Understanding Radioactive Decay and Decay Rates

Radioactive decay is a first-order process where unstable atomic nuclei transform into more stable ones by emitting particles or energy. The rate at which this decay occurs is characterized by several quantities, including the decay constant, half-life, and mean life.

Key Concepts: Half-life, Mean life, and Decay Constant

• Decay Constant ($\lambda$): This is the probability per unit time that a nucleus will decay. A larger decay constant means a faster decay.
• Half-life ($T_{1/2}$): This is the time it takes for half of the original radioactive nuclei in a sample to decay. It is related to the decay constant by the formula: $$\qquad T_{1/2} = \frac{\ln 2}{\lambda}$$
• Mean life ($\tau$): This is the average lifetime of a radioactive nucleus. It is the reciprocal of the decay constant: $$\qquad \tau = \frac{1}{\lambda}$$

From these definitions, we can see that half-life and mean life are inversely proportional to the decay constant. A larger decay constant corresponds to a shorter half-life and a shorter mean life, indicating faster decay.

Comparing Radioactive Elements X and Y

We are given two radioactive elements, X and Y. Let their decay constants be $\lambda_X$ and $\lambda_Y$ respectively. We are told:

1. The half-life period of element X is the same as the mean life of element Y. Mathematically, $$T_{1/2}(X) = \tau(Y)$$
2. Initially, both elements have the same number of atoms. Let the initial number of atoms be $N_0$, so $N_0(X) = N_0(Y) = N_0$.

Relating Decay Constants $\lambda_X$ and $\lambda_Y$

Using the relationships between half-life, mean life, and decay constant:

• For element X: $T_{1/2}(X) = \frac{\ln 2}{\lambda_X}$
• For element Y: $\tau(Y) = \frac{1}{\lambda_Y}$

Given that $T_{1/2}(X) = \tau(Y)$, we can write:

$$\frac{\ln 2}{\lambda_X} = \frac{1}{\lambda_Y}$$

Rearranging this equation to find the relationship between $\lambda_X$ and $\lambda_Y$:

$$\lambda_Y = \frac{\lambda_X}{\ln 2}$$

Since $\ln 2 \approx 0.693$, which is less than 1, the equation $\lambda_Y = \frac{\lambda_X}{\ln 2}$ tells us that $\lambda_Y$ is greater than $\lambda_X$.

$$\lambda_Y > \lambda_X$$

This is a crucial finding: Element Y has a larger decay constant than element X. A larger decay constant means a higher probability of decay per unit time.

Comparing Decay Rates

The decay rate (or activity), $A(t)$, of a radioactive sample at time $t$ is the number of decays per unit time. It is given by:

$$A(t) = \lambda N(t)$$

where $N(t)$ is the number of atoms remaining at time $t$, given by $N(t) = N_0 e^{-\lambda t}$. Thus, the decay rate can also be written as:

$$A(t) = \lambda N_0 e^{-\lambda t}$$

Initial Decay Rate (at t=0)

The initial decay rate is when $t=0$. $A(0) = \lambda N_0 e^{-\lambda \times 0} = \lambda N_0$.

• Initial decay rate of X: $A_X(0) = \lambda_X N_0$
• Initial decay rate of Y: $A_Y(0) = \lambda_Y N_0$

Since we found $\lambda_Y > \lambda_X$ and $N_0$ is the same for both, it follows that $A_Y(0) > A_X(0)$.

So, Y has a faster initial decay rate than X.

Decay Rate at Any Time t > 0

Let's compare the decay rates at any time $t > 0$:

• Decay rate of X: $A_X(t) = \lambda_X N_0 e^{-\lambda_X t}$
• Decay rate of Y: $A_Y(t) = \lambda_Y N_0 e^{-\lambda_Y t}$

Substitute $\lambda_Y = \lambda_X / \ln 2$ into the expression for $A_Y(t)$:

$$A_Y(t) = \left(\frac{\lambda_X}{\ln 2}\right) N_0 e^{-(\frac{\lambda_X}{\ln 2}) t}$$

Now compare $A_X(t)$ and $A_Y(t)$: We want to know if $A_Y(t)$ is greater than, less than, or equal to $A_X(t)$. Let's look at their ratio:

$$\frac{A_Y(t)}{A_X(t)} = \frac{(\frac{\lambda_X}{\ln 2}) N_0 e^{-(\frac{\lambda_X}{\ln 2}) t}}{\lambda_X N_0 e^{-\lambda_X t}} = \frac{1}{\ln 2} \frac{e^{-(\frac{\lambda_X}{\ln 2}) t}}{e^{-\lambda_X t}} = \frac{1}{\ln 2} e^{\lambda_X t - \frac{\lambda_X t}{\ln 2}} = \frac{1}{\ln 2} e^{\lambda_X t (1 - \frac{1}{\ln 2})}$$

Since $\ln 2 \approx 0.693$, $\frac{1}{\ln 2} \approx 1.44$. Also, $1 - \frac{1}{\ln 2} \approx 1 - 1.44 = -0.44$.

So, $\frac{A_Y(t)}{A_X(t)} = \frac{1}{\ln 2} e^{-\lambda_X t (\frac{1}{\ln 2} - 1)}$.

Since $\frac{1}{\ln 2} > 1$, $(\frac{1}{\ln 2} - 1)$ is positive. Therefore, the exponent $-\lambda_X t (\frac{1}{\ln 2} - 1)$ is negative for $t > 0$. This means $e^{-\lambda_X t (\frac{1}{\ln 2} - 1)}$ is a positive number less than 1 for $t > 0$. At $t=0$, the exponent is 0, and $e^0 = 1$.

So, the ratio $\frac{A_Y(t)}{A_X(t)} = \frac{1}{\ln 2} \times (\text{a number that is 1 at } t=0 \text{ and less than 1 for } t > 0)$.

Since $\frac{1}{\ln 2} \approx 1.44$, which is greater than 1, and the exponential term $e^{-\lambda_X t (\frac{1}{\ln 2} - 1)}$ is always positive (equal to 1 at $t=0$ and decreasing towards 0 as $t \to \infty$), let's evaluate the ratio at $t=0$ and as $t \to \infty$.

• At $t=0$: $\frac{A_Y(0)}{A_X(0)} = \frac{1}{\ln 2} e^0 = \frac{1}{\ln 2} \approx 1.44$. This is greater than 1, confirming $A_Y(0) > A_X(0)$.
• As $t \to \infty$: The exponent $\lambda_X t (1 - \frac{1}{\ln 2})$ becomes $\infty \times (\text{negative number}) = -\infty$. So $e^{\lambda_X t (1 - \frac{1}{\ln 2})} \to 0$. The ratio $\frac{A_Y(t)}{A_X(t)} \to \frac{1}{\ln 2} \times 0 = 0$. This means as time approaches infinity, the decay rates for both elements approach zero, but the ratio indicates how they compare at finite times.

We need to determine if $\frac{A_Y(t)}{A_X(t)} > 1$ for all $t$. This is equivalent to checking if $\frac{1}{\ln 2} e^{\lambda_X t (1 - \frac{1}{\ln 2})} > 1$. Since $1 - \frac{1}{\ln 2} < 0$, let $1 - \frac{1}{\ln 2} = -c$ where $c = \frac{1}{\ln 2} - 1 > 0$. The inequality is $\frac{1}{\ln 2} e^{-c \lambda_X t} > 1$, or $e^{-c \lambda_X t} > \ln 2$. Since $c \lambda_X t \ge 0$, $e^{-c \lambda_X t} \le 1$. Is it always greater than $\ln 2$? $\ln 2 \approx 0.693$. At $t=0$, $e^0 = 1$, and $1 > \ln 2$ is true. For $t > 0$, $e^{-c \lambda_X t}$ decreases from 1. It will remain greater than $\ln 2$ as long as $-c \lambda_X t > \ln(\ln 2)$, which means $c \lambda_X t < -\ln(\ln 2)$. Since $\ln(\ln 2)$ is $\ln(0.693...)$, which is a negative number, $-\ln(\ln 2)$ is a positive number. So, $t < \frac{-\ln(\ln 2)}{c \lambda_X}$. This shows there is a time beyond which $A_Y(t) < A_X(t)$.

Ah, let's re-read the options and the standard interpretation of "faster rate". When comparing radioactive decay, a substance with a larger decay constant ($\lambda$) is generally considered to decay "faster" because its activity $A(t) = \lambda N(t)$ is proportional to $\lambda$, and its half-life and mean life are shorter. The statement "Y will decay at a faster rate than X" often refers to the intrinsic property governed by the decay constant, or that its initial rate is higher and it decays more rapidly towards zero than X. Given $\lambda_Y > \lambda_X$, the substance Y is indeed more 'active' or decays more quickly. The question likely implies comparing the magnitudes of the decay constants or initial rates, or perhaps the time it takes for a certain fraction to decay.

Let's look at the options again based on $\lambda_Y > \lambda_X$ and $A_Y(0) > A_X(0)$.

Comparing the options:

• A. X and Y have the same decay rate initially. False, $A_Y(0) > A_X(0)$.
• B. X and Y decay at the same rate always. False, their decay constants are different, leading to different rates over time.
• C. Y will decay at a faster rate than X. This is consistent with $\lambda_Y > \lambda_X$ and $A_Y(0) > A_X(0)$. While $A_Y(t)$ might become less than $A_X(t)$ at very large times, the term "faster rate" in this context typically refers to the larger decay constant or initial rate, which dictates how quickly the sample reduces over time. Y reduces its number of atoms faster than X because of its larger decay constant.
• D. X will decay at a faster rate than Y. False, as $\lambda_Y > \lambda_X$.

Based on the standard interpretation where a larger decay constant signifies a faster decay, option C is the correct conclusion derived from $\lambda_Y > \lambda_X$. The rate $A(t) = \lambda N(t)$ is always proportional to $\lambda$ at any given fraction of remaining nuclei $N(t)/N_0$. Since $\lambda_Y > \lambda_X$, for any given number of remaining nuclei, the decay rate of Y is proportionally higher than X.

Summary of Analysis:

• Given $T_{1/2}(X) = \tau(Y)$.
• Using $T_{1_2} = \frac{\ln 2}{\lambda}$ and $\tau = \frac{1}{\lambda}$, we get $\frac{\ln 2}{\lambda_X} = \frac{1}{\lambda_Y}$.
• This implies $\lambda_Y = \frac{\lambda_X}{\ln 2}$. Since $\ln 2 < 1$, $\lambda_Y > \lambda_X$.
• The decay rate is $A = \lambda N$. Since $\lambda_Y > \lambda_X$, element Y decays faster.
• Initial decay rate $A_X(0) = \lambda_X N_0$ and $A_Y(0) = \lambda_Y N_0$. Since $N_0$ is same and $\lambda_Y > \lambda_X$, $A_Y(0) > A_X(0)$.

Option C states that Y will decay at a faster rate than X, which aligns with $\lambda_Y > \lambda_X$ and the initial decay rates. This is the standard interpretation in such comparison questions.

Final Conclusion

Based on the relationship derived from the given information, element Y has a larger decay constant ($\lambda_Y > \lambda_X$), meaning it decays more rapidly than element X. Therefore, Y will decay at a faster rate than X.

Radioactive Decay Revision Table

Additional Information on Radioactive Decay

Radioactive decay is a spontaneous process that follows exponential decay. The number of radioactive nuclei $N$ at time $t$ is given by the equation:

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

where $N_0$ is the initial number of nuclei and $\lambda$ is the decay constant. This equation shows that the number of nuclei decreases exponentially with time.

The decay rate or activity $A(t)$ is the rate of change of the number of nuclei (with a negative sign because N is decreasing), or equivalently, the number of decays per unit time:

$$A(t) = -\frac{dN}{dt} = \lambda N_0 e^{-\lambda t} = \lambda N(t)$$

The decay constant $\lambda$ is a fundamental property of a specific radioactive isotope. A larger $\lambda$ means the exponential term $e^{-\lambda t}$ decreases faster, leading to a quicker decrease in the number of nuclei and the decay rate over time. This is why a larger decay constant is associated with a faster decay process.

The half-life $T_{1/2}$ is a more intuitive measure of how quickly a substance decays. A shorter half-life means the substance decays away faster. Since $T_{1/2} = \frac{\ln 2}{\lambda}$, a larger $\lambda$ corresponds to a shorter $T_{1/2}$.

The mean life $\tau = \frac{1}{\lambda}$ is another way to characterize the decay speed. A larger $\lambda$ corresponds to a shorter $\tau$.

In this question, $T_{1/2}(X) = \tau(Y)$. This directly leads to $\frac{\ln 2}{\lambda_X} = \frac{1}{\lambda_Y}$, which means $\lambda_Y = \frac{\lambda_X}{\ln 2}$. Since $\ln 2 \approx 0.693$, $\frac{1}{\ln 2} \approx 1.44$. So $\lambda_Y \approx 1.44 \lambda_X$, clearly showing $\lambda_Y > \lambda_X$. Therefore, element Y decays intrinsically faster than element X.