91Ó°ÊÓ

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

$Q=[M_A+M_B-M_C-M_D]×931.5MeV/u$

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

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

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:

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

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;

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

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;$$\begin{gathered} E_b=[6×1.007825u+6×1.008665u-12.000000]×931.5MeV/u \\ =[0.098940u]×931.5MeV/u \\ =92.163MeV \end{gathered}$$

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.