(a) Calculate the minimum energy required to remove one proton from the nucleus${}_{\mathbf{6}}{}^{\mathbf{12}}\mathit{C}$. This is called the proton-removal energy. (Hint: Find the difference between the mass of a ${}_{\mathbf{6}}{}^{\mathbf{12}}\mathit{C}$nucleus and the mass of a proton plus the mass of the nucleus formed when a proton is removed from ${}_{\mathbf{6}}{}^{\mathbf{12}}\mathit{C}$. (b) How does the proton-removal energy for ${}_{\mathbf{6}}{}^{\mathbf{12}}\mathit{C}$compare to the binding energy per nucleon for ${}_{\mathbf{6}}{}^{\mathbf{12}}\mathit{C}$, calculated using Eq. (43.10)?

The reaction energy Q is defined as follows if initial particles A and B interact to form end particles C and D:

$Q=[M_A+M_B-M_C-M_D]\times 931.5MeV/u$

The binding energy of a nucleus with Z protons and N neutrons is given by the equation;

$E_b=[Z{M}_H+N{m}_n-{}_Z{}^{A}M]\times 931.5MeV/u$

Given;

The atomic mass of ${}_1{}^{1}H$is $M\left({}_1{}^{1}H\right)=1.007825u$.

The atomic mass of localid="1663932528906" ${}_6{}^{12}C$is $M\left({}_6{}^{12}C\right)=12.000000u$.

Neutron mass is $m_n=1.008665u$.

(a)The proton removal reaction has the following equation:

${}_6{}^{12}C\to {}_Z{}^{A}X+{}_1{}^{1}p$

the number of nucleons and the atomic number are both conserved

$$\begin{gathered} 6=Z+0 \\ Z=6 \\ 12=A+1 \\ A=11 \end{gathered}$$

The element with Z=5 in the periodic table is boron.

As a result, the isotope produced is localid="1663932207696" ${}_5{}^{11}B$and its atomic mass $M\left({}_5{}^{11}B\right)=11.009305u$.

By substituting the hydrogen atom for the proton in the equation, the number of electrons on both sides is identical, and the computations cancel out.

Therefore, putting the values of masses the mass difference is;

$$\begin{gathered} ∆M=\left[12.000000u-11.009305u-1.007825u\right] \\ =[-0.017130u] \end{gathered}$$

So, the energy required to remove the proton is;

$$\begin{gathered} Q=∆M\times 931.5MeV/u \\ =(0.017130u)\times 931.5MeV/u \\ =15.957MeV \end{gathered}$$

Hence, the minimum energy required to remove one proton from the nucleus is $15.957MeV$.

(a) The number of protons in the ${}_6{}^{12}C$nucleus is $Z=6$while the number of neutrons is $N=12-6=6$.

as a result of putting the values together, the binding energy is;$$\begin{gathered} E_b=[6\times 1.007825u+6\times 1.008665u-12.000000]\times 931.5MeV/u \\ =[0.098940u]\times 931.5MeV/u \\ =92.163MeV \end{gathered}$$

The binding energy per nucleon is calculated by dividing the number of nucleons;

$$\begin{gathered} E_{bpern}=\frac{E_b}{A} \\ =\frac{92.163MeV}{12} \\ =7.680MeV \end{gathered}$$

when compared to the energy of proton removal

$$\begin{gathered} \frac{Q}{E_{bpern}}=\frac{15.957MeV}{7.680MeV} \\ =2.08 \end{gathered}$$

Hence, the energy required to remove a proton is twice that required to bind a nucleon.