Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

In the nuclear decay given below: $_{Z}^{A}X \rightarrow _{Z+1}^{A}Y \rightarrow _{Z-1}^{A-4}B^* \rightarrow _{Z-1}^{A-4}B$ The particles emitted in the sequence are:

\beta , \alpha , \gamma

\gamma , \beta , \alpha

\beta , \gamma , \alpha

\alpha , \beta , \gamma

Step-by-Step Solution

First Decay ( $X \rightarrow Y$ ): The mass number ( $A$ ) remains the same, while the atomic number ( $Z$ ) increases by 1 ( $Z \rightarrow Z+1$ ). This indicates the conversion of a neutron into a proton with the emission of an electron (\beta particle). This is Beta ( $\beta^-$ ) decay.
Second Decay ( $Y \rightarrow B^*$ ):The mass number decreases by 4 ( $A \rightarrow A-4$ ) and the atomic number decreases by 2 ( $(Z+1) - 2 = Z-1$ ). This corresponds to the emission of a helium nucleus ( $_{2}^{4}\text{He}$ ). This isAlpha ( $\alpha$ ) decay. 3.Third Decay ( $B^* \rightarrow B$ ):There is no change in either the mass number or the atomic number, but the nucleus transitions from an excited state (implied by the context of sequential decay ending in the same nucleus) to a lower energy state. This releases energy in the form of a high-energy photon. This isGamma ( $\gamma$ ) decay. 4.Sequence: The order is $\beta, \alpha, \gamma$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started