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 need to analyze the sequence of events described in the question regarding the stable nuclide, neutron absorption, beta decay, and subsequent alpha decay. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Initial Nucleide
Let’s denote the initial stable nuclide 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 mass number becomes \( A + 1 \) (since the neutron adds 1 to the mass number), and the atomic number remains \( Z \) (neutrons have no charge). Thus, the new nuclide can be represented as:
- Atomic number: \( Z \)
- Mass number: \( A + 1 \)

### Step 3: Emission of a Beta Particle
The nuclide then undergoes beta decay, which involves the emission of a beta particle (an electron). In beta decay, a neutron is converted into a proton, which increases the atomic number by 1 while keeping the mass number the same. Therefore, after beta decay, we have:
- New atomic number: \( Z + 1 \)
- New mass number: \( A + 1 \)

Let’s denote this new nuclide as \( Y \):
- \( Y \) has atomic number \( Z + 1 \) and mass number \( A + 1 \).

### Step 4: Spontaneous Splitting into Alpha Particles
The nuclide \( Y \) then splits spontaneously into two alpha particles. Each alpha particle has:
- Atomic number: 2
- Mass number: 4

Since two alpha particles are produced, the total atomic number and mass number after the decay will be:
- Total atomic number from alpha particles: \( 2 \times 2 = 4 \)
- Total mass number from alpha particles: \( 2 \times 4 = 8 \)

### Step 5: Balancing the Atomic Number and Mass Number
Now we can set up the equations based on the conservation of atomic number and mass number:
1. For atomic number:
\[ Z + 1 = 4 \]
Solving for \( Z \):
\[ Z = 3 \]

2. For mass number:
\[ A + 1 = 8 \]
Solving for \( A \):
\[ A = 7 \]

### Step 6: Identify the Nucleide
Now we have determined that the atomic number \( Z = 3 \) and mass number \( A = 7 \). The nuclide with atomic number 3 is lithium (\( \text{Li} \)), and its mass number is 7. Therefore, the nuclide is:
\[ \text{Li}_7^3 \]

### Conclusion
The stable nuclide that fits the description given in the question is \( \text{Li}_7^3 \).

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