`""_(92)"U"^(235)` undergoes successive disintegrations with the end product of `""_(82)"Pb"^(203)`. The number of `alpha` and `beta` particles emitted are

Options

A. `alpha=6, beta=4`
B. `alpha=6, beta=0`
C. `alpha=8, beta=6`
D. `alpha=3, beta=3`

Answer

C

Explanation

Let number of `alpha` particles decayed be x and number of `beta` particles decayed be y.
Then equation for the decay is given by
`""_(92)U^(235)toxalpha_(2)^(4)+ybeta_(-1)^(0)+Pb_(82)^(203)`
Equating the mass number on both sides
`235=4x+203" "...(i)`
Equating atomic number on both sides
`92=2x-y+82" "...(ii)`
Solving Eqs. (i) and (ii), we get
`x=8, y=6`
`:.8alpha` particles and `6beta` particles are emitted in disintegration.