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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.
Solution
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.
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