Question

If a ^b aX species emits firstly a positron, then two α and two β and at last one α is also after intially it fainally convets into stable ^c dY species so correct relation will be

• A. c = b – 12, d = a – 5
• B. a = c – 8, d = b – 1
• C. a = c – 6, d = b – 0
• D. a = c – 4, a = b – 2

Answer 1

Answer: (A)

c = b – 12, d = a – 5

Answer 2

The correct option is(A): c = b – 12, d = a – 5. Initially, the species a^b emits a positron, which reduces the atomic number (a) by 1. Then, it emits two alpha particles (2α), reducing the atomic number (a) by 8. It also emits two beta particles (2β), reducing the atomic number (a) by 2. Finally, it emits one more alpha particle (1α), reducing the atomic number (a) by 4. So, the overall change in atomic number (a) is -1 - 8 - 2 - 4 = -15, which is equivalent to d = a - 5. Since the species ends up as a stable c^d, the final atomic number (c) must be 12 less than the initial atomic number (b), which is represented by c = b - 12.