Under certain circumstances, a nucleus can decay by emitting aparticle more massive than an α-particle. Consider the followingdecay processes:223 209 1488 82 6 Ra Pb C 223 219 488 86 2 Ra Rn He → +Calculate the Q-values for these decays and determine that bothare energetically allowed.
Given: The decay processes are,
R 88 223 a → P 88 223 b + C 6 14 R 88 223 a → R 86 219 n + H 2 4 e
Case 1: For decay of C 6 14
The Q -value of the reaction is given as,
Q = ( m 1 - m 2 - m 3 ) c 2
Where, the mass of R 88 223 a is m 1 , the mass of P 88 223 b is m 2 and the mass of C 6 14 is m 3 .
Since, 1 u = 931.5 MeV/ c 2
By substituting the given values in the above expression, we get
Q = ( 223.01850 − 208.98107 − 14.00324 ) c 2 = ( 0.03419 c 2 ) u = 0.03419 × 931.5 = 31.848 MeV
Case 2: For decay of H 2 4 e
The Q -value of the reaction is given as,
Q = ( m ′ 1 - m ′ 2 - m ′ 3 ) c 2
Where, the mass of R 88 223 a is m ′ 1 , the mass of R 82 219 n is m ′ 2 and the mass of H 2 4 e is m ′ 3 .
By substituting the given values in the above expression, we get
Q = ( 223.01850 − 219.00948 − 4.00260 ) c 2 = ( 0.00642 c 2 ) u = 0.00642 × 931.5 = 5.98 MeV
Thus, the Q value of first nuclear reaction is 31.848 MeV and the Q value of second nuclear reaction is 5.98 MeV .
Since, both values are positive so the both reactions are energetically allowed.
Given: The decay processes are,
Case 1: For decay of
The
Where, the mass of
Since,
By substituting the given values in the above expression, we get
Case 2: For decay of
The
Where, the mass of
By substituting the given values in the above expression, we get
Thus, the
Since, both values are positive so the both reactions are energetically allowed.
