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Atoms and Nuclei question

2023 · Shift 1 · Q48
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Atoms and Nuclei question

2023 · Shift 1 · Q48

JEE AdvancedPhysicsAtoms and NucleiMCQ+3 / −1
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) 92238U→91234 Pa{ }_{92}^{238} U \rightarrow{ }_{91}^{234} \mathrm{~Pa}92238​U→91234​ Pa (1) one α\alphaα particle and one β+\beta^{+}β+ particle
(Q) 82214 Pb→82210 Pb{ }_{82}^{214} \mathrm{~Pb} \rightarrow{ }_{82}^{210} \mathrm{~Pb}82214​ Pb→82210​ Pb (2) three β−\beta^{-}β− particles and one α\alphaα particle
(R) 81210Tl→82206 Pb{ }_{81}^{210} \mathrm{Tl} \rightarrow{ }_{82}^{206} \mathrm{~Pb}81210​Tl→82206​ Pb (3) two β−\beta^{-}β− particles and one α\alphaα particle
(S) 91228 Pa→88224Ra{ }_{91}^{228} \mathrm{~Pa} \rightarrow{ }_{88}^{224} \mathrm{Ra}91228​ Pa→88224​Ra (4) one α\alphaα particle and one β−\beta^{-}β− particle
(5) one α\alphaα particle and two β+\beta^{+}β+ particles
  1. A
    P→4,Q→3,R→2,S→1P \rightarrow 4, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 1P→4,Q→3,R→2,S→1
  2. B
    P→4,Q→1,R→2,S→5P \rightarrow 4, Q \rightarrow 1, R \rightarrow 2, S \rightarrow 5P→4,Q→1,R→2,S→5
  3. C
    P→5,Q→3,R→1,S→4P \rightarrow 5, Q \rightarrow 3, R \rightarrow 1, S \rightarrow 4P→5,Q→3,R→1,S→4
  4. D
    P→5,Q→1,R→3,S→2P \rightarrow 5, Q \rightarrow 1, R \rightarrow 3, S \rightarrow 2P→5,Q→1,R→3,S→2
View written solutionFree

Correct answer: A

  1. Use decay rules

For radioactive decay:

  • An α\alphaα particle emission changes: A→A−4,Z→Z−2A \to A-4, \quad Z \to Z-2A→A−4,Z→Z−2
  • A β−\beta^-β− particle emission changes: A→A,Z→Z+1A \to A, \quad Z \to Z+1A→A,Z→Z+1
  • A β+\beta^+β+ particle emission changes: A→A,Z→Z−1A \to A, \quad Z \to Z-1A→A,Z→Z−1

We compare the initial and final nuclei in each case.

  1. Process P

92238U→91234Pa^{238}_{92}U \to {}^{234}_{91}Pa92238​U→91234​Pa

Change in mass number: 238→234⇒ΔA=−4238 \to 234 \Rightarrow \Delta A=-4238→234⇒ΔA=−4 So, one α\alphaα particle must be emitted.

After one α\alphaα emission: Z:92→90Z: 92 \to 90Z:92→90 But final atomic number is 919191, so we need one increase in ZZZ. That is achieved by one β−\beta^-β− emission.

Hence, P→one α and one β−P \to \text{one }\alpha\text{ and one }\beta^-P→one α and one β− So, P→4P \to 4P→4

  1. Process Q

82214Pb→82210Pb^{214}_{82}Pb \to {}^{210}_{82}Pb82214​Pb→82210​Pb

Change in mass number: 214→210⇒ΔA=−4214 \to 210 \Rightarrow \Delta A=-4214→210⇒ΔA=−4 So, one α\alphaα particle is emitted.

After one α\alphaα emission: Z:82→80Z: 82 \to 80Z:82→80 But final atomic number is 828282, so we need increase by 222. Each β−\beta^-β− increases ZZZ by 111. Therefore, we need two β−\beta^-β− particles.

Hence, Q→one α and two β−Q \to \text{one }\alpha\text{ and two }\beta^-Q→one α and two β− So, Q→3Q \to 3Q→3

  1. Process R

81210Tl→82206Pb^{210}_{81}Tl \to {}^{206}_{82}Pb81210​Tl→82206​Pb

Change in mass number: 210→206⇒ΔA=−4210 \to 206 \Rightarrow \Delta A=-4210→206⇒ΔA=−4 So, one α\alphaα particle is emitted.

After one α\alphaα emission: Z:81→79Z: 81 \to 79Z:81→79 But final atomic number is 828282, so net increase needed is +3+3+3. Each β−\beta^-β− gives +1+1+1 in ZZZ. So, three β−\beta^-β− particles are needed.

Hence, R→one α and three β−R \to \text{one }\alpha\text{ and three }\beta^-R→one α and three β− So, R→2R \to 2R→2

  1. Process S

91228Pa→88224Ra^{228}_{91}Pa \to {}^{224}_{88}Ra91228​Pa→88224​Ra

Change in mass number: 228→224⇒ΔA=−4228 \to 224 \Rightarrow \Delta A=-4228→224⇒ΔA=−4 So, one α\alphaα particle is emitted.

After one α\alphaα emission: Z:91→89Z: 91 \to 89Z:91→89 But final atomic number is 888888, so we need one more decrease in ZZZ. That is achieved by one β+\beta^+β+ emission.

Hence, S→one α and one β+S \to \text{one }\alpha\text{ and one }\beta^+S→one α and one β+ So, S→1S \to 1S→1

  1. Final matching

P→4,Q→3,R→2,S→1P \to 4, \quad Q \to 3, \quad R \to 2, \quad S \to 1P→4,Q→3,R→2,S→1

This corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So, the derived answer agrees with the stored answer.

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