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Question: <sub>86</sub>A<sup>222</sup>®<sub>84</sub>B<sup>210</sup>. In this reaction how many a and b particl...

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

3a, 4b

Explanation

Solution

na = 2222104\frac { 222 - 210 } { 4 }= 124\frac { 12 } { 4 } = 3

nb = 2na – Z + Z' = 2 × 3 – 86 + 84 = 6 – 2 = 4