Back to Directory
NEET PHYSICSEasy

A radioactive nucleus ZAX{}_{Z}^{A}\mathrm{X} undergoes spontaneous decay in the sequence ZX→Z−1B→Z−3C→Z−2D{}_{Z}\mathrm{X} \rightarrow {}_{Z-1}\mathrm{B} \rightarrow {}_{Z-3}\mathrm{C} \rightarrow {}_{Z-2}\mathrm{D}, where ZZ is the atomic number of element X\mathrm{X}. The possible decay particles in the sequence are:

A

β+,α,β−\beta^+, \alpha, \beta^-

B

β−,α,β+\beta^-, \alpha, \beta^+

C

α,β−,β+\alpha, \beta^-, \beta^+

D

α,β+,β−\alpha, \beta^+, \beta^-

Step-by-Step Solution

  1. Analyze Step 1 (XZ→BZ−1X_Z \rightarrow B_{Z-1}): The atomic number decreases by 1 (Z→Z−1Z \rightarrow Z-1). This change is characteristic of β+\beta^+ decay (positron emission), where a proton converts into a neutron, reducing the nuclear charge by 1 unit .
  2. Analyze Step 2 (BZ−1→CZ−3B_{Z-1} \rightarrow C_{Z-3}): The atomic number decreases by 2 (from Z−1Z-1 to Z−3Z-3). This change corresponds to α\alpha decay, where an \alpha particle (24He{}_{2}^{4}\mathrm{He}) is emitted, reducing the atomic number by 2 .
  3. Analyze Step 3 (CZ−3→DZ−2C_{Z-3} \rightarrow D_{Z-2}): The atomic number increases by 1 (from Z−3Z-3 to Z−2Z-2). This change is characteristic of β−\beta^- decay (electron emission), where a neutron converts into a proton, increasing the nuclear charge by 1 unit .
  4. Conclusion: The sequence of particles emitted is β+,α,β−\beta^+, \alpha, \beta^-.
Practice Mode Available

Master this Topic on Sushrut

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

Get Started