Respuesta :
The 17 Cl 37 nucleus has 37 nucleons. Thus, making the energy per nucleon is 1.6944 x 10^-19 J/37 nucleons. Energy = 5.09x10^-11 J. So, (5.09x10^-11)/37 nucleons you get a 1.38x10^-12 J/nucleon as an answer.
I hope this helps you with your problem.
Answer: The binding energy per nucleon is [tex]1.4\times 10^{-12}J[/tex]
Explanation:
Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.
We are given a nucleus having representation: [tex]_{17}^{37}\textrm{Cl}[/tex]
Number of protons = 17
Number of neutrons = 37 - 17 = 20
To calculate the mass defect of the nucleus, we use the equation:
[tex]\Delta m=[(n_p\times m_p)+(n_n\times m_n)-M[/tex]
where,
[tex]n_p[/tex] = number of protons = 17
[tex]m_p[/tex] = mass of one proton = 1.00728 amu
[tex]n_n[/tex] = number of neutrons = 20
[tex]m_n[/tex] = mass of one neutron = 1.00866 amu
M = nuclear mass = 36.9566 amu
Putting values in above equation, we get:
[tex]\Delta m=[(17\times 1.00728)+(20\times 1.00866)]-36.9566\\\\\Delta m=0.34036amu[/tex]
To calculate the binding energy of the nucleus, we use the equation:
[tex]E=\Delta mc^2\\E=(0.34036u)\times c^2[/tex]
[tex]E=(0.34036u)\times (931.5MeV)[/tex] (Conversion factor: [tex]1u=931.5MeV/c^2[/tex] )
[tex]E=317.04MeV=507.264\times 10^{-13}J[/tex] (Conversion factor: [tex]1MeV=1.6\times 10^{-13}J[/tex] )
Number of nucleons in [tex]_{17}^{37}\textrm{Cl}[/tex] atom = 37
To calculate the binding energy per nucleon, we divide the binding energy by the number of nucleons, we get:
[tex]\text{Binding energy per nucleon}=\frac{\text{Binding energy}}{\text{Nucleons}}[/tex]
[tex]\text{Binding energy per nucleon}=\frac{507.26\times 10^{-13}J}{37}=1.4\times 10^{-12}J[/tex]
Hence, the binding energy per nucleon is [tex]1.4\times 10^{-12}J[/tex]
