QJEEM04Ch0f9e: JEE Main 2004 Chemistry - Class 12 JEEM04Ch0f9e Solution: Consider the following nuclear reactions 92 238 M → Y X N + 2 2 …

We know that, $\alpha$ - decay formula is -

$${}_Z^{A}X\stackrel{-\alpha }{\to }{}_{Z-2}^{A-4}Y$$

And the formula for $\beta$ is -

$${}_Z^{A}X\stackrel{-\beta }{\to }{}_{Z-1}^{A}Y$$

Now for the given above question we see that, -

$$\begin{array}{rl}& {}_{92}^{238}M\stackrel{-2\alpha }{\to }{}_{92-4}^{238-8}N\\ \\ & {}_Y^{X}X\stackrel{-2{\beta }^{+}}{\to }{}_{86}^{230}L\end{array}$$

Therefore we can now calculate the number of neutrons in $L$ , -

$$L=230-86$$

$$L=144$$

So, as we can see the no. of neutrons present in $\mathrm{L}$ is 144 .