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

Nuclear Decay Reactions and Calculations 

Correct Answers:

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

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

From the reaction equation:

\(238 = 234 + 4n_1 \Rightarrow n_1 = 1\)

\(92 = 91 + 2n_1 - n_2 \Rightarrow n_2 = 1\)

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

From the reaction equation:

\(214 = 210 + 4n_1 \Rightarrow n_1 = 1\)

\(82 = 82 + 2n_1 - n_2 \Rightarrow n_2 = 2\)

(R) Reaction: \(_{81}^{210}\textrm{Tl} \rightarrow_{82}^{206}\textrm{Pb} + n_1\,\, {_{2}^{4}\textrm{He}} + n_{2-1}^{0}e\)

From the reaction equation:

\(210 = 206 + 4n_1 \Rightarrow n_1 = 1\)

\(81 = 82 + 2n_1 - n_2 \Rightarrow n_2 = 3\)

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

From the reaction equation:

\(228 = 224 + 4n_1 \Rightarrow n_1 = 1\)

\(91 = 88 + 2n_1 - n_2 \Rightarrow n_2 = -1\)