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 $$\frac{N_B(t)}{N_B(0)}$$ with respect to time $$t$$?

We need to determine the correct graphical representation for the time evolution of the ratio $$\frac{N_B(t)}{N_B(0)}$$ for a sequential radioactive decay process.

1. Identify the Given Parameters

From the problem , the sequential disintegration scheme is represented as $$A \xrightarrow{\lambda} B \xrightarrow{\lambda} C$$, with the following initial conditions at time $$t = 0$$:

• Initial population of nuclei $$A$$ = $$N_A(0)$$
• Initial population of nuclei $$B$$ = $$N_B(0)$$, where $$N_A(0) = 2N_B(0)$$
• Decay constant for both atoms = $$\lambda$$

2. Formulate the Rate Equations

The rate of change of the population of nuclei $$B$$ over time depends simultaneously on its production from $$A$$ and its own decay into $$C$$:

$$\frac{dN_B}{dt} = \lambda N_A - \lambda N_B$$

Since substance $$A$$ undergoes standard isolated exponential decay, its population at any time $$t$$ is given by $$N_A(t) = N_A(0)e^{-\lambda t}$$. Substituting this into the differential equation yields:

$$\frac{dN_B}{dt} + \lambda N_B = \lambda N_A(0)e^{-\lambda t}$$

3. Solve for the Population Ratio $$\frac{N_B(t)}{N_B(0)}$$

Using an integrating factor $$I.F. = e^{\int \lambda dt} = e^{-\lambda t}$$, the general solution for the population of $$B$$ over time simplifies to:

$$N_B(t) = \left( N_B(0) + \lambda N_A(0)t \right) e^{-\lambda t}$$

Substituting the given relation $$N_A(0) = 2N_B(0)$$ into the expression:

$$N_B(t) = \left( N_B(0) + 2\lambda N_B(0)t \right) e^{-\lambda t} = N_B(0)(1 + 2\lambda t)e^{-\lambda t}$$

Dividing by $$N_B(0)$$ gives the final time-dependent ratio function:

$$\frac{N_B(t)}{N_B(0)} = (1 + 2\lambda t)e^{-\lambda t}$$

4. Analyze the Critical Points of the Function

To determine the nature of the curve graph, we examine its behavior at critical boundaries and find its local extrema by taking the first derivative:

• At Initial Time ($t = 0$):
  Evaluating the ratio at the origin:

  $$\frac{N_B(0)}{N_B(0)} = (1 + 0)e^{0} = 1$$

• Finding the Maximum Point ($$\frac{d}{dt} = 0$):
  Applying the product rule to differentiate the ratio function with respect to $$t$$:

  $$\frac{d}{dt}\left[\frac{N_B(t)}{N_B(0)}\right] = 2\lambda e^{-\lambda t} - \lambda(1 + 2\lambda t)e^{-\lambda t} = \lambda(1 - 2\lambda t)e^{-\lambda t}$$

  Setting the derivative equal to zero to find the turnaround peak point:

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

Since the derivative is positive ($>0$) for $$t < \frac{1}{2\lambda}$$, the curve initially rises up to a peak value at $$t = \frac{1}{2\lambda}$$, and then exponentially decays towards zero as $$t \to \infty$$.

Conclusion

The correct figure shows the ratio starting at 1, increasing to a maximum value at time $$t = \frac{1}{2\lambda}$$, and then smoothly decreasing towards zero.