₉₂U²³⁵ undergoes successive disintegrations with the end product of ₈₂P²⁰³

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.

admin:

Related Questions

  1. The ratio of radii of planets A and B is K₁ and ratio of accelerations
  2. A light emitting diode(LED) has a voltage drop of 2 volt across it
  3. The radius of curvature of the convex face of a plano-convex lens is 12 cm
  4. In Young’s experiment, the ratio of maximum to minimum intensities of the fringe
  5. If the angular momentum of any rotating body increase by 200 % , then the increase