A radioactive nucleus (initial mass number A and atomic number Z) emits 3 alpha-particles and 2 posi

Q: Why does the number of neutrons increase during positron emission?

In positron emission ($\beta^+$ decay), a proton inside the nucleus converts into a neutron to emit the positron ($p^+ \rightarrow n^0 + e^+$). Therefore, the nucleus loses a proton but gains a neutron.

Q: Does an alpha particle emission affect both protons and neutrons equally?

Yes! An alpha particle is a helium nucleus ($^4_2\text{He}$), which consists of exactly 2 protons and 2 neutrons. So, emitting one alpha particle reduces both the proton count and the neutron count by 2.

Q: Why is the initial number of neutrons $A - Z$?

The mass number $A$ represents the total number of nucleons (protons + neutrons) in the nucleus. The atomic number $Z$ represents just the protons. Subtracting the protons from the total gives you the neutrons: $A - Z$.

Q: What is the difference between $\beta^+$ and $\beta^-$ decay?

They are opposites! In $\beta^+$ (positron) decay, a proton turns into a neutron. In $\beta^-$ (electron) decay, a neutron turns into a proton. Always read the question carefully to see which beta decay is occurring.

Q: Why do we multiply the alpha particle changes by 3?

The problem states that the nucleus emits $3\alpha$-particles. Since one alpha particle removes 2 protons and 2 neutrons, three of them will remove $3 \times 2 = 6$ protons and $3 \times 2 = 6$ neutrons in total.

Question

A radioactive nucleus (initial mass number $A$ and atomic number $Z$) emits $3\alpha$-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Options:

(A) $\frac{A - Z - 8}{Z - 4}$

(B) $\frac{A - Z - 4}{Z - 8}$

(C) $\frac{A - Z - 12}{Z - 4}$

(D) $\frac{A - Z - 4}{Z - 2}$

Sigma Percentile · Accepted Answer

### The Alchemy of the Nucleus: Tracking Protons and Neutrons Through Decay

Nuclear decay problems can sometimes feel like a chaotic accounting exercise. Particles are flying out, elements are transmuting, and mass numbers are shifting. But if we break the process down and track the fundamental building blocks—protons and neutrons—separately, the chaos turns into simple arithmetic.

Let's dive into this fascinating journey of a radioactive nucleus undergoing multiple decays.

### The Starting Line: Protons and Neutrons

Imagine we have a radioactive nucleus. We are given its atomic number $Z$ and its mass number $A$.

The atomic number $Z$ is the identity of the element; it tells us exactly how many protons are in the nucleus. The mass number $A$ is the total count of nucleons (protons plus neutrons). Therefore, the initial number of neutrons is simply the total mass minus the protons:

$$N_{	ext{initial}} = A - Z$$

$$P_{	ext{initial}} = Z$$

### The Alpha Emission: Shedding Weight

The first event in our nuclear story is the emission of $3\alpha$-particles.

An alpha particle is essentially a helium nucleus, denoted as $^4_2	ext{He}$. It is a heavy, tightly bound cluster consisting of exactly **2 protons** and **2 neutrons**.

Since our nucleus emits three of these heavy clusters, we must multiply the losses by three. The nucleus loses a total of $3 	imes 2 = 6$ protons and $3 	imes 2 = 6$ neutrons. Let's update our ledger:

$$P_{	ext{intermediate}} = Z - 6$$

$$N_{	ext{intermediate}} = (A - Z) - 6$$

### The Positron Emission: The Proton's Sacrifice

Next, the nucleus undergoes positive beta decay, emitting 2 positrons ($\beta^+$). This is where many students make a critical error.

A positron is the antimatter counterpart of an electron. It has a positive charge but negligible mass. How does a nucleus emit a positron? It happens when a proton decides to sacrifice its positive charge to become a neutral neutron:

$$p^+ \longrightarrow n^0 + e^+ + 
u_e$$

Because 2 positrons are emitted, **2 protons are destroyed**, but they don't just vanish from existence—they are **reborn as 2 neutrons**. This means our proton count decreases by 2, while our neutron count simultaneously increases by 2. Let's calculate the final state:

$$P_{	ext{final}} = (Z - 6) - 2 = Z - 8$$

$$N_{	ext{final}} = (A - Z - 6) + 2 = A - Z - 4$$

### The Final Tally: Calculating the Ratio

We have successfully navigated the decay sequence. The question asks for the final ratio of the number of neutrons to the number of protons.

By simply dividing our final neutron count by our final proton count, we arrive at the elegant solution:

$$	ext{Ratio} = \frac{N_{	ext{final}}}{P_{	ext{final}}} = \frac{A - Z - 4}{Z - 8}$$

**The key takeaway?** Never try to calculate the mass number and atomic number simultaneously in your head. Always split the problem into two independent tracks: one for protons and one for neutrons. This method is foolproof and guarantees you won't fall into the traps set by beta decays!

Animated Solution for Physics - Atoms and Nuclei: A radioactive nucleus (initial mass number $A$ and atomic number $Z$ ) emits $3 α$ -particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Select Answer:

Visualized Solution

Initial State of the Nucleus

Understanding Alpha Decay

Effect of $3 α$ Emissions

Understanding Positron Emission

Effect of $2 β^{+}$ Emissions

Final Nucleus Composition

Calculating the Final Ratio

The Way Forward

The Sigma Insight: Radioactivity

The Alchemy of the Nucleus

The Starting Line

The Alpha Emission

The Positron Emission

The Final Tally

Similar Questions

A nucleus with $Z = 92$ emits the following in a sequence $α, α, β^{-}, β^{-}, α, α, α, α; β^{-}, β^{-}, α, β^{+}, β^{+}, α$ . The $Z$ of the resulting nucleus is

Two radioactive substances $X$ and $Y$ originally have $N_{1}$ and $N_{2}$ nuclei, respectively. Half-life of $X$ is half of the half-life of $Y$ . After three half-lives of $Y$ , number of nuclei of both are equal. The ratio $\frac{N _{1}}{N _{2}}$ will be equal to

A uranium nucleus (atomic number $92$ , mass number $238$ ) emits an alpha particle and the resulting nucleus emits $β$ -particle. What are the atomic number and mass number of the final nucleus?

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

In a radioactive decay chain reaction, $_{90}^{230} Th$ nucleus decays into $_{84}^{214} Po$ nucleus. The ratio of the number of $α$ to number of $β^{-}$ particles emitted in this process is_________.

Half lives of two radioactive nuclei $A$ and $B$ are $10$ minutes and $20$ minutes, respectively. If initially a sample has equal number of nuclei, then after $60$ minutes, the ratio of decayed numbers of nuclei $A$ and $B$ will be

Half-lives of two radioactive elements $A$ and $B$ are $20 min$ and $40 min$ , respectively. Initially, the samples have equal number of nuclei. After $80 min$ , the ratio of decayed numbers of $A$ and $B$ nuclei will be

Two radioactive materials $A$ and $B$ have decay constants $10 λ$ and $λ$ , respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $A$ to that of $B$ will be $1/ e$ after a time

A radioactive nucleus $A$ with a half-life $T$ , decays into a nucleus $B$ . At $t = 0$ , there is no nucleus $B$ . After sometime $t$ , the ratio of the number of $B$ to that of $A$ is $0.3$ . Then, $t$ is given by