Question

₈₆A²²²®₈₄B²¹⁰. In this reaction how many a and b particles are emitted-

• A. 6a, 3b
• B. 3a, 4b
• C. 4a, 3b
• D. 3a, 6b

Answer 1

Answer: (B)

3a, 4b

Answer 2

n_a = $\frac { 222 - 210 } { 4 }$= $\frac { 12 } { 4 }$ = 3

n_b = 2n_a – Z + Z' = 2 × 3 – 86 + 84 = 6 – 2 = 4