Question

In list - I some conversions are shown and in the list- II emitted particles are shown. Match the column

| List-I| | List- II
---|---|---|---
(P)| $_{92}^{238}\textrm{U}$ → $_{91}^{234}\textrm{Pa}$| (1)| one α particle and one β+ particle
(Q)| $_{82}^{214}\textrm{Pb}$ → $_{82}^{210}\textrm{Pb}$| (2)| Three β- particle and 1 α particle
(R)| $_{81}^{210}\textrm{Tl}$ → $_{82}^{206}\textrm{Pb}$| (3)| Two β- particle and 1 α particle
(S)| $_{81}^{228}\textrm{Pa}$→ $_{88}^{224}\textrm{Ra}$| (4)| one α particle and one β- particle
| | (5)| one α particle and two β+ particle

Answer 1

The correct answer is : (P)$\rightarrow$(4),(Q)$\rightarrow$(3),(R)$\rightarrow$(2),(S)$\rightarrow$(1)

(P) $_{92}^{238}\textrm{U}\rightarrow_{91}^{234}\textrm{Pa}+n_1\,\,{_{2}^{4}\textrm{He}}+n_{2-1}^{0}e$
$238=234+4n_1\Rightarrow n_1=1$
$92=91+2n_1-n_2\Rightarrow n_2=1$

(Q) $_{82}^{214}\textrm{Pb}\rightarrow_{82}^{210}\textrm{Pb}+n_1\,\,{_{2}^{4}\textrm{He}}+n_{2-1}^{0}e$
$214=210+4n_1\Rightarrow n_1=1$
$82=82+2n_1-n_2\Rightarrow n_2=2$

(R) $_{81}^{210}\textrm{Tl}\rightarrow_{82}^{206}\textrm{Pb}+n_1\,\,{_{2}^{4}\textrm{He}}+n_{2-1}^{0}e$
$210=206+4n_1\Rightarrow n_1=1$
$81=82+2n_1-n_2\Rightarrow n_2=3$

(S) $_{91}^{228}\textrm{Pa}\rightarrow_{88}^{224}\textrm{Ra}+n_1\,\,{_{2}^{4}\textrm{He}}+n_{2-1}^{0}e$
$228=224+4n_1\Rightarrow n_1=1$
$91=88+2n_1-n_2\Rightarrow n_2=-1$