Question

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

\[\ce{^223_88Ra -> ^209_82Pb + ^14_6C}\]

\[\ce{^223_88 Ra -> ^219_86 Rn + ^4_2He}\]

Calculate the Q-values for these decays and determine that both are energetically allowed.

Answer 1

Take a `""_6^14"C"` emission nuclear reaction:

\[\ce{^223_88Ra -> ^209_82Pb + ^14_6C}\]

We know that:

Mass of `""_88^223"Ra"` m₁ = 223.01850 u

Mass of `""_82^209"Pb"` m₂ = 208.98107 u

Mass of `""_6^14"C"`, m₃ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = (m₁ − m₂ − m₃) c²

= (223.01850 − 208.98107 − 14.00324) c²

= (0.03419 c²) u

But 1 u = 931.5 MeV/c²

∴ Q = 0.03419 × 931.5

= 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now take a `""_2^4"He"` emission nuclear reaction:

\[\ce{^223_88 Ra -> ^219_86 Rn + ^4_2He}\]

We know that:

Mass of `""_88^223"Ra"`, m₁ = 223.01850

Mass of `""_82^219"Rn"`  m₂ = 219.00948

Mass of `""_2^4"He"`, m₃ = 4.00260

Q-value of this nuclear reaction is given as:

Q = (m₁ − m₂ − m₃) c²

= (223.01850 − 219.00948 − 4.00260) C²

= (0.00642 c²) u

= 0.00642 × 931.5 = 5.98 MeV

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.