In the nuclear process, 6^11C - 5^11B + beta^+ + X, X stands for ........ .

Q: Why can't particle X be a photon ($\gamma$)?

While a photon also has zero mass and zero charge, $\beta^+$ decay specifically requires the emission of a neutrino to conserve lepton number. A proton converting into a neutron and a positron leaves the lepton number unbalanced without a neutrino.

Q: What exactly is a positron ($\beta^+$)?

A positron is the antimatter counterpart of an electron. It has the exact same mass as an electron but carries a positive charge ($+1e$).

Q: What is the difference between a neutrino and an antineutrino?

A neutrino ($\nu$) is emitted during $\beta^+$ decay (positron emission) to balance the anti-lepton, while an antineutrino ($\bar{\nu}$) is emitted during $\beta^-$ decay (electron emission) to balance the standard lepton. They are antimatter counterparts of each other.

Q: Does the neutrino have mass?

For a long time, neutrinos were thought to be completely massless. However, modern experiments have shown they have a very tiny, non-zero mass, though it is often approximated as zero in basic nuclear equations.

Q: Why does Carbon-11 undergo $\beta^+$ decay?

Carbon-11 has 6 protons and 5 neutrons. It is proton-rich compared to stable Carbon-12 (6 protons, 6 neutrons). To achieve stability, it converts a proton into a neutron via $\beta^+$ decay, becoming Boron-11.

Question

In the nuclear process, $_6^{11}	ext{C} ightarrow _5^{11}	ext{B} + \beta^+ + X$, $X$ stands for ........ .

Sigma Percentile · Accepted Answer

### Unmasking the Ghost Particle: Beta Plus Decay Explained

Imagine you are a nuclear detective, and you've just been handed a crime scene report. A Carbon-11 nucleus has spontaneously decayed into a Boron-11 nucleus, emitting a positron ($\beta^+$) and a mysterious, unidentified particle $X$. Our mission is to use the fundamental laws of physics to unmask particle $X$.

### The Law of Conservation of Mass Number

In any nuclear reaction, the total number of nucleons (protons plus neutrons) must remain constant before and after the decay. Let's look at our reactants and products:

- **Reactants:** Carbon-11 has a mass number of $11$.
- **Products:** Boron-11 also has a mass number of $11$. The emitted positron ($\beta^+$) is an antimatter electron, which has a mass number of $0$.

Since $11 = 11 + 0 + A_X$, the mass number of our mystery particle $X$ must be exactly $0$.

### The Law of Conservation of Charge

Next, we must ensure that the total electric charge is conserved. This is represented by the atomic number ($Z$) in our nuclear equation:

- **Reactants:** Carbon has an atomic number of $6$ (meaning it has $6$ protons).
- **Products:** Boron has an atomic number of $5$. The positron carries a positive charge of $+1$.

Let's do the math: $5 + 1 = 6$. The charge on the right side perfectly matches the charge on the left side! This implies that our mystery particle $X$ must be electrically neutral, meaning its charge is $0$.

### The Missing Piece: Lepton Number Conservation

So far, we know that particle $X$ has zero mass number and zero charge. You might be tempted to guess that it's a photon ($\gamma$-ray). However, there is a catch! We must also conserve the **lepton number**.

In $\beta^+$ decay, a proton inside the nucleus transforms into a neutron and a positron:

$$p ightarrow n + e^+ + 
u_e$$

The positron ($e^+$) is an anti-lepton, which has a lepton number of $-1$. To keep the total lepton number balanced at $0$ (since a proton is a baryon, not a lepton), the reaction must emit a standard lepton with a lepton number of $+1$.

This elusive, nearly massless, and electrically neutral particle is the **electron neutrino** ($
u_e$). It acts as the perfect counterweight to the positron, ensuring all quantum accounting books are perfectly balanced.

### Final Conclusion

By applying the conservation of mass, charge, and lepton number, we have successfully identified the mystery particle. The complete nuclear reaction is:

$$_6^{11}	ext{C} ightarrow _5^{11}	ext{B} + \beta^+ + 
u$$

Therefore, the unknown particle $X$ is a **Neutrino**.

Animated Solution for Physics - Atoms and Nuclei: In the nuclear process, $_{6}^{11} C \to_{5}^{11} B + β^{+} + X$ , $X$ stands for ........ .

Visualized Solution

$_{6}^{11} C \to_{5}^{11} B + β^{+} + X$

$Conservation of Mass Number$

$Conservation of Charge$

$Identifying Particle X$

$Final Conclusion$

The Sigma Insight: Radioactivity

Unmasking the Ghost Particle

The Law of Conservation of Mass Number

The Law of Conservation of Charge

The Missing Piece

Final Conclusion

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Animated Solution for Physics - Atoms and Nuclei: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the nuclear process, 611​C→511​B+β++X, X stands for ........ .