Question 15.9: Find the energy of the excited states of 10 5 B correspondin...

Substituting the values, E_{i} = 10.02 MeV, m_{p} = 1.007276 u, and m(^{10}_{ 5} B) = 10.012937 u, into Eq. (15.22), the equation becomes

E = E_{i} \left( 1 −\frac{ m_{p}}{ m^{∗}_{ A}} \right)− E_{f} \left(1 + \frac{m_{p} }{m^{∗}_{ A}} \right)+\frac{ 2m_{p}}{ m^{∗}_{ A}} (E_{i}E_{f} )^{ 1/2} \cos θ .                          (15.22)

E = 9.0120 \ MeV − 1.1006 \ MeV \ E_{f} .

We may now substitute the energies E_{f } given along the bottom of Fig. 15.14 to obtain the corresponding values of E given in Table 15.6.