Question

In the nuclear process, ₁₁²²Na → ₁₀²²Ne + e⁺ + ν_e, then X is

• A. neutrino
• B. anti-neutrino
• C. electron
• D. positron
• E. neutron

Hint:

In beta-plus decay, a proton decays into a neutron, emitting a positron and a neutrino. The neutrino ensures conservation of energy and momentum.

Answer 1

The Correct Option is A

The given nuclear process is:

Step 1: Conservation of Mass Number
The mass number must be conserved in the reaction. The mass number of the reactant ( ₁₁²²Na ) is 22. The mass number of the products must also sum to 22:

Mass number of ₁₀²²Ne = 22, Mass number of e⁺ = 0, Mass number of X = 0

Thus, the mass number is conserved:
22 = 22 + 0 + 0

Step 2: Conservation of Atomic Number
The atomic number (or charge) must also be conserved. The atomic number of the reactant ( ₁₁²²Na ) is 11. The atomic numbers of the products must sum to 11:

Atomic number of ₁₀²²Ne = 10, Atomic number of e⁺ = +1, Atomic number of X = 0

Thus, the atomic number is conserved:
11 = 10 + 1 + 0

Step 3: Identification of X
In beta-plus decay, a proton in the nucleus is converted into a neutron, a positron (e⁺), and a neutrino (ν). The neutrino is emitted to conserve energy, momentum, and other quantum numbers. Therefore, X must be a neutrino.

Final Answer:
The particle X is a neutrino.

Hence, the correct answer is (A) neutrino.