Question

For the given reaction, the particle \(\text {X}\) is:
\(_6C^{11}→\ _5B^{11}+β^++X\)

• A. Neutron
• B. Anti neutrino
• C. Neutrino
• D. Proton

Answer 1

The Correct Option is C

The given nuclear reaction is: 

\(_6C^{11} → _5B^{11} + β^+ + X\)

Let us analyze this reaction to determine the identity of the particle \(X\).

1. The reaction begins with carbon-11 (\(_6C^{11}\)), which is an isotope of carbon with 6 protons and 5 neutrons.
2. Boron-11 (\(_5B^{11}\)) is the product observed post-reaction, containing 5 protons and 6 neutrons.
3. The positron (\(β^+\)) is emitted, indicating a type of beta decay in which a proton in the nucleus is converted into a neutron, emitting a positron and a neutrino.

Analyzing the conservation laws:

• Conservation of atomic number (protons): Initial carbon has 6 protons, and after losing a positron, it has 5 protons, matching boron's 5 protons. Thus, proton count is conserved.
• Conservation of nucleon number (total protons and neutrons): Carbon and boron both have 11 nucleons, hence nucleon count is conserved.

In this reaction:

• A positron is emitted, usually accompanied by the emission of a neutrino, specifically an electron neutrino, to conserve lepton number.

 

Based on standard particle physics, during beta-plus decay:

• Neutrinos (option: Neutrino) are emitted to conserve lepton number and other quantum properties consistent with observed reactions of this form.

Hence, the correct identity of the particle \(X\) is Neutrino.