Calculate number of `alpha-and beta`- particles emitted when `""_(92)^(238)U` changes into radioactive `""_(82)^(206)Pb`

Options

A. 5, 8
B. 8, 6
C. 8, 9
D. 4, 6

Answer

B

Explanation

Suppose the number of `alpha`-particles emitted by .a. and `beta`- particles be .b.
`""_(92)^(238)U to ""_(82)^(206)Pb + a""_(2)^(4)He + b ""_(-1)^(0)e`. Balancing the mass number on both sides of the above equation.
`238 = 206 + 4a + 0 xx b therefore a = 8` Hence number of `alpha`-particles emitted = 8.
On balancing the above equation with respect to atomic number on both sides
`92 = 82 + 2a + b (-1) = 82 + 2 xx 8 - b = 98-b " "therefore b = 6`
So the number of `beta`- particles emitted = 6