JEE Advanced 2023 — Paper 1 — PHY-IV solution

JEE Advanced2023Paper 1PHY-IV

List-I shows different radioactive decay processes and List-II provides possible emitted particles. Match each entry in List-I with an appropriate entry from List-II, and choose the correct option.

List-I	List-II
(P) $_{92}^{238} U \to_{91}^{234} Pa$	(1) one $α$ particle and one $β^{+}$ particle
(Q) $_{82}^{214} Pb \to_{82}^{210} Pb$	(2) three $β^{-}$ particles and one $α$ particle
(R) $_{81}^{210} Tl \to_{82}^{206} Pb$	(3) two $β^{-}$ particles and one $α$ particle
(S) $_{91}^{228} Pa \to_{88}^{224} Ra$	(4) one $α$ particle and one $β^{-}$ particle
	(5) one $α$ particle and two $β^{+}$ particles

A
P $\to$ 4, Q $\to$ 3, R $\to$ 2, S $\to$ 1
B
P $\to$ 4, Q $\to$ 1, R $\to$ 2, S $\to$ 5
C
P $\to$ 5, Q $\to$ 3, R $\to$ 1, S $\to$ 4
D
P $\to$ 5, Q $\to$ 1, R $\to$ 3, S $\to$ 2

Solution

$_{Z_{1}}^{A_{1}} Z \to_{Z_{2}}^{A_{2}} Y + N_{1}_{2}^{4} He + N_{2}_{1}^{0} e + N_{3}_{- 1}^{0} e$

Conservation of charge

$Z_{1} = Z_{2} + 2 N_{1} + N_{2} - N_{3} \dots (i)$

Conservation of nucleons

$A_{1} = A_{2} + 4 N_{1}$

$N_{1} = \frac{A _{1} - A _{2}}{4} \dots (ii)$

From (i) and (ii),

$N_{2} - N_{3} = Z_{1} - Z_{2} - (\frac{A _{1} - A _{2}}{2})$

(P) $_{92}^{238} U \to_{91}^{234} Pa$

$N_{1} = \frac{238 - 234}{4} = 1 \to 1 α$

$N_{2} - N_{3} = (92 - 91) - (\frac{4}{2}) = - 1 \to 1 β^{-}$

(Q) $_{82}^{214} Pb \to_{82}^{210} Pb$

$N_{1} = \frac{214 - 210}{4} = 1 \to 1 α$

$N_{2} - N_{3} = (82 - 82) - (\frac{4}{2}) = - 2 \to 2 β^{-}$

(R) $_{81}^{210} Tl \to_{82}^{206} Pb$

$N_{1} = \frac{210 - 206}{4} = 1 \to 1 α$

$N_{2} - N_{3} = (81 - 82) - \frac{4}{2} = - 3 \to 3 β^{-}$

(S) $_{91}^{228} Pa \to_{88}^{224} Ra$

$N_{1} = \frac{228 - 224}{4} = 1 \to 1 α$

$N_{2} - N_{3} = (91 - 88) - \frac{4}{2} = 1 \to 1 β^{+}$

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