In a radioactive decay chain, the initial nucleus is ²³²₉₀ Th. At the end there are 6 α -particles and 4 β -particles which are emitted. If the end nucleus is AZ X, A and Z are given by: (a) A = 208; Z = 80 (b) A = 208; Z = 82 (c) A = 200; Z = 81 (d) A = 202; Z = 80

Answer: (b). 6 α: A = 232 - 24 = 208, Z = 90 - 12 = 78. 4 β⁻: Z = 78 + 4 = 82, A unchanged. So the answer is (b) A = 208; Z = 82.

In a radioactive decay chain, the initial nucleus is ²³²₉₀ Th. At the… | JEE Main 2019 (12 Jan, Shift 2)

Q23JEE Main 2019 · 12 JAN · SHIFT 2Easy

Radioactive Decays: Alpha, Beta, Gamma

In a radioactive decay chain, the initial nucleus is ${}^{232}_{90}$ Th. At the end there are 6 $\alpha$ -particles and 4 $\beta$ -particles which are emitted. If the end nucleus is ${}^{A}_{Z}$ X, A and Z are given by:

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Answer: (b)

6 $\alpha$ : $A = 232 - 24 = 208$ , $Z = 90 - 12 = 78$ .
4 $\beta^-$ : $Z = 78 + 4 = 82$ , $A$ unchanged.
So the answer is (b) A = 208; Z = 82.

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