Question

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).

Answer 1

Explanation:

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Answer 2

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