The number of beta particles emitter by radioactive sustance is twice the number of alpha particles emitter by it. The resulting daughter is an

Options

A. Isobar of parent
B. isomer of parent
C. isoton of parent
D. isotope of parent

Explanation

To solve the problem, we need to analyze the decay process of the radioactive substance that emits alpha and beta particles. Let's break down the steps:

### Step 1: Understand the Decay Process
1. **Identify the initial nucleus**: Let’s denote the initial nucleus as \( A \) (mass number) and \( Z \) (atomic number), represented as \( AXZ \).
2. **Alpha decay**: An alpha particle consists of 2 protons and 2 neutrons. When an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2. Thus, after emitting one alpha particle, the new nucleus will be \( A-4 \) and \( Z-2 \), represented as \( A-4, Z-2 \).

### Step 2: Beta Decay
1. **Beta decay**: A beta particle is an electron emitted from the nucleus when a neutron converts into a proton. This process increases the atomic number by 1 while keeping the mass number the same.
2. **Number of beta particles emitted**: According to the problem, the number of beta particles emitted is twice the number of alpha particles emitted. Since we emitted 1 alpha particle, we will emit 2 beta particles.
3. **After emitting 2 beta particles**: Each beta decay increases the atomic number by 1. Therefore, after emitting 2 beta particles, the atomic number will increase by 2. The new nucleus will be \( A-4 \) (mass number remains the same) and \( Z-2+2 \) (atomic number increases by 2), which simplifies to \( A-4, Z \).

### Step 3: Resulting Daughter Nucleus
1. **Final representation**: The resulting daughter nucleus after the decay process will be represented as \( A-4, Z \).
2. **Conclusion**: The daughter nucleus is a new element with the same atomic number \( Z \) but a mass number \( A-4 \).

### Final Answer
The resulting daughter nucleus is \( A-4, Z \).

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