Animated Solution for Physics - Atoms and Nuclei: In the uranium radioactive series, the initial nucleus is ${}_{92}^{238}\text{U}$ and the final nucleus is ${}_{82}^{206}\text{Pb}$. When the uranium nucleus decays to lead, the number of $\alpha$-particles emitted is ... and the number of $\beta$-particles emitted is ..... .

The journey of a radioactive nucleus is one of the most fascinating phenomena in modern physics. In this problem, we are looking at the granddaddy of all decay chains: the Uranium series. We start with the heavy, unstable Uranium-238 nucleus and end up with the stable Lead-206 nucleus. But how does it get there? It emits a series of $\alpha$ and $\beta$ particles. Our goal is to find out exactly how many of each are emitted.

Let's first understand what these particles are. An $\alpha$-particle is essentially a Helium nucleus, denoted as ${}_2^4\text{He}$. When a nucleus emits an $\alpha$-particle, it loses 4 units of mass number ($A$) and 2 units of atomic number ($Z$).

On the other hand, a $\beta$-particle is a high-speed electron, denoted as ${}_{-1}^0\text{e}$. Emitting a $\beta$-particle doesn't change the mass number at all, but it increases the atomic number by 1 (since a neutron turns into a proton).

We can write the entire decay process as a single nuclear equation: Here, $n_1$ is the number of $\alpha$-particles and $n_2$ is the number of $\beta$-particles.

In any nuclear reaction, the total mass number must be conserved. Let's balance the mass numbers on both sides of our equation. The initial mass is 238. The final mass is 206 from Lead, plus $4\times n_1$ from the $\alpha$-particles. The $\beta$-particles contribute nothing to the mass number.

This is a straightforward linear equation. Let's solve for $n_1$:

So, exactly 8 $\alpha$-particles are emitted in this decay series.

Now that we know the number of $\alpha$-particles, we can find the number of $\beta$-particles by conserving the atomic number (charge). The initial atomic number is 92. The final atomic number is 82 from Lead, plus $2\times n_1$ from the $\alpha$-particles, minus $1\times n_2$ from the $\beta$-particles.

We already found that $n_1=8$. Let's substitute this value into our equation:

Solving for $n_2$, we get:

Thus, 6 $\beta$-particles are emitted.

The Uranium-238 nucleus undergoes a long cascade of decays, emitting a total of 8 $\alpha$-particles and 6 $\beta$-particles before finally finding stability as Lead-206. This beautiful balance of mass and charge is a cornerstone of nuclear physics!