Saved Bookmarks
| 1. |
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ""_(20)^(41)Ca and ""_(13)^(27)Al from the following data: m(""_(20)^(40)Ca) = 39.962591u m(""_(20)^(41)Ca) = 40.962278 u m(""_(13)^(26)Al) = 25.986895 u m(""_(13)^(27)Al) = 26.981541 u. |
|
Answer» Solution :Neutron SEPARATION energy `S_n` of a nucleus `""_Z^AX` is given by `S_n = {m_N(""_Z^(A-1)X) + m_n - m_N(""_Z^AX)}c^2` `=[{m_N(""_Z^(A-1)X) + Zm_e} - m_n - {m_N(""Z^AX) + Zm_c}]c^2` `[m(""_Z^(A-1)X) + m_N - m(""_Z^AX)]c^2` `:. S_n(""_(20)^(41)Ca) = [m(""_(20)^(40)Ca) + m_N -m(""_(20)^(41)Ca)]c^2` `= [39.962591 + 1.008665 - 40.962278]c^2` `= 0.008978 xx 931 = 8.363 MeV` Similarly `S_n (""_(13)^(27)Al) = [m(""_Z^(A-1)X) + m_N - m(""Z^AX)]c^2 = 13.06 MeV`. |
|