Q.Define the term 'decay constant' of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10000 disintegrations/s and 5,000 disintegrations/s after 20 hr. and 30 hr. respectively from start. Calculate the half life and initial number of nuclei at $t = 0$.

The key idea here is Radioactive Decay Kinetics, specifically the exponential decay law for activity and the relationship between decay constant and half-life.

The decay constant ($\lambda$) of a radioactive sample is defined as the probability per unit time for a single nucleus to undergo radioactive decay. It is a characteristic constant for a given radioactive nuclide, independent of external conditions, and its unit is typically s$^{-1}$ or hr$^{-1}$.

Here's the essential reasoning:

1. The rate of disintegration $R(t)$ at time $t$ is given by $R(t) = R_0 e^{-\lambda t}$, where $R_0$ is the initial rate of disintegration and $\lambda$ is the decay constant.

   Given $R_1 = 10000$ disintegrations/s at $t_1 = 20$ hr and $R_2 = 5000$ disintegrations/s at $t_2 = 30$ hr.

   Dividing the expressions for $R_1$ and $R_2$:

   $\frac{R_1}{R_2} = \frac{R_0 e^{-\lambda t_1}}{R_0 e^{-\lambda t_2}} = e^{-\lambda (t_1 - t_2)}$

   $\frac{10000}{5000} = e^{-\lambda (20 \text{ hr} - 30 \text{ hr})}$

   $2 = e^{10\lambda}$

   Taking the natural logarithm of both sides:

   $\ln(2) = 10\lambda \implies \lambda = \frac{\ln(2)}{10}$ hr$^{-1}$.

2. The half-life ($T_{1/2}$) is related to the decay constant by the formula $T_{1/2} = \frac{\ln(2)}{\lambda}$.

   Substituting the value of $\lambda$:

   $T_{1/2} = \frac{\ln(2)}{\frac{\ln(2)}{10 \text{ hr}}} = 10$ hr.

3. To find the initial number of nuclei ($N_0$), first calculate the initial rate of disintegration ($R_0$). Using $R_1 = R_0 e^{-\lambda t_1}$:

   $10000 = R_0 e^{-\left(\frac{\ln(2)}{10}\right) \times 20}$

   $10000 = R_0 e^{-2\ln(2)} = R_0 (e^{\ln(2)})^{-2} = R_0 (2)^{-2} = \frac{R_0}{4}$ …

Radioactive decay is a fundamental process where unstable atomic nuclei spontaneously transform into more stable forms, emitting radiation in the process. This transformation is probabilistic, meaning we cannot predict when a single nucleus will decay, but for a large sample, the rate of decay follows a predictable pattern. This pattern is crucial for applications like radioactive dating, where we determine the age of ancient artifacts or geological formations by measuring the remaining amount of a radioactive isotope.

The core idea is that the rate of disintegration (or activity) of a radioactive sample is directly proportional to the number of radioactive nuclei present at that moment. As nuclei decay, their number decreases, and consequently, the rate of disintegration also decreases over time. This relationship is described by an exponential decay law. By measuring the activity at different times, or knowing the current activity and the initial activity, we can determine the decay constant and thus the half-life of the sample, which tells us how quickly it decays.

Definition of Decay Constant

The decay constant ($\lambda$) is a characteristic property of a particular radioactive isotope. It represents the probability per unit time that a nucleus will decay. It is a measure of the intrinsic instability of a radioactive nucleus. A larger decay constant implies a higher probability of decay per unit time, meaning the substance decays more rapidly and has a shorter half-life. Its unit is typically s$^{-1}$ or (time unit)$^{-1}$.

Step-by-Step Solution

1. Understand the given information:

   We are given the rate of disintegration (activity) at two different times:

   • At $t_1 = 20$ hours, the activity $A_1 = 10000$ disintegrations/s (dps).
   • At $t_2 = 30$ hours, the activity $A_2 = 5000$ disintegrations/s (dps). We need to find the half-life ($T_{1/2}$) and the initial number of nuclei ($N_0$) at $t=0$.

2. Recall the radioactive decay law for activity:

   The activity $A$ of a radioactive sample at time $t$ is given by:

   $A = A_0 e^{-\lambda t}$

   where $A_0$ is the initial activity at $t=0$, and $\lambda$ is the decay constant.

   Using the given data, we can write two equations:

   • For $t_1 = 20$ hr: $10000 = A_0 e^{-\lambda (20 \text{ hr})}$ \quad (Equation 1)
   • For $t_2 = 30$ hr: $5000 = A_0 e^{-\lambda (30 \text{ hr})}$ \quad (Equation 2)

3. Calculate the decay constant ($\lambda$):

   To find $\lambda$, we can divide Equation 1 by Equation 2. This eliminates $A_0$:

Taking the natural logarithm on both sides:

So, the decay constant is $\frac{\ln 2}{10}$ per hour.

> [!WARNING]
> Be careful with units. Since the time difference was in hours, $\lambda$ is in (hours)$^{-1}$. If we need $\lambda$ in s$^{-1}$ later (e.g., for calculating $N_0$ from dps), we must convert it.

4. Calculate the half-life ($T_{1/2}$):

The half-life is the time it takes for half of the radioactive nuclei in a sample to decay. It is related to the decay constant by the formula:

> [!FORMULA]

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

Substitute the value of $\lambda$ we just found:

The half-life of the radioactive nucleus is 10 hours. …