Animated Solution for Physics - Atoms and Nuclei: A radioactive element decays by $\beta$-emission. A detector records $n$ beta particles in $2\text{ s}$ and in next $2\text{ s}$ it records $0.75n$ beta particles. Find mean life of the radioactive element. Given $\ln |2|=0.6931,\ln |3|=1.0986$.

The problem of radioactive decay often feels like trying to predict the unpredictable. We know that individual nuclei decay randomly, but as a massive group, they follow a beautifully precise mathematical law. In this problem, we are acting as nuclear detectives, using the counts from a beta-particle detector to uncover the hidden "mean life" of a radioactive element.

Imagine a sample of a radioactive element sitting in front of a detector. As the nuclei decay, they emit $\beta$-particles. The detector is counting these particles.

The fundamental law of radioactive decay tells us that the number of active nuclei at any time $t$ is $N(t)=N_0{e}^{-\lambda t}$. However, the detector doesn't count the active nuclei; it counts the decayed nuclei! Every time a nucleus decays, a $\beta$-particle is emitted. Therefore, the number of particles recorded in time $t$ is exactly equal to the number of nuclei that have decayed:

We are given two pieces of data: 1. In the first $2\text{ s}$, the detector records $n$ particles. 2. In the next $2\text{ s}$, it records $0.75n$ particles.

Let's translate this into math. For the first $2\text{ s}$ ($t=2$):

For the total time of $4\text{ s}$ ($t=4$), the total number of particles recorded is the sum of the particles from the first interval and the second interval. That is $n+0.75n=1.75n$.

We now have a system of two equations. To solve for the decay constant $\lambda$, we can divide the second equation by the first to eliminate both $N_0$ and $n$:

This looks like a messy exponential equation, but a simple substitution turns it into basic algebra. Let $x=e^{-2\lambda }$. Then $e^{-4\lambda }=(e^{-2\lambda }{)}^2=x^2$.

Notice that the numerator is a difference of squares! We can factor it as $(1-x)(1+x)$.

Since $x=e^{-2\lambda }$ and $\lambda >0$, $x$ cannot be exactly $1$. Thus, we can safely cancel the $(1-x)$ terms:

Now we substitute back $x=e^{-2\lambda }$:

Taking the natural logarithm of both sides:

We are given the values $\ln (2)=0.6931$ and $\ln (3)=1.0986$. Let's plug them in:

The question asks for the mean life ($t_{\text{mean}}$) of the radioactive element. The mean life is simply the reciprocal of the decay constant $\lambda$: