Question 66: Atoms & Nuclei

Question

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

Accepted Answer

In the given nuclear reaction :
$${}_{{Z_1}}{Z^{{A_1}}} 	o {}_{{Z_2}}{Y^{{A_2}}} + {N_1}\,\,\left( {{}_2{H^4}} ight) + {N_2}\,\,\left( {{}_1{e^0}} ight) + {N_3}\,\,\left( {{}_{ - 1}{e^0}} ight)$$
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
$$
\mathrm{N}_2-\mathrm{N}_3=\mathrm{Z}_1-\mathrm{Z}_2-\left(\frac{\mathrm{A}_1-\mathrm{A}_2}{2}ight)
$$
$\begin{aligned} & 	ext { (P) }{ }_{92} \mathrm{U}^{238} ightarrow{ }_{91} \mathrm{~Pa}^{234} \\ & \mathrm{~N}_1=\frac{238-234}{4}=1 ightarrow 1 \alpha \\ & \mathrm{N}_2-\mathrm{N}_3=(92-91)-\left(\frac{4}{2}ight)=-1 ightarrow 1 \beta^{-}\end{aligned}$
$\begin{aligned} & 	ext { (Q) }{ }_{82} \mathrm{~Pb}^{214} ightarrow { }_{82} \mathrm{~Pb}^{210} \\ & \quad \mathrm{~N}_1=\frac{214-210}{4}=1 ightarrow 1 \alpha \\ & \mathrm{N}_2-\mathrm{N}_3=(82-82)-\left(\frac{4}{2}ight)=-2 ightarrow 2 \beta^{-}\end{aligned}$
$\begin{aligned} & (\mathrm{R}){ }_{81} \mathrm{~T} \ell^{210} ightarrow { }_{82} \mathrm{~Pb}^{206} \\ & \mathrm{~N}_1=\frac{210-206}{4}=1 ightarrow 1 \alpha \\ & \mathrm{N}_2-\mathrm{N}_3=(81-83)-\frac{4}{2}=-3 ightarrow 3 \beta^{-}\end{aligned}$
$\begin{aligned} & 	ext { (S) }{ }_{91} \mathrm{~Pa}^{228} ightarrow_{88} \mathrm{Ra}^{224} \\ & \mathrm{~N}_1=\frac{228-224}{4}=1 ightarrow 1 \alpha \\ & \mathrm{N}_2-\mathrm{N}_3=(91-88)-\frac{4}{2}=1 \beta^{+}\end{aligned}$

Answer

$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 1$

Answer

$P \rightarrow 4, Q \rightarrow 1, R \rightarrow 2, S \rightarrow 5$

Answer

$P \rightarrow 5, Q \rightarrow 3, R \rightarrow 1, S \rightarrow 4$

Answer

$P \rightarrow 5, Q \rightarrow 1, R \rightarrow 3, S \rightarrow 2$

Atoms & Nuclei

Choose the correct answer: