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)

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

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)

Answer 1

To calculate the binding energy of the last neutron in ${}^{13}C$, we need to use the given equation: $E_{S}=[M_{n}+M(^{A-1}Z)-M(^{\wedge }Z)]^{1331}Mev/amu$ where $E_{S}$ is the energy required to remove a neutron from the nucleus without any kinetic energy. In this case, the nucleus is ${}^{13}C$, which means it has 6 protons and 7 neutrons. When a neutron is removed, the residual nucleus is ${}^{12}C$, which has 6 protons and 6 neutrons. Substituting the values into the equation, we have: $E_{S}=[M_{n}+M(^{12}C)-M(^{13}C)]^{1331}Mev/amu$ Now, we need to find the values of $M_{n}$, $M(^{12}C)$, and $M(^{13}C)$. $M_{n}$ is the mass of a neutron, which is approximately 1.008665 amu. $M(^{12}C)$ is the mass of the ${}^{12}C$ nucleus, which can be calculated by adding the masses of 6 protons and 6 neutrons: $M(^{12}C) = 6 \times M_{p} + 6 \times M_{n} = 6 \times 1.007825 + 6 \times 1.008665 = 12.0004 \text{ amu}$ $M(^{13}C)$ is the mass of the ${}^{13}C$ nucleus, which can be calculated by adding the masses of 6 protons and 7 neutrons: $M(^{13}C) = 6 \times M_{p} + 7 \times M_{n} = 6 \times 1.007825 + 7 \times 1.008665 = 13.0127 \text{ amu}$ Now, we can substitute these values into the equation: $E_{S}=[1.008665 + 12.0004 - 13.0127]^{1331}Mev/amu$ $E_{S}=[13.009065 - 13.0127]^{1331}Mev/amu$ $E_{S}=-0.003625 \text{ amu}^{1331}Mev/amu$ 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 ${}^{13}C$ is 0.003625 amu.

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