Animated Solution for Physics - Atoms and Nuclei: Activities of three radioactive substances $A$, $B$ and $C$ are represented by the curves $A$, $B$ and $C$ in the figure. Then, their half-lives $T_{1/2}(A):T_{1/2}(B):T_{1/2}(C)$ are in the ratio

Graphical problems in radioactivity are incredibly elegant because they allow us to visualize exponential decay as simple straight lines. In this problem, we are given a graph of $\ln R$ versus time $t$ for three different radioactive substances: A, B, and C. Our goal is to find the ratio of their half-lives.

To do this, we first need to understand the mathematical relationship governing the graph.

The fundamental law of radioactive decay states that the activity $R$ at any time $t$ is given by:

where $R_0$ is the initial activity and $\lambda$ is the decay constant.

Because dealing with exponential curves visually is tricky, we linearize the equation by taking the natural logarithm on both sides:

If we compare this to the standard equation of a straight line, $y=mx+c$, we can immediately see the physical significance of the graph's features: - The y-axis represents $y=\ln R$. - The x-axis represents $x=t$. - The y-intercept $c=\ln R_0$. - The slope $m=-\lambda$.

This means the decay constant $\lambda$ is simply the negative of the slope of the line!

Let's extract the slopes for each curve directly from the given coordinates.

For Substance A: The line starts at $(0,6)$ and ends at $(10,0)$. Therefore, ${\lambda }_A=\frac{6}{10}$.

For Substance B: The line starts at $(0,6)$ and ends at $(5,0)$. Therefore, ${\lambda }_B=\frac{6}{5}$.

For Substance C: The line starts at $(0,2)$ and ends at $(5,0)$. Therefore, ${\lambda }_C=\frac{2}{5}$.

We know that the half-life $T_{1/2}$ is inversely proportional to the decay constant:

Now, we can write the half-lives for all three substances: - $T_{1/2}(A)=\frac{10}{6}\ln 2$ - $T_{1/2}(B)=\frac{5}{6}\ln 2$ - $T_{1/2}(C)=\frac{5}{2}\ln 2$

Finally, we take their ratio:

To clear the denominators, we can multiply the entire ratio by 6:

Dividing by 5, we get our final, beautifully simplified ratio:

This perfectly matches option (d).

As a bonus insight, notice how curves A and B start at the exact same point on the y-axis. This tells us that despite having different half-lives, substances A and B had the exact same initial activity $R_0$ at $t=0$.