Consider the following nuclear reactions:
`""_(92)""^(238)Mto.""_(y)""^(x)N+2.""_(2)""^(4)He, ""_(x)""^(y)Nto ""_(B)""^(A)L+2beta""^(+)`
The number of neutrons in the element L is

Options

A. 146
B. 144
C. 140
D. 142

Explanation

To find the number of neutrons in the element L from the given nuclear reactions, we will follow the steps outlined in the video transcript.

### Step-by-Step Solution:

1. **Identify the first nuclear reaction:**
The first reaction is:
\[ _{92}^{238}M \to _{y}^{x}N + 2 \, _{2}^{4}He \]
Here, \(M\) is a uranium isotope (U-238), which decays into \(N\) and two helium nuclei (alpha particles).

2. **Balance the mass number (A):**
The mass number before the reaction is 238 (from \(M\)), and we have two helium nuclei with a mass number of 4 each, contributing a total of 8. Therefore, we can write:
\[ 238 = x + 8 \]
Solving for \(x\):
\[ x = 238 - 8 = 230 \]

3. **Balance the atomic number (Z):**
The atomic number before the reaction is 92 (from \(M\)), and the two helium nuclei contribute a total of 4. Therefore, we can write:
\[ 92 = y + 4 \]
Solving for \(y\):
\[ y = 92 - 4 = 88 \]

4. **Determine the properties of element N:**
Thus, we find that:
\[ N = _{88}^{230}N \]

5. **Identify the second nuclear reaction:**
The second reaction is:
\[ _{x}^{y}N \to _{B}^{A}L + 2 \beta^+ \]
Here, \(N\) decays into \(L\) and two beta particles.

6. **Balance the mass number (A) for element L:**
The mass number of \(N\) is 230, and since beta decay does not change the mass number, we can write:
\[ 230 = A + 0 \]
Thus, \(A = 230\).

7. **Balance the atomic number (Z) for element L:**
The atomic number of \(N\) is 88, and since each beta particle contributes +1 to the atomic number, we can write:
\[ 88 = B + 2 \]
Solving for \(B\):
\[ B = 88 - 2 = 86 \]

8. **Determine the number of neutrons in element L:**
The number of neutrons \(N_n\) in element \(L\) can be calculated using the formula:
\[ N_n = A - B \]
Substituting the values:
\[ N_n = 230 - 86 = 144 \]

### Final Answer:
The number of neutrons in the element \(L\) is **144**.