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Physics · Nuclear Physics

JEE Advanced 2023 — Paper 1 — Question 30

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-ILIST-II
P) 92238U→91234 Pa{ }_{92}^{238} \mathrm{U} \rightarrow{ }_{91}^{234} \mathrm{~Pa}1) one α\alpha particle and one β+\beta^{+}particle
Q) 82214 Pb→82210 Pb{ }_{82}^{214} \mathrm{~Pb} \rightarrow{ }_{82}^{210} \mathrm{~Pb}2) three β−\beta^{-}particles and one α\alpha particle
R) 81210 Tℓ→82206 Pb{ }_{81}^{210} \mathrm{~T} \ell \rightarrow{ }_{82}^{206} \mathrm{~Pb} & (3)(3)3) two β−\beta^{-}particles and one α\alpha particle
S) 91228 Pa→88224Ra{ }_{91}^{228} \mathrm{~Pa} \rightarrow{ }_{88}^{224} \mathrm{Ra}4) one α\alpha particle and one β−\beta^{-}particle
5) one α\alpha particle and two β+\beta^{+}particles
  1. Option A:

    P→4,Q→3,R→2, S→1\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1

    Correct
  2. Option B:

    P→4,Q→1,R→2, S→5\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5

  3. Option C:

    P→5,Q→3,R→1, S→4\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 1, \mathrm{~S} \rightarrow 4

  4. Option D:

    P→5,Q→1,R→3, S→2\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 2

Answer: A

Step-by-step solution

Let x=x= No of α\alpha particles &y=\& y= No of β−\beta^{-}particles (if y=+vey=+v e ) == No of β+\beta^{+}particles (if y=−vey=-\mathrm{ve} ) (P) 238−4x=234⇒x=1238-4 x=234 \Rightarrow x=1 (one α\alpha particle) and, 92−2x+y=9192-2 x+y=91 ⇒y=1\Rightarrow y=1 (one β−\beta^{-}particle) (Q) 214−4x=210⇒x=1214-4 x=210 \Rightarrow x=1 (one α\alpha particle) and, 82−2x+y=8282-2 \mathrm{x}+\mathrm{y}=82 ⇒y=2\Rightarrow \mathrm{y}=2 (two β−\beta^{-}particle) (R) 210−4x=206⇒x=1210-4 x=206 \Rightarrow x=1 (one α\alpha particle) and, 81−2x+y=8281-2 \mathrm{x}+\mathrm{y}=82 ⇒y=3\Rightarrow \mathrm{y}=3 (three β−\beta^{-}particle) (S) 228−4x=224⇒x=1228-4 x=224 \Rightarrow x=1 (one α\alpha particle) and, 91−2x+y=8891-2 x+y=88 ⇒y=−1\Rightarrow y=-1 (one β+\beta^{+}particle)

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Physics
Chapter
Nuclear Physics
Topic
Laws of Radioactive Decay
List-I shows different radioactive decay processes and List-II… | JEE Advanced 2023 PYQ with Solution · DhiX AI