Consider the following radioactive decay process
₈₄²¹⁸A→(α) A₁→(β -) A₂→(γ) A₃→(α) A₄→(β +) A₅→(γ) A₆
The mass number and the atomic number of A₆ are given by :

Solution

Physics sir · 3 steps

Track A and Z one decay at a time: only alpha changes the mass number, and gamma changes nothing.

1. The three rules. An α decay lowers the mass number A by 4 and the atomic number Z by 2. A β⁻ decay raises Z by 1 and a β⁺ decay lowers Z by 1, and neither changes A. A γ decay changes neither.

   Decay	Change in A	Change in Z	Nucleus (A, Z)
   start	none	none	A: (218, 84)
   α	-4	-2	A₁: (214, 82)
   β⁻	0	+1	A₂: (214, 83)
   γ	0	0	A₃: (214, 83)
   α	-4	-2	A₄: (210, 81)
   β⁺	0	-1	A₅: (210, 80)
   γ	0	0	A₆: (210, 80)

2. Quick check. Net change in A: two alphas give 218 - 8 = 210. Net change in Z: -2 + 1 - 2 - 1 = -4, so 84 - 4 = 80.
3. So. A₆ has mass number 210 and atomic number 80, which is option (b).

The idea to keep. Count the alphas for the mass number (4 each). For the atomic number, alpha takes 2, β⁻ adds 1, β⁺ takes 1. Gamma only releases energy, so skip it.

Answer from the NTA final answer key. Solution by PrepWiser.