The isotope ¹²₅B having a mass 12.014 u undergoes ????-decay to ¹²₆C. ¹²₆C has an excited state of the nucleus
(¹²₆C*) at 4.041 MeV above its ground state. If ¹²₅B decays to ¹²₆C* , the maximum kinetic energy of the ????-
particle in units of MeV is
(1 u = 931.5 MeV/c², where c is the speed of light in vacuum).
Answers
Answered by
0
Explanation:
what is your question dear pls ask completely
Answered by
0
12
B→
6
12
C+β
−
+γ
−
Q=(m
5
12
B−m
6
12
C)c
2
=(m
5
12
B−(m
6
12
C+Δm))c
2
=(m
5
12
B−m
6
12
C)c
2
−Δmc
2
=0.014×931.5−4.041
=9 MeV
Attachments:
Similar questions