net rate of change of $N_B$ is the difference between its rate of production and its rate of decay.

$$ \frac{dN_B}{dt} = \lambda_A N_A - \lambda_B N_B $$

Here, $\lambda_A N_A$ represents the rate at which $B$ is being formed from $A$. Since $A$ undergoes standard exponential decay, $N_A = N_0 e^{-\lambda_A t}$. The term $\lambda_B N_B$ represents the rate at which $B$ is decaying into $C$.

### Analyzing the Graph's Journey

Let's trace the journey of $N_B$ from the very beginning. At $t = 0$, the problem states there are no $B$ atoms in the sample. So, $N_B = 0$. This means our graph must strictly start from the origin.

In the initial phase, there is an abundance of $A$ atoms and zero $B$ atoms. Therefore, the production rate $\lambda_A N_A$ is at its maximum, while the decay rate $\lambda_B N_B$ is zero. The number of $B$ atoms will start to rise rapidly.

As time goes on, $N_B$ increases, which means its decay rate $\lambda_B N_B$ also increases. Meanwhile, the pool of $A$ atoms is depleting, so the production rate $\lambda_A N_A$ is dropping. Eventually, these two rates will perfectly balance each other.

$$ \lambda_A N_A = \lambda_B N_B $$

At this exact moment, $\frac{dN_B}{dt} = 0$. The graph reaches its **maximum peak**.

### The Final Phase

After this peak, the parent nucleus $A$ is mostly exhausted. The production of new $B$ atoms slows down to a trickle. However, we now have a large population of $B$ atoms, so they are decaying into $C$ very quickly.

The decay rate now dominates the production rate. The number of $B$ atoms will start to fall. As $t 	o \infty$, the supply of $A$ becomes negligible, and $B$ simply undergoes standard exponential decay, asymptotically approaching zero.

Therefore, the correct graph must start at the origin, rise to a peak, and then decay exponentially. This perfectly matches the curve shown in option (b).

Select Answer:

Visualized Solution

$A \to B \to C$

$\frac{d N _{B}}{d t} = λ_{A} N_{A} - λ_{B} N_{B}$

$t = 0 ⟹ N_{B} = 0$

$\frac{d N _{B}}{d t} = 0$

$t \to \infty ⟹ N_{B} \to 0$

\text{Final Graph}

The Sigma Insight: Radioactivity

The Phenomenon of Sequential Decay

The Mathematics of Growth and Decay

Analyzing the Graph's Journey

The Final Phase

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