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 step-by-step, we need to analyze the nuclear reactions described in the question.

### Step 1: Define the initial nuclide
Let the initial stable nuclide be represented as \( X \) with atomic number \( Z \) and mass number \( A \).

### Step 2: Absorption of a neutron
When the nuclide \( X \) absorbs a neutron, the new nuclide can be represented as:
\[ X + n \rightarrow Y \]
Here, \( n \) is the neutron with atomic number \( 0 \) and mass number \( 1 \). The mass number of \( Y \) becomes \( A + 1 \) (since we add one neutron), and the atomic number remains \( Z \).

### Step 3: Emission of a beta particle
Next, the nuclide \( Y \) emits a beta particle (\( \beta \)). The emission of a beta particle increases the atomic number by 1 (due to the conversion of a neutron into a proton) while the mass number remains the same. Thus, we have:
\[ Y \rightarrow Y' + \beta \]
Here, the new nuclide \( Y' \) has:
- Atomic number: \( Z + 1 \)
- Mass number: \( A + 1 \)

### Step 4: Spontaneous splitting into two alpha particles
The nuclide \( Y' \) then splits spontaneously into two alpha particles. Each alpha particle (\( \alpha \)) has an atomic number of \( 2 \) and a mass number of \( 4 \). Therefore, we can represent this reaction as:
\[ Y' \rightarrow 2\alpha \]
The total mass number and atomic number on the right side will be:
- Total mass number: \( 2 \times 4 = 8 \)
- Total atomic number: \( 2 \times 2 = 4 \)

### Step 5: Setting up equations
From the above reactions, we can set up equations based on the conservation of mass number and atomic number:
1. Mass number:
\[ A + 1 = 8 \]
Solving this gives:
\[ A = 7 \]

2. Atomic number:
\[ Z + 1 = 4 \]
Solving this gives:
\[ Z = 3 \]

### Step 6: Identifying the nuclide
Now we have determined that the initial nuclide \( X \) has:
- Mass number \( A = 7 \)
- Atomic number \( Z = 3 \)

The nuclide with atomic number 3 is Lithium, and its mass number is 7. Therefore, the nuclide is:
\[ \text{Lithium-7} \quad (^{7}_{3}\text{Li}) \]

### Conclusion
The final answer is that the nucleide is Lithium-7 (Li-7).