Let N and N0 be the number of radioactive nuclei in a sample at time t and at time t = 0, respectively. Then the ratio $\left(\frac{N}{N_0}\right)$ is equal to _____ where λ is the disintegration constant or decay constant.

Question

Let N and N0 be the number of radioactive nuclei in a sample at time t and at time t = 0, respectively. Then the ratio $\left(\frac{N}{N_0}\right)$ is equal to _____ where &lambda; is the disintegration constant or decay constant.

Prepp Team · Accepted Answer

Radioactive Decay Law Explained

Radioactive decay is a fundamental process in nuclear physics where an unstable atomic nucleus transforms into a more stable nucleus by emitting radiation. This process is spontaneous and follows a predictable mathematical pattern over time. The question focuses on understanding the relationship between the initial number of radioactive nuclei and the number remaining after a certain period.

The problem defines $N$ as the number of radioactive nuclei in a sample at time $t$ and $N_0$ as the initial number of radioactive nuclei in the sample at time $t = 0$. It also introduces $\lambda$ as the disintegration constant, also known as the decay constant. We need to find the ratio $\left(\frac{N}{N_0}\right)$.

Number of Radioactive Nuclei: The Decay Law

The law of radioactive decay states that the rate at which radioactive nuclei disintegrate is directly proportional to the number of nuclei present at that given instant. This can be expressed as a differential equation:

\[\frac{dN}{dt} = -\lambda N\]

Where:

$\frac{dN}{dt}$ represents the rate of change of the number of nuclei with respect to time.
$N$ is the number of radioactive nuclei present at time $t$.
$\lambda$ is the disintegration constant, a characteristic constant for each radioactive isotope.
The negative sign indicates that the number of radioactive nuclei decreases over time as they decay.

Deriving the Ratio $\left(\frac{N}{N_0}\right)$

To find the expression for $N$ as a function of time, we can rearrange and integrate the differential equation:

First, separate the variables:

\[\frac{dN}{N} = -\lambda dt\]

Next, integrate both sides. We integrate $N$ from its initial value $N_0$ (at $t=0$) to $N$ (at time $t$), and $t$ from $0$ to $t$:

\[\int_{N_0}^{N} \frac{dN}{N} = \int_{0}^{t} -\lambda dt\]

Performing the integration:

\[[\ln N]_{N_0}^{N} = [-\lambda t]_{0}^{t}\]

Applying the limits of integration:

\[\ln N - \ln N_0 = (-\lambda t) - (-\lambda \cdot 0)\]

\[\ln \left(\frac{N}{N_0}\right) = -\lambda t\]

Finally, to solve for the ratio $\left(\frac{N}{N_0}\right)$, we take the exponential of both sides:

\[\frac{N}{N_0} = e^{-\lambda t}\]

This equation is the fundamental law of radioactive decay, showing the exponential decrease of radioactive nuclei over time.

Disintegration Constant and its Significance

The disintegration constant $\lambda$ is crucial for understanding the decay process. It represents the probability per unit time that a nucleus will decay. A larger value of $\lambda$ indicates that the substance decays more rapidly and therefore has a shorter half-life.

Units of $\lambda$: The units of $\lambda$ are typically inverse time units, such as s^-1, min^-1, or year^-1. This ensures that the exponent $-\lambda t$ is dimensionless, as exponents must always be dimensionless quantities.

Matching with the Options

We derived the ratio $\left(\frac{N}{N_0}\right)$ to be $e^{-\lambda t}$. Let's compare this with the given options:

Option Number	Expression	Matches Derived Formula?
1	$e^{-(2\lambda)t}$	No
2	$e^{-(\lambda)t}$	Yes
3	$e^{-(\frac{\lambda}{2})t}$	No
4	$e^{-(\frac{\lambda}{4})t}$	No

As observed, option 2, $e^{-(\lambda)t}$, perfectly matches our derived formula for the ratio of radioactive nuclei at time $t$ to the initial number of nuclei.