By the succesive disintegration of ₉₂U²³⁸, the final product obtained is ₈₂Pb²⁰⁶, then

By the succesive disintegration of ₉₂U²³⁸, the final product obtained is ₈₂Pb²⁰⁶, then how many number of α and β-particles are emitted?

Options

(a) 6 and 8
(b) 8 and 6
(c) 12 and 6
(d) 8 and 12

Correct Answer:

8 and 6

Explanation:

The number of α-particles, n₁ = Change in mass number / 4

n₁ = [(238 – 206) / 4] = 32 / 4 = 8

Now, number of β-particles, n₂ = 82 – (92 – 2 n₁)

= 82 – (92 – 2 × 8) = 82 – 76 = 6

admin:

Related Questions

  1. If a small sphere is let fall vertically in a large quantity of still liquid of density
  2. A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter
  3. Two metal wires of identical dimensions are connected in series. If σ₁ and σ₂ are
  4. For a satellite escape velocity is 11 kms⁻¹.If the satellite is launched at an angle
  5. Three unequal resistors in parallel are equivalent to a resistance 1 Ω.