1. First nuclear reaction

Given: ${}_{92}^{238}M \to {}_{Y}^{X}N + 2\,{}_{2}^{4}He$

Each $\alpha$-particle is ${}_{2}^{4}He$.

Using conservation of mass number: $238 = X + 2\times 4$ $238 = X + 8$ $X = 230$

Using conservation of atomic number: $92 = Y + 2\times 2$ $92 = Y + 4$ $Y = 88$

So, ${}_{88}^{230}N$

2. Second nuclear reaction

Given: ${}_{Y}^{X}N \to {}_{B}^{A}L + 2\beta^+$

We found $X=230$ and $Y=88$, so: ${}_{88}^{230}N \to {}_{B}^{A}L + 2\beta^+$

A positron $\beta^+$ has:

• mass number $0$
• atomic number $+1$

Using conservation of mass number: $230 = A + 2\times 0$ $A = 230$

Using conservation of atomic number: $88 = B + 2\times (+1)$ $88 = B + 2$ $B = 86$

So, ${}_{86}^{230}L$

3. Number of neutrons in $L$

Number of neutrons $= A - B$

$= 230 - 86 = 144$

4. Option check

The correct option is: $\boxed{144}$ which is Option B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, they agree.