List-I shows different radioactive decay proc | vedclass

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(A) For a decay process ${ }_{Z_1}^{A_1} X \rightarrow { }_{Z_2}^{A_2} Y + N_{\alpha} { }_{2}^{4} He + N_{\beta^-} { }_{-1}^{0} e + N_{\beta^+} { }_{1

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$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 1$

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(A) For a decay process ${ }_{Z_1}^{A_1} X \rightarrow { }_{Z_2}^{A_2} Y + N_{\alpha} { }_{2}^{4} He + N_{\beta^-} { }_{-1}^{0} e + N_{\beta^+} { }_{1}^{0} e$:$1$. Conservation of mass number: $A_1 = A_2 + 4 N_{\alpha} \implies N_{\alpha} = \frac{A_1 - A_2}{4}$.$2$. Conservation of atomic number: $Z_1 = Z_2 + 2 N_{\alpha} - N_{\beta^-} + N_{\beta^+}$.$(P)$ ${ }_{92}^{238} U \rightarrow { }_{91}^{234} Pa$: $N_{\alpha} = \frac{238-234}{4} = 1$. $92 = 91 + 2(1) - N_{\beta^-} + N_{\beta^+} \implies N_{\beta^-} - N_{\beta^+} = 1$. Matches $(4)$: $1 \alpha, 1 \beta^-$.$(Q)$ ${ }_{82}^{214} Pb \rightarrow { }_{82}^{210} Pb$: $N_{\alpha} = \frac{214-210}{4} = 1$. $82 = 82 + 2(1) - N_{\beta^-} + N_{\beta^+} \implies N_{\beta^-} - N_{\beta^+} = 2$. Matches $(3)$: $1 \alpha, 2 \beta^-$.$(R)$ ${ }_{81}^{210} Tl \rightarrow { }_{82}^{206} Pb$: $N_{\alpha} = \frac{210-206}{4} = 1$. $81 = 82 + 2(1) - N_{\beta^-} + N_{\beta^+} \implies N_{\beta^-} - N_{\beta^+} = 3$. Matches $(2)$: $1 \alpha, 3 \beta^-$.$(S)$ ${ }_{91}^{228} Pa \rightarrow { }_{88}^{224} Ra$: $N_{\alpha} = \frac{228-224}{4} = 1$. $91 = 88 + 2(1) - N_{\beta^-} + N_{\beta^+} \implies N_{\beta^-} - N_{\beta^+} = -1$. Matches $(1)$: $1 \alpha, 1 \beta^+$.Correct mapping: $P \rightarrow 4, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 1$.

List-$I$	List-$II$
$(P)$ ${ }_{92}^{238} U \rightarrow{ }_{91}^{234} Pa$	$(1)$ $1 \alpha$ and $1 \beta^{+}$
$(Q)$ ${ }_{82}^{214} Pb \rightarrow{ }_{82}^{210} Pb$	$(2)$ $3 \beta^{-}$ and $1 \alpha$
$(R)$ ${ }_{81}^{210} Tl \rightarrow{ }_{82}^{206} Pb$	$(3)$ $2 \beta^{-}$ and $1 \alpha$
$(S)$ ${ }_{91}^{228} Pa \rightarrow{ }_{88}^{224} Ra$	$(4)$ $1 \alpha$ and $1 \beta^{-}$
	$(5)$ $1 \alpha$ and $2 \beta^{+}$

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