JEE Advance - Physics (2023 - Paper 1 Online - No. 14)
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) ${ }_{92}^{238} U \rightarrow{ }_{91}^{234} \mathrm{~Pa}$ | (1) one $\alpha$ particle and one $\beta^{+}$particle |
| (Q) ${ }_{82}^{214} \mathrm{~Pb} \rightarrow{ }_{82}^{210} \mathrm{~Pb}$ | (2) three $\beta^{-}$particles and one $\alpha$ particle |
| (R) ${ }_{81}^{210} \mathrm{Tl} \rightarrow{ }_{82}^{206} \mathrm{~Pb}$ | (3) two $\beta^{-}$particles and one $\alpha$ particle |
| (S) ${ }_{91}^{228} \mathrm{~Pa} \rightarrow{ }_{88}^{224} \mathrm{Ra}$ | (4) one $\alpha$ particle and one $\beta^{-}$particle |
| (5) one $\alpha$ particle and two $\beta^{+}$particles |
$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 1$
$P \rightarrow 4, Q \rightarrow 1, R \rightarrow 2, S \rightarrow 5$
$P \rightarrow 5, Q \rightarrow 3, R \rightarrow 1, S \rightarrow 4$
$P \rightarrow 5, Q \rightarrow 1, R \rightarrow 3, S \rightarrow 2$
Explanation
In the given nuclear reaction :
$${}_{{Z_1}}{Z^{{A_1}}} \to {}_{{Z_2}}{Y^{{A_2}}} + {N_1}\,\,\left( {{}_2{H^4}} \right) + {N_2}\,\,\left( {{}_1{e^0}} \right) + {N_3}\,\,\left( {{}_{ - 1}{e^0}} \right)$$
we have the conservation of charge expressed as :
$ Z_1 = Z_2 + 2N_{1} + N_{2} - N_{3} $ ...............(i)
and the conservation of nucleons (mass number) is given by :
$ \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
$$ \mathrm{N}_2-\mathrm{N}_3=\mathrm{Z}_1-\mathrm{Z}_2-\left(\frac{\mathrm{A}_1-\mathrm{A}_2}{2}\right) $$
$\begin{aligned} & \text { (P) }{ }_{92} \mathrm{U}^{238} \rightarrow{ }_{91} \mathrm{~Pa}^{234} \\\\ & \mathrm{~N}_1=\frac{238-234}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(92-91)-\left(\frac{4}{2}\right)=-1 \rightarrow 1 \beta^{-}\end{aligned}$
$\begin{aligned} & \text { (Q) }{ }_{82} \mathrm{~Pb}^{214} \rightarrow { }_{82} \mathrm{~Pb}^{210} \\\\ & \quad \mathrm{~N}_1=\frac{214-210}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(82-82)-\left(\frac{4}{2}\right)=-2 \rightarrow 2 \beta^{-}\end{aligned}$
$\begin{aligned} & (\mathrm{R}){ }_{81} \mathrm{~T} \ell^{210} \rightarrow { }_{82} \mathrm{~Pb}^{206} \\\\ & \mathrm{~N}_1=\frac{210-206}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(81-83)-\frac{4}{2}=-3 \rightarrow 3 \beta^{-}\end{aligned}$
$\begin{aligned} & \text { (S) }{ }_{91} \mathrm{~Pa}^{228} \rightarrow_{88} \mathrm{Ra}^{224} \\\\ & \mathrm{~N}_1=\frac{228-224}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(91-88)-\frac{4}{2}=1 \beta^{+}\end{aligned}$
$${}_{{Z_1}}{Z^{{A_1}}} \to {}_{{Z_2}}{Y^{{A_2}}} + {N_1}\,\,\left( {{}_2{H^4}} \right) + {N_2}\,\,\left( {{}_1{e^0}} \right) + {N_3}\,\,\left( {{}_{ - 1}{e^0}} \right)$$
we have the conservation of charge expressed as :
$ Z_1 = Z_2 + 2N_{1} + N_{2} - N_{3} $ ...............(i)
and the conservation of nucleons (mass number) is given by :
$ \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
$$ \mathrm{N}_2-\mathrm{N}_3=\mathrm{Z}_1-\mathrm{Z}_2-\left(\frac{\mathrm{A}_1-\mathrm{A}_2}{2}\right) $$
$\begin{aligned} & \text { (P) }{ }_{92} \mathrm{U}^{238} \rightarrow{ }_{91} \mathrm{~Pa}^{234} \\\\ & \mathrm{~N}_1=\frac{238-234}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(92-91)-\left(\frac{4}{2}\right)=-1 \rightarrow 1 \beta^{-}\end{aligned}$
$\begin{aligned} & \text { (Q) }{ }_{82} \mathrm{~Pb}^{214} \rightarrow { }_{82} \mathrm{~Pb}^{210} \\\\ & \quad \mathrm{~N}_1=\frac{214-210}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(82-82)-\left(\frac{4}{2}\right)=-2 \rightarrow 2 \beta^{-}\end{aligned}$
$\begin{aligned} & (\mathrm{R}){ }_{81} \mathrm{~T} \ell^{210} \rightarrow { }_{82} \mathrm{~Pb}^{206} \\\\ & \mathrm{~N}_1=\frac{210-206}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(81-83)-\frac{4}{2}=-3 \rightarrow 3 \beta^{-}\end{aligned}$
$\begin{aligned} & \text { (S) }{ }_{91} \mathrm{~Pa}^{228} \rightarrow_{88} \mathrm{Ra}^{224} \\\\ & \mathrm{~N}_1=\frac{228-224}{4}=1 \rightarrow 1 \alpha \\\\ & \mathrm{N}_2-\mathrm{N}_3=(91-88)-\frac{4}{2}=1 \beta^{+}\end{aligned}$
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