In a given reaction : zX^A → (z+1)Y^A → (z-1)K^(A-4) → (z-1)K^(A-4), the radioactive radiations are emitted in the sequence of –

(a) α, β, γ
(b) γ, α, β
(c) β, α, γ
(d) γ, β, α

Question

In a given reaction : zX^A → (z+1)Y^A → (z-1)K^(A-4) → (z-1)K^(A-4), the radioactive radiations are emitted in the sequence of –

(a) α, β, γ
(b) γ, α, β
(c) β, α, γ
(d) γ, β, α

Accepted Answer

>

> [!TLDR]
> Track how Z and A change at each step of the decay chain: an increase in Z with same A is β⁻ decay; a drop of 2 in Z with a drop of 4 in A is α decay; no change in Z or A but a distinct final state is γ emission.
>
> **Why / steps:**
> 1. $_{Z}X^{A}	o {}_{Z+1}Y^{A}$: mass number unchanged, atomic number increases by 1 → this is **β⁻ decay** (a neutron converts to a proton, emitting an electron and an antineutrino).
> 2. $_{Z+1}Y^{A}	o {}_{Z-1}K^{A-4}$: atomic number drops by 2, mass number drops by 4 → this is **α decay** (loss of a helium-4 nucleus). …

Q.In a given reaction : zX^A → (z+1)Y^A → (z-1)K^(A-4) → (z-1)K^(A-4), the radioactive radiations are emitted in the sequence of –

Concept understanding — Alpha Beta Gamma Decay

Alpha Beta Gamma Decay

Why Nuclei Decay at All

Alpha Decay — The Nucleus Ejects a Helium Core

Beta Decay — A Neutron Turns Into a Proton (or Vice Versa)

Gamma Decay — The Nucleus Sheds Excess Energy

Q.In a given reaction : zX^A → (z+1)Y^A → (z-1)K^(A-4) → (z-1)K^(A-4), the radioactive radiations are emitted in the sequence of –

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