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

Step 1: Understanding the Concept:
The binding energy of a nucleus is the energy required to completely separate its constituent protons and neutrons.
It is equivalent to the mass defect ($\Delta m$) of the nucleus multiplied by $c^2$.
Step 2: Key Formula or Approach:
Mass defect: $\Delta m = \left[ Z \cdot m_p + (A - Z) \cdot m_n \right] - M_{\text{nucleus}}$
Binding Energy: $BE = \Delta m \times 931.5 \text{ MeV/u}$
Step 3: Detailed Explanation:
For the $^{12}_{6}\text{C}$ nucleus, the number of protons $Z = 6$ and the number of neutrons $A - Z = 12 - 6 = 6$.
By definition of the atomic mass unit, the mass of a Carbon-12 atom is exactly $12.0000\text{ u}$.
Calculate the sum of the masses of individual nucleons:
Mass of 6 protons $= 6 \times 1.0078\text{ u} = 6.0468\text{ u}$
Mass of 6 neutrons $= 6 \times 1.0087\text{ u} = 6.0522\text{ u}$
Total mass of constituent nucleons:
\[ M_{\text{constituents}} = 6.0468\text{ u} + 6.0522\text{ u} = 12.0990\text{ u} \]
Now, find the mass defect $\Delta m$:
\[ \Delta m = M_{\text{constituents}} - M_{\text{nucleus}} \]
\[ \Delta m = 12.0990\text{ u} - 12.0000\text{ u} = 0.0990\text{ u} \]
Calculate the Binding Energy:
\[ BE = 0.0990\text{ u} \times 931.5 \text{ MeV/u} \]
\[ BE \approx 92.2185 \text{ MeV} \]
Rounding to match the closest available standard option, it evaluates to $92.19\text{ MeV}$.
Step 4: Final Answer:
The binding energy of $^{12}_{6}\text{C}$ is 92.19 MeV.