E_(S)=[M_(n)+M(^A-1Z)-M(^wedge Z)]^1331Mev/amu The energy E_(s) is just sufficient to remove a neutron from the nucleus without ding it with any kinetic energy. However, if this procedure is reversed and a on with no kinctic energy is absorbed by the nucleus A-1Z the energy E_(s) is sed in the process. nple 2.9 Calculate the binding energy of the last neutron in {}^13C Solution. If the neutron is removed from (}^13C the residual nucleus is {)^12C The bind- ing energy or separation energy is then computed from Eq. (245) as follows: (2.45)

To calculate the binding energy of the last neutron in , we need to use the given equation: where is the energy required to remove a neutron from the nucleus without any kinetic energy.In this case, the nucleus is , which means it has 6 protons and 7 neutrons. When a neutron is removed, the residual nucleus is , which has 6 protons and 6 neutrons.Substituting the values into the equation, we have: Now, we need to find the values of , , and . is the mass of a neutron, which is approximately 1.008665 amu. is the mass of the nucleus, which can be calculated by adding the masses of 6 protons and 6 neutrons: is the mass of the nucleus, which can be calculated by adding the masses of 6 protons and 7 neutrons: Now, we can substitute these values into the equation: Since the energy is negative, it means that the energy is released when a neutron is absorbed by the nucleus. Therefore, the binding energy of the last neutron in is 0.003625 amu.