Consider the following nuclear reactions $$ \begin{aligned} & { }_{92}^{238} \mathrm{M} \rightarrow{ }_{\mathrm{y}}^{\mathrm{x}} \mathrm{N}+{ }_2^4 \mathrm{He} \\ & { }_{\mathrm{y}}^{\mathrm{N}} \mathrm{N} \rightarrow{ }_{\mathrm{B}}^{\mathrm{A}} \mathrm{L}+2 \beta^{+} \end{aligned} $$ The number of neutrons in the element $L$ is

• A

  142
• B

  146
• C

  140
• Correct answer
  D

  144

Hence, the correct answer is option: 'D'.