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Q.

What is the respective number of α and β particles emitted in the following radioactive decay

 90X200→80Y168

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a

6 and 8

b

8 and 8

c

6 and 6

d

8 and 6

answer is D.

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Detailed Solution

 90X200→80Y168+n12He4+n2

−1β0

Mass No. balance

200=168+n14

⇒n1=8

At. No. balance

90=80+n1(2)+n2(−1)

⇒n2=6

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