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:

We need to determine the approximate half-life ($$t_{1/2}$$) of an unknown radioactive material using the given linear graph between the natural logarithm of the decay rate ($$\ln R$$) and time ($$t$$).

1. Establish the Governing Radioactivity Equation

The decay rate (or activity) $$R$$ of a radioactive substance at any given time $$t$$ follows the exponential decay law:

$$R = R_0 e^{-\lambda t}$$

Where $$R_0$$ is the initial decay rate at $$t = 0$$, and $$\lambda$$ is the radioactive decay constant.

Taking the natural logarithm ($\ln$) on both sides of the equation:

$$\ln R = \ln(R_0 e^{-\lambda t})$$

$$\ln R = \ln R_0 + \ln(e^{-\lambda t})$$

$$\ln R = -\lambda t + \ln R_0$$

Comparing this with the standard equation of a straight line ($$y = mx + c$$), where $$y = \ln R$$ and $$x = t$$:

• Slope ($$m$$): $$-\lambda$$
• y-intercept ($$c$$): $$\ln R_0$$

2. Calculate the Decay Constant ($$\lambda$$) from the Graph

From the provided graph, we can easily identify two key points where the straight line intercepts the axes:

• When $$t = 0\text{ s}$$, $$\ln R = 6$$  $$\implies (0, 6)$$
• When $$t = 40\text{ s}$$, $$\ln R = 0$$  $$\implies (40, 0)$$

Now, we compute the slope ($$m$$) of this straight line:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 6}{40 - 0} = -\frac{6}{40} = -0.15\text{ s}^{-1}$$

Since the slope of our line is equal to $$-\lambda$$:

$$-\lambda = -0.15 \implies \lambda = 0.15\text{ s}^{-1}$$

3. Calculate the Half-Life ($$t_{1/2}$$)

The standard formula linking the half-life period to the decay constant is:

$$t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$$

Substituting the value of $$\lambda = 0.15\text{ s}^{-1}$$ into this relation:

$$t_{1/2} = \frac{0.693}{0.15} = \frac{69.3}{15} \approx 4.62\text{ sec}$$

Final Answer: 4.62 sec (Option D)