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(B) The given nuclear reaction sequence is: $_{92}X^{238}$ $\xrightarrow{-\alpha} Y$ $\xrightarrow{-\beta} Z$ $\xrightarrow{-\beta} L$ $\xrightarrow{-

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$4$

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(B) The given nuclear reaction sequence is: $_{92}X^{238}$ $\xrightarrow{-\alpha} Y$ $\xrightarrow{-\beta} Z$ $\xrightarrow{-\beta} L$ $\xrightarrow{-n\alpha} _{84}M^{218}$.$1$. An $\alpha$-decay decreases the mass number by $4$ and atomic number by $2$.$2$. $A$ $\beta$-decay does not change the mass number but increases the atomic number by $1$.Let the initial nucleus be $_{92}X^{238}$ and the final nucleus be $_{84}M^{218}$.The total change in mass number is $\Delta A = 238 - 218 = 20$.Since $\beta$-decay does not change the mass number,the entire change in mass number is due to $\alpha$-decays.Each $\alpha$-particle carries a mass number of $4$.Total $\alpha$-particles emitted = $1$ (first step) + $n$ (last step) = $1 + n$.Therefore,$(1 + n) 	imes 4 = 20$.$1 + n = 5$.$n = 4$.

In the sequence of the following nuclear reactions $_{92}X^{238}$ $\xrightarrow{-\alpha} Y$ $\xrightarrow{-\beta} Z$ $\xrightarrow{-\beta} L$ $\xrightarrow{-n\alpha} _{84}M^{218}$. The value of $n$ will be:

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