Verified Solution • 2025–26 Session moderate

QJEEAdv23Ph327d: JEE Advanced 2023 Physics - Class 12 JEEAdv23Ph327d Solution: List-I shows different radioactive decay processes and List-II…

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} (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) 81210Tl→82206 Pb{ }_{81}^{210} \mathrm{Tl} \rightarrow{ }_{82}^{206} \mathrm{~Pb} (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

Multiple Choice Options

  • A P→4,Q→3,R→2,S→1P \rightarrow 4, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 1
  • B P→4,Q→1,R→2,S→5P \rightarrow 4, Q \rightarrow 1, R \rightarrow 2, S \rightarrow 5
  • C P→5,Q→3,R→1,S→4P \rightarrow 5, Q \rightarrow 3, R \rightarrow 1, S \rightarrow 4
  • D P→5,Q→1,R→3,S→2P \rightarrow 5, Q \rightarrow 1, R \rightarrow 3, S \rightarrow 2

Step-by-Step Verified Solution

In the given nuclear reaction :

Z1ZA1→Z2YA2+N1  (2H4)+N2  (1e0)+N3  (−1e0){}_{{Z_1}}{Z^{{A_1}}} \to {}_{{Z_2}}{Y^{{A_2}}} + {N_1}\,\,\left( {{}_2{H^4}} \right) + {N_2}\,\,\left( {{}_1{e^0}} \right) + {N_3}\,\,\left( {{}_{ - 1}{e^0}} \right)

we have the conservation of charge expressed as :

Z1=Z2+2N1+N2−N3Z_1 = Z_2 + 2N_{1} + N_{2} - N_{3} ...............(i)

and the conservation of nucleons (mass number) is given by :

A1amp;=A2+4N1N1amp;=A1−A24..........(ii)\begin{aligned}A_1 &= A_2 + 4N_{1} \\\\N_{1} &= \frac{A_1 - A_2}{4} ..........(ii)\end{aligned}

From (i) and (ii), we get

N2−N3=Z1−Z2−(A1−A22)\mathrm{N}_2-\mathrm{N}_3=\mathrm{Z}_1-\mathrm{Z}_2-\left(\frac{\mathrm{A}_1-\mathrm{A}_2}{2}\right)

amp; (P) 92U238→91 Pa234amp; N1=238−2344=1→1αamp;N2−N3=(92−91)−(42)=−1→1β−\begin{aligned} & \text { (P) }{ }_{92} \mathrm{U}^{238} \rightarrow{ }_{91} \mathrm{~Pa}^{234} \\\\ & \mathrm{~N}_1=\frac{238-234}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(92-91)-\left(\frac{4}{2}\right)=-1 \rightarrow 1 \beta^{-}\end{aligned}

amp; (Q) 82 Pb214→82 Pb210amp; N1=214−2104=1→1αamp;N2−N3=(82−82)−(42)=−2→2β−\begin{aligned} & \text { (Q) }{ }_{82} \mathrm{~Pb}^{214} \rightarrow { }_{82} \mathrm{~Pb}^{210} \\\\ & \quad \mathrm{~N}_1=\frac{214-210}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(82-82)-\left(\frac{4}{2}\right)=-2 \rightarrow 2 \beta^{-}\end{aligned}

amp;(R)81 Tℓ210→82 Pb206amp; N1=210−2064=1→1αamp;N2−N3=(81−83)−42=−3→3β−\begin{aligned} & (\mathrm{R}){ }_{81} \mathrm{~T} \ell^{210} \rightarrow { }_{82} \mathrm{~Pb}^{206} \\\\ & \mathrm{~N}_1=\frac{210-206}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(81-83)-\frac{4}{2}=-3 \rightarrow 3 \beta^{-}\end{aligned}

amp; (S) 91 Pa228→88Ra224amp; N1=228−2244=1→1αamp;N2−N3=(91−88)−42=1β+\begin{aligned} & \text { (S) }{ }_{91} \mathrm{~Pa}^{228} \rightarrow_{88} \mathrm{Ra}^{224} \\\\ & \mathrm{~N}_1=\frac{228-224}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(91-88)-\frac{4}{2}=1 \beta^{+}\end{aligned}

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