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 solve the problem of identifying the possible decay particles in the sequence, we need to analyze the changes in the atomic number (Z) at each step of decay:

1. Step 1: \(_{Z}^{A}\textrm{X}\rightarrow_{Z-1}\textrm{B}\) 
   The atomic number decreases by 1, while the mass number remains the same. This suggests the emission of a positron, which corresponds to \(\beta^+\) decay. In beta-plus decay, a proton is converted into a neutron, thus reducing the atomic number by one.
2. Step 2: \(_{Z-1}^{A}\textrm{B}\rightarrow_{Z-3}\textrm{C}\)
   The atomic number decreases by 2, and the mass number decreases by 4. This indicates an alpha particle (\(\alpha\)) emission. An alpha particle consists of 2 protons and 2 neutrons, so its emission decreases the atomic number by 2 and the mass number by 4.
3. Step 3: \(_{Z-3}^{A-4}\textrm{C}\rightarrow_{Z-2}\textrm{D}\)
   The atomic number increases by 1, while the mass number remains unchanged. This corresponds to beta-minus decay (\(\beta^-\)), where a neutron is converted into a proton, increasing the atomic number by one.

Thus, following this step-by-step reasoning of decay processes, the sequence of decay particles is: \(\beta^+\), \(\alpha\), \(\beta^-\).

Therefore, the correct answer is: β⁺, α, β⁻.