Answer to Question #167246 in Atomic and Nuclear Physics for Mesut

Question

a.     Boron 10 has an atomic number of 5 and a nuclear mass of 10.01294 u. Calculate the binding energy per nucleon for a boron 10 atom in MeV.         

      The mass of a proton is 1.00728 u and the mass of a neutron is 1.00867 u.

b.    Boron 10 can be produced in two different ways:

(1)  By a single beta minus decay from beryllium 10

(2)  By a single beta plus decay from carbon 10.

Carbon 10 has a nuclear mass of 10.01685 u and beryllium 10 has a nuclear mass of 10.01353 u.

Explain which beta particle would have the highest velocity.  

Answer 1

a)

$$E=\frac{M-Zm_p-(A-Z)m_n}{A}931.5\\E=\frac{(10.01294)-5(1.00728)-5(1.00867)}{10}931.5\\=-6.22\frac{MeV}{nucleon}$$

b) 1)

$$K_{C}=(10.01685-10.01294)931.5=3.64\ MeV$$

2)

$$K_{Be}=(10.01353-10.01294)931.5=0.55\ MeV$$

Answer: (2)  By a single beta plus decay from carbon 10.