Animated Solution for Physics - Atoms and Nuclei: In a radioactive decay chain, ${}_{90}^{232}\text{Th}$ nucleus decays to ${}_{82}^{212}\text{Pb}$ nucleus. Let $N_{\alpha }$ and $N_{\beta }$ be the number of $\alpha$ and $\beta$- particles respectively, emitted in this decay process. Which of the following statements is (are) true?

Radioactive decay chains are like nature's alchemy, transforming heavy, unstable elements into lighter, more stable ones through a series of particle emissions. In this problem, we are looking at the decay of Thorium-232 into Lead-212. This transformation doesn't happen in a single step; it involves the emission of multiple alpha ($\alpha$) and beta ($\beta$) particles. Let's break down the physics behind this process to find out exactly how many of each particle are emitted.

We start with a parent nucleus, Thorium, denoted as ${}_{90}^{232}\text{Th}$. The superscript 232 is the mass number ($A$), which is the total number of protons and neutrons. The subscript 90 is the atomic number ($Z$), which is the number of protons. The final product is Lead, ${}_{82}^{212}\text{Pb}$.

During this decay, the nucleus emits $\alpha$ particles and $\beta$ particles. An $\alpha$ particle is essentially a helium nucleus, ${}_2^4\text{He}$, meaning it carries away a mass of 4 units and a charge of 2 units. A $\beta$ particle is a high-speed electron, ${}_{-1}^0\text{e}$, which carries away virtually zero mass but effectively increases the atomic number of the nucleus by 1 (since a neutron turns into a proton).

Because $\beta$ particles have a mass number of 0, any change in the total mass number of the nucleus must be entirely due to the emission of $\alpha$ particles. This is our golden key to solving the problem.

First, let's find the total change in the mass number:

Since each $\alpha$ particle reduces the mass number by 4, we can easily find the number of $\alpha$ particles ($N_{\alpha }$) by dividing the total mass change by 4:

So, exactly 5 $\alpha$ particles are emitted in this decay chain.

Now that we know 5 $\alpha$ particles are emitted, let's see how they affect the atomic number. Each $\alpha$ particle reduces the atomic number by 2. Therefore, 5 $\alpha$ particles should reduce the atomic number by:

If only $\alpha$ particles were emitted, the final atomic number would be:

However, the actual atomic number of the final Lead nucleus is 82. There is a discrepancy here! The atomic number is higher than expected by $82-80=2$ units.

This is where the $\beta$ particles come into play. Each $\beta$ emission increases the atomic number by 1. To make up for the difference of 2 units, the nucleus must have emitted exactly 2 $\beta$ particles.

Through our step-by-step analysis, we have determined that the decay chain from ${}_{90}^{232}\text{Th}$ to ${}_{82}^{212}\text{Pb}$ involves the emission of 5 $\alpha$ particles and 2 $\beta$ particles.

Looking at the given options, statement (a) $N_{\alpha }=5$ and statement (c) $N_{\beta }=2$ are the correct choices.