What is the respective number of `alpha` and `beta` particles emitted in the following radioactive decay
`._90 X^200 rarr ._80 Y^168`.

Options

A. 8 and 8
B. 8 and 6
C. 6 and 6
D. 6 and 8

Answer

B

Explanation

In this nuclear reaction, mass number is decreased by 200 - 168 = 32 and the atomic number is decreased by 90-80=10
If x and y denote the number of `alpha " and " beta` particles emitted, then for mass number , 4x = 32
`therefore` x = 8 = no. of particles [ mass number decreases by 4 in a emission ]
and for atomic number `[-8xx2+ y xx1 = -10]`
`therefore` - 16 + y = -10 `therefore` y = -10 + 16 = 6
Thus x = 8 and y=6.