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Question: Answered & Verified by Expert
In the nuclear decay given below
$$
{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \longrightarrow{ }_{\mathrm{Z}+1}^{\mathrm{A}} \mathrm{Y} \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \mathrm{~B}^* \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \mathrm{~B} \text {, }
$$
the particles emitted in the sequence are
PhysicsNuclear PhysicsNEETNEET 2009 (Screening)
Options:
  • A $\beta, \alpha, \gamma$
  • B $\gamma, \beta, \alpha$
  • C $\beta, \gamma, \alpha$
  • D $\alpha, \beta, \gamma$
Solution:
1713 Upvotes Verified Answer
The correct answer is: $\beta, \alpha, \gamma$
Key Idea In a nuclear reaction conservation of charge number and mass number must hold good.
Alpha particles are positively charged particles with charge $+2 e$ and mass $4 \mathrm{~m}$. Emission of an $\alpha$-particle reduces the mass of the radionuclide by 4 and its atomic number by 2. $\beta$-particles are negatively charged particles with rest mass as well as charge same as that of electrons. $\gamma$-particles carry no charge and mass.
Radioactive transition will be as follows
$$
\begin{aligned}
& { }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \longrightarrow{ }_{\mathrm{Z}+1}{ }^{\mathrm{A}} \mathrm{Y}+\beta_{-1}^0 \\
& { }_{\mathrm{Z}+1}^{\mathrm{A}} \mathrm{Y} \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \beta+\alpha_2^4 \\
& { }_{\mathrm{Z}+1}^{\mathrm{A}-4} \beta \longrightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}-4} \mathrm{\beta}+\gamma_0^0 \\
&
\end{aligned}
$$

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