Question

Find binding energy of \(^{12}_{6}C\). Given mass of proton \(m_p = 1.0078\,u\) and mass of neutron \(m_n = 1.0087\,u\).

• A. \(92.19\,\text{MeV}\)
• B. \(80.20\,\text{MeV}\)
• C. \(85.19\,\text{MeV}\)
• D. \(100.19\,\text{MeV}\)

Hint:

Binding energy depends on mass defect: \[ \Delta m = Zm_p + (A-Z)m_n - M \] \[ BE = \Delta m \times 931.5\,\text{MeV} \]

Answer 1

The Correct Option is A

Concept: Binding energy: \[ BE = \left[Zm_p + (A-Z)m_n - M_{\text{atom}}\right]c^2 \] Energy equivalent: \[ 1u = 931.5\,\text{MeV} \]
Step 1: Substitute values. For \(^{12}_{6}C\): \[ Z = 6,\quad A = 12 \] \[ BE = [6m_p + 6m_n - 12] \times 931.5 \]
Step 2: Calculate mass defect. \[ 6(1.0078) + 6(1.0087) - 12 \] \[ = 6.0468 + 6.0522 - 12 \] \[ = 0.099 \]
Step 3: Calculate binding energy. \[ BE = 0.099 \times 931.5 \] \[ BE = 92.19\,\text{MeV} \] \[ \boxed{BE = 92.19\,\text{MeV}} \]