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Physics Question on Atoms

In the nuclear decay given below ZAX→ Z+1AY→ Z−1A−AB∗ → Z−1A−AB,^A_Z X \rightarrow \, ^A _{Z+1} Y \rightarrow \, ^{A-A}_{Z-1}B * \, \rightarrow \, ^{A-A}_{Z-1}B, the particles emitted in the sequence are

A

γ,β,α\gamma,\beta,\alpha

B

β,γ,α\beta,\gamma,\alpha

C

α,β,γ\alpha,\beta,\gamma

D

β,α,γ\beta,\alpha,\gamma

Answer

β,α,γ\beta,\alpha,\gamma

Explanation

Solution

The correct option is(D): β,α,γ\beta,\alpha,\gamma

ZXA⟶βZ+1YA⟶αZ−1BA−4⟶γZ−1BA−4Z X ^{ A} \stackrel{\beta}{\longrightarrow} Z +1 ^{Y ^{ A}} \stackrel{\alpha}{\longrightarrow} Z -1 ^{B ^{ A -4}} \stackrel{\gamma}{\longrightarrow}Z -1 ^{B ^{ A -4}}
First X decays by β\beta emission emitting v‾\overline{v}, antineutrino simultaneously. Y emits α\alpha resulting in the excited level of B which in tum emits a γ\gamma ray.
∴β,α,γ\therefore \beta,\alpha,\gamma is the answer.