Question

Consider the following nuclear reaction:

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

${}_y^{y}N\to {}_B^{A}L+2{\beta }^{+}$

The number of neutrons in the element $L$ is:

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

Answer 1

The correct option is D $146$

We can write the following equations from the reaction:

$x+4=238\Rightarrow x=234$

$y+2=92\Rightarrow y=90$

$A+2X_0=x\Rightarrow A=x=234$

$B+2X_1=y\Rightarrow B=90-2=88$

Number of neutrons in $L=A-B=146$