In a radioactive decay chain, the initial nucleus is 232_90 Th. At the end, there are 6 alpha-partic

Q: Why does beta decay increase the atomic number?

During beta decay, a neutron inside the nucleus converts into a proton and an electron. The electron is emitted as a beta particle, leaving an extra proton in the nucleus. Since the atomic number is the count of protons, it increases by 1.

Q: Does beta decay affect the mass number?

No, it does not. The mass number is the total count of protons and neutrons. Since a neutron simply turns into a proton, the total count of these nucleons remains exactly the same.

Q: What happens if a positron ($\beta^+$) is emitted instead?

Positron emission is the opposite of standard beta decay. A proton converts into a neutron and a positron. This decreases the atomic number by 1, while the mass number still remains unchanged.

Q: Can we do the calculation in any order? What if beta decays happened first?

Yes, the net change in mass and atomic numbers depends only on the total number of particles emitted, not the sequence of emission. The final values of $A$ and $Z$ will be exactly the same.

Q: How do we know which element the final nucleus is?

The atomic number $Z$ uniquely identifies the chemical element. In this case, $Z=82$ corresponds to Lead (Pb). So the final stable nucleus is an isotope of Lead.

Question

In a radioactive decay chain, the initial nucleus is $^{232}_{90}	ext{Th}$. At the end, there are $6\alpha$-particles and $4\beta$-particles which are emitted. If the end nucleus is $^{A}_{Z}	ext{X}$, $A$ and $Z$ are given by

Options:

(A) $A = 202; Z = 80$

(B) $A = 208; Z = 82$

(C) $A = 200; Z = 81$

(D) $A = 208; Z = 80$

Sigma Percentile · Accepted Answer

In the fascinating world of nuclear physics, radioactive decay is nature's way of transforming unstable elements into more stable ones. This problem takes us on a journey through a radioactive decay chain, starting with a heavy Thorium nucleus and ending up with a completely different element. Let's break down the mechanics of this transformation step by step.

### The Mechanics of Alpha Decay

Our journey begins with the initial nucleus, Thorium-232, denoted as $^{232}_{90}	ext{Th}$. The problem states that this nucleus undergoes $6\alpha$ decays. But what exactly is an alpha particle?

An alpha particle is essentially a Helium nucleus, represented as $^{4}_{2}	ext{He}$. It consists of two protons and two neutrons. Therefore, every time a nucleus emits an alpha particle, it loses a significant chunk of its mass and charge. Specifically, its mass number $A$ decreases by $4$, and its atomic number $Z$ decreases by $2$.

Since our Thorium nucleus emits $6$ alpha particles, we can calculate the total change in its mass and atomic numbers:

$$ \Delta A = 6 	imes (-4) = -24 $$
$$ \Delta Z = 6 	imes (-2) = -12 $$

Applying these changes to our initial Thorium nucleus, we get an intermediate nucleus, let's call it $	ext{Y}$:

$$ A' = 232 - 24 = 208 $$
$$ Z' = 90 - 12 = 78 $$

So, after the alpha decays, we are left with the nucleus $^{208}_{78}	ext{Y}$.

### The Subtlety of Beta Decay

Next, the problem tells us that the nucleus undergoes $4\beta$ decays. A beta particle is a high-energy electron emitted from the nucleus, represented as $_{-1}^{0}	ext{e}$.

During beta decay, a neutron inside the nucleus transforms into a proton and an electron. The electron is ejected as the beta particle. Because the mass of an electron is negligible compared to protons and neutrons, the total mass number $A$ of the nucleus remains completely unchanged. However, because we gained a proton, the atomic number $Z$ increases by $1$.

Let's calculate the total change for $4$ beta decays:

$$ \Delta A = 4 	imes 0 = 0 $$
$$ \Delta Z = 4 	imes (+1) = +4 $$

### The Final Destination

Now, we apply these changes to our intermediate nucleus $^{208}_{78}	ext{Y}$ to find the final nucleus $	ext{X}$:

$$ A = 208 + 0 = 208 $$
$$ Z = 78 + 4 = 82 $$

Our final nucleus is $^{208}_{82}	ext{X}$.

By simply keeping track of the mass and charge conservation, we have successfully navigated the decay chain. The final mass number is $A = 208$ and the atomic number is $Z = 82$, which perfectly matches option (b). This elegant bookkeeping is a fundamental skill in nuclear physics, allowing us to predict the outcomes of complex nuclear reactions.

Animated Solution for Physics - Atoms and Nuclei: In a radioactive decay chain, the initial nucleus is $_{90}^{232} Th$ . At the end, there are $6 α$ -particles and $4 β$ -particles which are emitted. If the end nucleus is $_{Z}^{A} X$ , $A$ and $Z$ are given by

Select Answer:

Visualized Solution

$Understanding α Decay$

$Effect of 6 α Decays$

$Understanding β Decay$

$Effect of 4 β Decays$

$Final Answer$

The Sigma Insight: Radioactivity

The Mechanics of Alpha Decay

The Subtlety of Beta Decay

The Final Destination

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