Question

Calculate the nuclear binding energy per nucleon of ${}_6^{12}C$ from the following data:
Given mass of proton $\left(m_p\right)=1.00758$ amu
mass of neutron $\left(m_n\right)=1.00893$ amu
mass of electron $\left(m_e\right)=0.00055$ amu
Isotopic mass of ${}^{12}C=12.00000$ amu. Take $1amu=931$ MeV
Plan from the mass defect $△m$ (amu), binding energy $=△m\times 931$ MeV
binding energy per nucleon$=\frac{△m\times 931}{numberofnucleons}$

• A. $7.94$ MeV
• B. $6.34$ MeV
• C. $8.84$ MeV
• D. None of these

Answer 1

The correct option is A $7.94$ MeV

${}_6^{12}C$ contains 6 protons, 6 neutrons and 6 electrons hence, calculated atomic mass of ${}_6^{12}C$
$=6m_p+6m_n+6m_e$
$=6\times 1.00758+6\times 1.00893+6\times 0.00055$
$=12.10236$
$\therefore$ mass defect $=12.10236-12.00000$
$=0.10236$ amu
binding energy$=0.10236\times 931=95.30$ MeV
binding energy per nucleon $=\frac{95.30}{12}=7.94$ MeV