A certain stable nucleide, after absorbing a neutron, emits `beta`-particle and the new nucleide splits spontaneously into two `alpha`- particles. The nucleide is

Options

A. `""_2^4He`
B. `""_3^7Li`
C. `""_4^6Be`
D. `""_3^6Li`

Answer

To solve the problem, we will analyze the nuclear reactions step by step.

### Step-by-Step Solution:

1. **Understanding the Initial Reaction**:
We have a stable nuclide that absorbs a neutron. Let's denote the initial nuclide as \( X \) with atomic number \( Z \) and mass number \( A \). When it absorbs a neutron, the new nuclide becomes \( X + n \) (where \( n \) is the neutron).

2. **Writing the Reaction**:
The reaction can be written as:
\[ X + n \rightarrow Y \]
where \( Y \) is the new nuclide formed after neutron absorption.

3. **Beta Decay**:
The nuclide \( Y \) emits a beta particle. In beta decay, a neutron in the nucleus is converted into a proton, resulting in an increase in the atomic number by 1 while the mass number remains unchanged. Therefore, the reaction becomes:
\[ Y \rightarrow Z + \beta^- \]
Here, \( Z \) is the new nuclide formed after beta decay.

4. **Splitting into Alpha Particles**:
The nuclide \( Z \) then splits spontaneously into two alpha particles. The reaction can be written as:
\[ Z \rightarrow 2 \alpha \]
Each alpha particle has an atomic number of 2 and a mass number of 4.

5. **Balancing the Mass and Atomic Numbers**:
Let's denote the atomic number and mass number of the original nuclide \( X \) as \( Z_X \) and \( A_X \) respectively. After absorbing a neutron, the new atomic number and mass number will be:
\[ Z_Y = Z_X \quad \text{and} \quad A_Y = A_X + 1 \]
After beta decay:
\[ Z_Z = Z_Y + 1 = Z_X + 1 \quad \text{and} \quad A_Z = A_Y = A_X + 1 \]
After splitting into two alpha particles:
\[ Z_Z = 2 \times 2 = 4 \quad \text{and} \quad A_Z = 2 \times 4 = 8 \]

6. **Setting Up Equations**:
From the above, we have:
\[ Z_X + 1 = 4 \quad \Rightarrow \quad Z_X = 3 \]
\[ A_X + 1 = 8 \quad \Rightarrow \quad A_X = 7 \]

7. **Identifying the Nucleide**:
The nuclide with atomic number 3 and mass number 7 is Lithium, represented as:
\[ ^7_3Li \]

### Final Answer:
The nucleide is Lithium-7, denoted as \( ^7_3Li \).

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