`._(92)U^(235)` undergoes successive disintegrations with the end product of `._(82)P^(203)` . The number of `alphaandbeta` particles emitted are

Options

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

Answer

A

Explanation

`._(92)U^(235)to` end product `._(82)P^(203)alphaandbeta` emitted. `DeltaA=235-203=32`.
`therefore` 8 alpha particles are emitted . The charge should be `92-16=76`.
But as the final charge is 82 , six `beta^(-)` particles had been emitted to make up the find atomic number Z=82. `therefore` 8 alpha particles and six `beta^(-)` have been emitted.