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

Question

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Answer 1

In this problem, we need to analyze the nuclear decay sequence and determine the particles emitted at each step. The sequence given is:

^A_Z X \rightarrow \, ^A _{Z+1} Y \rightarrow \, ^{A-A}_{Z-1}B * \, \rightarrow \, ^{A-A}_{Z-1}B

To solve this, we need to understand the nature of each type of decay particle:

Beta Decay (β): In beta decay, a neutron in the nucleus is converted into a proton, or a proton is converted into a neutron. When a neutron becomes a proton, a beta particle (which is an electron, \beta^-\)) is emitted, leading to an increase in the atomic number $Z$ by 1, while the mass number $A$ stays the same. This matches the first part of our sequence where ^A_Z X \rightarrow \, ^A _{Z+1} Y.
Alpha Decay (α): In alpha decay, the nucleus emits an alpha particle (which consists of 2 protons and 2 neutrons, essentially ^4_2He). This results in the reduction of both the atomic number $Z$ by 2 and the mass number $A$ by 4. This matches the second transition ^A _{Z+1} Y \rightarrow \, ^{A-A}_{Z-1}B * since the atomic number decreases by 2, and the mass number remains at $A-A$, considering the context, might be an error in notation, assuming $A-A$ signifies an unchanged $A$ after two decays.
Gamma Decay (γ): Gamma decay involves the emission of gamma photons and results in no change in atomic or mass numbers. It typically follows another form of decay to release excess energy, stabilizing the nucleus which explains the transition from an excited state denoted by $B^*$ to the ground state $B$.

Given the sequence, the correct order of decay is:

Beta decay ($ \beta $) resulting in: $ X \rightarrow Y $
Alpha decay ($ \alpha $) resulting in: $ Y \rightarrow B^* $
Gamma decay ($ \gamma $) resulting in: transition from the excited state $ B^* \rightarrow B $

Thus, the particles emitted in the sequence are \beta,\alpha,\gamma, which matches the correct option provided.

Answer 2

In this problem, we need to analyze the nuclear decay sequence and determine the particles emitted at each step. The sequence given is:

^A_Z X \rightarrow \, ^A _{Z+1} Y \rightarrow \, ^{A-A}_{Z-1}B * \, \rightarrow \, ^{A-A}_{Z-1}B

To solve this, we need to understand the nature of each type of decay particle:

Beta Decay (β): In beta decay, a neutron in the nucleus is converted into a proton, or a proton is converted into a neutron. When a neutron becomes a proton, a beta particle (which is an electron, \beta^-\)) is emitted, leading to an increase in the atomic number $Z$ by 1, while the mass number $A$ stays the same. This matches the first part of our sequence where ^A_Z X \rightarrow \, ^A _{Z+1} Y.
Alpha Decay (α): In alpha decay, the nucleus emits an alpha particle (which consists of 2 protons and 2 neutrons, essentially ^4_2He). This results in the reduction of both the atomic number $Z$ by 2 and the mass number $A$ by 4. This matches the second transition ^A _{Z+1} Y \rightarrow \, ^{A-A}_{Z-1}B * since the atomic number decreases by 2, and the mass number remains at $A-A$, considering the context, might be an error in notation, assuming $A-A$ signifies an unchanged $A$ after two decays.
Gamma Decay (γ): Gamma decay involves the emission of gamma photons and results in no change in atomic or mass numbers. It typically follows another form of decay to release excess energy, stabilizing the nucleus which explains the transition from an excited state denoted by $B^*$ to the ground state $B$.

Given the sequence, the correct order of decay is:

Beta decay ($ \beta $) resulting in: $ X \rightarrow Y $
Alpha decay ($ \alpha $) resulting in: $ Y \rightarrow B^* $
Gamma decay ($ \gamma $) resulting in: transition from the excited state $ B^* \rightarrow B $

Thus, the particles emitted in the sequence are \beta,\alpha,\gamma, which matches the correct option provided.

The Correct Option is D