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