Decay Factor Calculation from Logarithmic Activity Time Graph

From the radioactive decay law, the number of nuclei $N(t)$ remaining at time $t$ is given by:
$N(t) = N_0 e^{-\lambda t}$

The magnitude of the rate of decay (activity) at time $t$ is:
$\left|\frac{\mathrm{d}N(t)}{\mathrm{d}t}\right| = \lambda N(t) = \lambda N_0 e^{-\lambda t}$

Taking the natural logarithm of both sides:
$\ln\left|\frac{\mathrm{d}N(t)}{\mathrm{d}t}\right| = \ln(\lambda N_0) - \lambda t$

This represents a linear relation between $\ln\left|\frac{\mathrm{d}N(t)}{\mathrm{d}t}\right|$ and $t$, where the magnitude of the slope is the decay constant $\lambda$:
$\text{Slope} = -\lambda$

From the given graph, we identify points on the line:

• At $t = 3\text{ years}$, $\ln\left|\frac{\mathrm{d}N(t)}{\mathrm{d}t}\right| \approx 4.5$
• At $t = 7\text{ years}$, $\ln\left|\frac{\mathrm{d}N(t)}{\mathrm{d}t}\right| \approx 2.5$

Thus, the decay constant is:
$\lambda = -\frac{2.5 - 4.5}{7 - 3} = \frac{2}{4} = 0.5\text{ year}^{-1}$

The half-life $T_{1/2}$ of the element is:
$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.5} \approx 1.386\text{ years}$

We are given that after $t = 4.16\text{ years}$, the number of radioactive nuclei decreases by a factor of $p$:
$\frac{N_0}{N(t)} = e^{\lambda t} = p$

Taking natural logarithm on both sides:
$\ln p = \lambda t = 0.5 \times 4.16 = 2.08$

Since $\ln 2 \approx 0.693$:
$2.08 \approx 3 \times 0.693 = 3 \ln 2 = \ln(2^3) = \ln 8$

Therefore,
$p = 8$