For a certain radioactive process, the graph between ln R and t (sec) is obtained as shown in the fi

Q: Why do we take the natural logarithm of the decay rate?

Taking the natural logarithm transforms the exponential decay curve into a straight line. Straight lines are much easier to analyze visually, and their slope gives us direct access to the decay constant without complex curve fitting.

Q: What does the y-intercept of this graph represent?

The y-intercept represents $\ln R_0$, which is the natural logarithm of the initial decay rate at time $t = 0$.

Q: Why is the slope negative?

The slope is negative because the decay rate decreases over time. The negative slope mathematically represents $-\lambda$, confirming that $\lambda$ itself is a positive constant.

Q: Can I use $\log_{10}$ instead of $\ln$?

You can, but the slope would then be $-\frac{\lambda}{2.303}$. Using the natural logarithm ($\ln$) keeps the math cleaner because the base of the exponential decay is $e$, and $\ln(e) = 1$.

Q: What if the graph didn't pass exactly through the grid intersections?

In real-world experiments, you would draw a line of best fit and pick two points on that line that are far apart to calculate the most accurate average slope.

Question

For a certain radioactive process, the graph between $\ln R$ and $t$ (sec) is obtained as shown in the figure. Then, the value of half-life for the unknown radioactive material is approximately

Options:

(A) 9.15 s

(B) 6.93 s

(C) 2.62 s

(D) 4.62 s

Sigma Percentile · Accepted Answer

The study of radioactive decay often involves exponential functions, which can be tricky to analyze visually. However, by applying a simple mathematical transformation—the natural logarithm—we can turn a curved exponential graph into a straight line. This makes extracting crucial physical constants, like the half-life, incredibly straightforward. Let's dive into how this works!

### Decoding the Graph

We are given a graph plotted with the natural logarithm of the decay rate ($\ln R$) on the y-axis and time ($t$) on the x-axis. To understand what this line represents, we must start with the fundamental law of radioactive decay:
$$R = R_0 e^{-\lambda t}$$
where $R$ is the decay rate at time $t$, $R_0$ is the initial decay rate, and $\lambda$ is the decay constant.

To match our graph, we take the natural logarithm on both sides of the equation:
$$\ln R = \ln(R_0 e^{-\lambda t})$$
Using the properties of logarithms, this simplifies to:
$$\ln R = -\lambda t + \ln R_0$$

### Extracting the Physics

Notice the beautiful structure of this new equation. It perfectly mirrors the standard equation of a straight line, $y = mx + c$. 
Here, our y-variable is $\ln R$, our x-variable is $t$, the y-intercept $c$ is $\ln R_0$, and most importantly, the slope $m$ is exactly $-\lambda$.

This means that if we can find the slope of the line from the graph, we immediately know the decay constant! Let's pick two easy-to-read points from the graph:
1. The y-intercept: At $t = 0$, $\ln R = 6$. So, $(t_1, y_1) = (0, 6)$.
2. The x-intercept: At $t = 40$ s, $\ln R = 0$. So, $(t_2, y_2) = (40, 0)$.

Now, we calculate the slope $m$:
$$m = \frac{y_2 - y_1}{t_2 - t_1} = \frac{0 - 6}{40 - 0} = -\frac{6}{40}$$

Since we established that $m = -\lambda$, we have:
$$-\lambda = -\frac{6}{40} \implies \lambda = \frac{6}{40} = 0.15 	ext{ s}^{-1}$$

### The Final Calculation

The question asks for the half-life ($t_{1/2}$) of the material. The half-life is inversely proportional to the decay constant, given by the formula:
$$t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$$

Substituting our exact fractional value for $\lambda$ to avoid early rounding errors:
$$t_{1/2} = \frac{0.693}{\left(\frac{6}{40}ight)} = \frac{0.693 	imes 40}{6}$$

Performing the final arithmetic:
$$t_{1/2} = 0.1155 	imes 40 = 4.62 	ext{ s}$$

**The half-life of the unknown radioactive material is approximately 4.62 seconds.** 
By simply understanding the geometry of the graph, we unlocked the hidden physics of the radioactive sample!

Animated Solution for Physics - Atoms and Nuclei: For a certain radioactive process, the graph between $ln R$ and $t$ (sec) is obtained as shown in the figure. Then, the value of half-life for the unknown radioactive material is approximately

Select Answer:

Visualized Solution

Analyzing the Graph

Radioactive Decay Law

Logarithmic Form

Equation of Straight Line

Identifying Points on Graph

Calculating Slope

Finding Decay Constant

Half-Life Formula

Substituting $λ$

Final Calculation

Physical Interpretation

The Sigma Insight: Radioactivity

Decoding the Graph

Extracting the Physics

The Final Calculation

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