Question

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes: 22388Ra -→ 20982Pb + 146C 22388Ra -→ 21986Rn + 42He Calculate the Q-values for these decays and determine that both are energetically allowed.

Answer 1

Dear Student,

(I) In 88Ra223 -→ 82Pb209 + 6C14

Mass defect is calculated by -

∆m = m(Ra223) - m(Pb209) - m(C14)

∆m = 223.01850 - 208.98107 - 14.00324

∆m = 0.03419 u

Q-value is calculated by -

Q-value = ∆m × 931 MeV

Q-value = 0.03419 × 931 MeV

Q-value = 31.83 MeV

It is energetically allowed as Q-value for this reaction is positive.

(II) In 88Ra223 -→ 86Rn219 + 2He4

Mass defect is calculated by -

∆m = m(Ra223) - m(Rn119) - m(He4)

∆m = 223.01850 - 219.00948 - 4.00260

∆m = 0.00642 u

Q-value is calculated by -

Q-value = ∆m × 931 MeV

Q-value = 0.00642 × 931 MeV

Q-value = 5.977 MeV

It is energetically allowed as Q-value for this reaction is positive.

Thanks dear...