This problem involves radioactive decay and the interpretation of a graph to find the half-life and decay factor. Let's break it down step by step.

Step 1: Understanding the Graph

The student plots $\ln \left|\frac{\mathrm{dN}\left(t\right)}{\mathrm{dt}}\right|$ versus time $t$. The rate of radioactive decay is given by $\frac{\mathrm{dN}\left(t\right)}{\mathrm{dt}}$. For radioactive decay, this rate is proportional to the number of nuclei present at time $t$, expressed by the law:

$\left|\frac{\mathrm{dN}}{\mathrm{dt}}\right|=\lambda N$

where $\lambda$ is the decay constant.

Step 2: Expressing the Y-axis Quantity

Taking the natural logarithm of both sides:

$\ln \left|\frac{\mathrm{dN}}{\mathrm{dt}}\right|=\ln \left(\lambda N\right)=\ln \lambda +\ln N$

The number of nuclei $N$ at time $t$ is given by $N=N_0{e}^{-\lambda t}$, so $\ln N=\ln N_0-\lambda t$.

Substituting this in:

$\ln \left|\frac{\mathrm{dN}}{\mathrm{dt}}\right|=\ln \lambda +\ln N_0-\lambda t$

This is of the form $y=c+mx$, where $y=\ln \left|\frac{\mathrm{dN}}{\mathrm{dt}}\right|$, $x=t$, intercept $c=\ln \lambda +\ln N_0$, and slope $m=-\lambda$.

Step 3: Finding the Decay Constant from the Graph

The graph is a straight line with a negative slope. From the graph, when $t=0$, $\ln \left|\frac{\mathrm{dN}}{\mathrm{dt}}\right|=8$. When $t=4.16$ years, $\ln \left|\frac{\mathrm{dN}}{\mathrm{dt}}\right|=0$.

Using the slope formula:

$m=\frac{0-8}{4.16-0}=\frac{-8}{4.16}$

But $m=-\lambda$, so:

$-\lambda =\frac{-8}{4.16}$

$\lambda =\frac{8}{4.16}=\frac{800}{416}=\frac{100}{52}=\frac{25}{13}$ per year.

Step 4: Relating Decay Constant to Half-Life

The half-life $T_{1/2}$ is given by:

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

Substituting $\lambda =\frac{25}{13}$:

$T_{1/2}=\frac{0.693}{\frac{25}{13}}=0.693\times \frac{13}{25}=0.36036$ years.

This calculated half-life is for reference; we don't need its numerical value for the next step.

Step 5: Finding the Decay Factor p

The number of nuclei decreases by a factor of p after 4.16 years. This means $\frac{N}{N_0}=\frac{1}{p}$ at $t=4.16$ years.

Using the decay law:

$N=N_0{e}^{-\lambda t}$

$\frac{N}{N_0}=e^{-\lambda t}=\frac{1}{p}$

Taking natural logarithm:

$\ln \left(\frac{1}{p}\right)=-\lambda t$

$-\ln p=-\lambda t$

$\ln p=\lambda t$

Substitute $\lambda =\frac{8}{4.16}$ and $t=4.16$:

$\ln p=\left(\frac{8}{4.16}\right)\times 4.16=8$

Therefore, $p=e^8$

Final Answer

The value of p is $e^8$.

Related Topics and Formulae

Radioactive Decay Law: The fundamental law governing radioactive decay is $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda N$, which leads to the exponential solution $N=N_0{e}^{-\lambda t}$.

Half-Life: The time taken for half the radioactive nuclei to decay is $T_{1/2}=\frac{\ln 2}{\lambda }$.

Logarithmic Plots: Plotting the logarithm of a quantity that changes exponentially results in a straight line, whose slope is related to the rate constant. This is a powerful graphical method for analyzing exponential processes.