Question

A radioactive nucleus \(_{Z}^{A}\textrm{X}\) undergoes spontaneous decay in the sequence \(_{Z}^{A}\textrm{X}\)\(\rightarrow\)_Z-1B\(\rightarrow\)_Z-3C\(\rightarrow\)_Z-2D, where Z is the atomic number of element X. The possible decay particles in the sequence are

• A. β⁻, α, β⁺
• B. α, β⁻, β⁺
• C. α, β⁺, β⁻
• D. β⁺, α, β⁻

Answer 1

The Correct Option is D

To determine the possible decay particles in the sequence of spontaneous decay of the radioactive nucleus \(_{Z}^{A}\textrm{X}\) as it transitions to \(_{Z-1}^{A}\textrm{B}\), then \(_{Z-3}^{A}\textrm{C}\), and finally \(_{Z-2}^{A}\textrm{D}\), we need to analyze each step of the decay process and the types of particles typically emitted.

1. First Decay: \(_{Z}^{A}\textrm{X}\rightarrow \) \(_{Z-1}\textrm{B}\)

   This transition represents a decrease in the atomic number by 1, which is characteristic of a beta-plus decay (\(\beta^+\)). In beta-plus decay, a proton is converted into a neutron with the emission of a positron (\(\beta^+\)) and a neutrino.

2. Second Decay: \(_{Z-1}^{A}\textrm{B}\rightarrow \) \(_{Z-3}^{A}\textrm{C}\)

   This step involves a decrease in the atomic number by 2 and is generally due to an alpha (\(\alpha\)) decay. In alpha decay, the nucleus emits an alpha particle (composed of 2 protons and 2 neutrons), reducing the atomic number by 2.

3. Third Decay: \(_{Z-3}^{A}\textrm{C}\rightarrow \) \(_{Z-2}^{A}\textrm{D}\)

   The final step indicates an increase in the atomic number by 1, consistent with beta-minus decay (\(\beta^-\)). In beta-minus decay, a neutron is converted into a proton with the emission of an electron (\(\beta^-\)) and an antineutrino.

Combining these observations, the sequence of the decays involves the emission of a positron (\(\beta^+\)) in the first step, an alpha (\(\alpha\)) particle in the second step, and an electron (\(\beta^-\)) in the third step.

Therefore, the correct option is the sequence: \(\beta^+\), \(\alpha\), \(\beta^-\).