₉₂U²³⁵ undergoes successive disintegrations with the end product of ₈₂P²⁰³. The number of α and β-particles emitted are

Options

(a) α=8,β=6
(b) α=3,β=3
(c) α=6,β=4
(d) α=6,β=0

Correct Answer:

α=8,β=6

Explanation:

₉₂U²³⁵ → end product ₈₂P²°³ α and β emitted.

ΔA = 235 – 203 = 32

Therefore, 8 alpha particles are emitted. The charge should be 92 – 16 = 76.

But as the final charge is 82, six β⁻ particles had been emitted to make up the final atomic number Z = 82.

.·. 8 alpha particles and six β⁻ have been emitted.