Chemistry - Radioactivity Question with Solution | TestHub

To find the number of neutrons in element L, we analyze the two nuclear reactions:

1. Alpha decay:

$${}_{92}^{238}\mathrm{M}\to {}_{\mathrm{y}}^{\mathrm{x}}\mathrm{N}+{}_2^4\mathrm{He}$$

Conservation of mass number: $238=x+4\Rightarrow x=234$

Conservation of atomic number: $92=y+2\Rightarrow y=90$

So, element N is ${}_{90}^{234}\mathrm{N}$.

2. Positron emission (beta-plus decay):

$${}_{90}^{234}\mathrm{N}\to {}_{\mathrm{B}}^{\mathrm{A}}\mathrm{L}+2{\beta }^{+}$$

A positron is ${}_{+1}^0\mathrm{e}$ or ${}_{+1}^0\beta$.

Conservation of mass number: $234=A+2(0)\Rightarrow A=234$

Conservation of atomic number: $90=B+2(1)\Rightarrow B=88$

So, element L is ${}_{88}^{234}\mathrm{L}$.

The number of neutrons in L is the mass number (A) minus the atomic number (B):

Number of neutrons = $A-B=234-88=146$.

The final answer is $\boxed{146}$.