Unstable nuclei attain stability through disintegration. The nuclear stability is related to neutron proton ratio `(n//p)`. For stable nuclei `n//p` ratio lies close to unity for elements with low atmoic numbers (20 or less) but it is more than 1 for nuclei having higher atomic numbers. Nuclei having `n//p` ratio either very high or low undergo nuclear transformation. When `n//p` ratio is higher than required for stability, the nuclei have the tendency to emit `beta`-rays. while when `n//p` ratio is lower than required for stability, the nuclei either emits `alpha`-particles or a positron or capture `K`-electron.
For reaction `._(92)M^(238) rarr ._(y)N^(x) + 2 ._(2)He^(4), ._(y)N^(x) rarr ._(B)L^(A) + 2 ._(-1)e^(0)`
The number of neutrons in the element `L` is

Options

A. 140
B. 145
C. 138
D. 160

Explanation

To solve the problem, we need to analyze the given nuclear reaction step by step and determine the number of neutrons in the element L.

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction given is:
\[ _{92}M^{238} \rightarrow _{y}N^{x} + 2 \, _{2}He^{4} \]
This indicates that nucleus M (with atomic number 92 and mass number 238) undergoes alpha decay, releasing two helium nuclei (alpha particles).

2. **Calculate the Mass Number of N**:
- Each alpha particle has a mass number of 4. Since two alpha particles are emitted, the total mass number lost is:
\[ 2 \times 4 = 8 \]
- Therefore, the mass number \( x \) of nucleus N can be calculated as:
\[ x = 238 - 8 = 230 \]

3. **Calculate the Atomic Number of N**:
- Each alpha particle has an atomic number of 2. Thus, the total atomic number lost is:
\[ 2 \times 2 = 4 \]
- Therefore, the atomic number \( y \) of nucleus N can be calculated as:
\[ y = 92 - 4 = 88 \]

4. **Determine the Decay of N**:
- The next part of the reaction is:
\[ _{y}N^{x} \rightarrow _{B}L^{A} + 2 \, _{-1}e^{0} \]
Here, N undergoes beta decay, emitting two beta particles (electrons).

5. **Calculate the Atomic Number of L**:
- In beta decay, the atomic number increases by 1 for each emitted beta particle. Since there are two beta particles emitted, the atomic number \( B \) of nucleus L can be calculated as:
\[ B = 88 + 2 = 90 \]

6. **Determine the Mass Number of L**:
- The mass number does not change during beta decay. Therefore, the mass number \( A \) of nucleus L remains the same as that of N:
\[ A = 230 \]

7. **Calculate the Number of Neutrons in L**:
- The number of neutrons can be calculated using the formula:
\[ \text{Number of Neutrons} = \text{Mass Number} - \text{Atomic Number} \]
- Substituting the values we found:
\[ \text{Number of Neutrons} = 230 - 90 = 140 \]

### Final Answer:
The number of neutrons in the element L is **140**.

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