Consider the following radioactive decay process ²¹⁸₈₄A α A₁ β⁻ A₂ γ A₃ α A₄ β⁺ A₅ γ A₆ The mass number and the atomic number of A ₆ are given by: (a) 210 and 84 (b) 210 and 80 (c) 211 and 80 (d) 210 and 82

Answer: (b). α: (218, 84) → (214, 82); β⁻: (214, 83); γ: no change. α: (210, 81); β⁺: (210, 80); γ: no change. So A₆ has A = 210, Z = 80. So the answer is (b) 210 and 80.

Consider the following radioactive decay process ²¹⁸₈₄A α A₁ β⁻ A₂ γ… | JEE Main 2023 (24 Jan, Shift 1)

Q11JEE Main 2023 · 24 JAN · SHIFT 1Easy

Radioactive Decays: Alpha, Beta, Gamma

Consider the following radioactive decay process

${}^{218}_{84}A \xrightarrow{\alpha} A_{1} \xrightarrow{\beta^{-}} A_{2} \xrightarrow{\gamma} A_{3} \xrightarrow{\alpha} A_{4} \xrightarrow{\beta^{+}} A_{5} \xrightarrow{\gamma} A_{6}$

The mass number and the atomic number of A ${}_{6}$ are given by:

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Answer: (b)

$\alpha$ : $(218, 84) \to (214, 82)$ ; $\beta^-$ : $(214, 83)$ ; $\gamma$ : no change.
$\alpha$ : $(210, 81)$ ; $\beta^+$ : $(210, 80)$ ; $\gamma$ : no change.
So $A_6$ has $A = 210$ , $Z = 80$ .
So the answer is (b) 210 and 80.

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Radioactive Decays: Alpha, Beta, Gamma question from JEE Main 2023 (24 Jan, Shift 1)

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