A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine?
[Given : N = No. of bacteria, t = time, bacteria growth follows $$I^{st}$$ order kinetics.]

According to the problem, the bacterial growth follows first-order kinetics.

The differential rate law for first-order growth is:

$$\frac{dN}{dt} = kN$$

Integrating this expression from $$t = 0$$ (where the initial number of bacteria is $$N_0$$) to time $$t$$:

$$\int_{N_0}^{N} \frac{1}{N} \, dN = k \int_{0}^{t} dt$$
$$\ln\left(\frac{N}{N_0}\right) = kt \implies \frac{N}{N_0} = e^{kt}$$

• Graph Analysis: A plot of $$\frac{N}{N_0}$$ versus time $$t$$ will be an exponentially increasing curve.
• At $$t = 0$$, $$\frac{N}{N_0} = e^0 = 1$$. Thus, the graph begins at $1$ on the vertical axis and curves upwards.

After Applying the Medicine (Bacterial Decay)

The problem states that the rate of bacterial decay ($$r$$) is directly proportional to the square of the existing number of bacteria ($$N$$).

The rate equation is given by:

$$r = -\frac{dN}{dt} = k'N^2$$

• Graph Analysis: We are plotting the decay rate $$r$$ versus the number of bacteria $$N$$.
• Since $$r = k'N^2$$, the relationship follows the equation of a parabola ($$y = ax^2$$).
• When $$N = 0$$, the rate $$r = 0$$, meaning the curve passes through the origin $$(0,0)$$ and curves upwards increasingly as $$N$$ increases.

Correct option is B