Mindyard | α, β⁻, β⁺ , α,β⁺,β⁻, , β⁺, α, β⁻ , β⁻, α, β⁺

Question Details

Options

α, β⁻, β⁺

α,β⁺,β⁻,

β⁺, α, β⁻

β⁻, α, β⁺

Show Answer

Correct Answer :

Option C

β⁺, α, β⁻

Solution :

The correct answer is β⁺, α, β⁻.

Based on the provided image, we have a radioactive nucleus X undergoing a series of spontaneous decays. We are given the following sequence of transformations for the atomic numbers:

$_{Z}^{A} X \to_{Z - 1} B \to_{Z - 3} C \to_{Z - 2} D$

To determine the decay particles, we must analyze the change in the atomic number (Z) at each step. Let's break down the sequence step-by-step:

Step 1: The nucleus goes from Z to Z - 1. A decrease in the atomic number by 1 indicates the emission of a positron (β⁺ particle). In β⁺ decay, a proton is converted into a neutron, reducing the atomic number by 1 while the mass number A remains unchanged.

$_{Z}^{A} X \to_{Z - 1}^{A} B + β^{+} + ν$

Step 2: The nucleus transitions from Z - 1 to Z - 3. The atomic number decreases by 2. This is the hallmark of alpha (α) decay, in which an alpha particle (a helium nucleus with 2 protons and 2 neutrons) is emitted.

$_{Z - 1}^{A} B \to_{Z - 3}^{A - 4} C +_{2}^{4} α$

Step 3: Finally, the nucleus changes from Z - 3 to Z - 2. The atomic number increases by 1. This occurs during beta-minus (β⁻) decay, where a neutron is converted into a proton, emitting an electron (β⁻ particle).

$_{Z - 3}^{A - 4} C ��_{Z - 2}^{A - 4} D + β^{-} + \bar{ν}$