Question

The mass of a proton is $1.0073\text{u}$ and that of a neutron is $1.0087\text{u}$. The binding energy of $_2^4\text{He}$ is approximately (Given: helium nucleus mass = $4.0015\text{u}$)}

• A. $24.8 \text{ MeV}$
• B. $24.4 \text{ MeV}$
• C. $2.48 \text{ MeV}$
• D. $2.84 \text{ MeV}$
• E. $28.4 \text{ MeV}$

Hint:

Mass defect is always positive for a stable nucleus. In calculations, using $1 \text{ u} = 931 \text{ MeV}$ is usually sufficient for competitive exams.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The binding energy is the energy equivalent of the "mass defect"—the difference between the sum of the masses of individual unbound nucleons and the actual mass of the assembled nucleus.
Step 2: Key Formula or Approach:
Calculate mass defect: \(\Delta m = [Z \cdot m_p + (A - Z) \cdot m_n] - M_{\text{nucleus}}\).
Convert mass defect to energy: Binding Energy = \(\Delta m \times 931.5 \text{ MeV/u}\).
Step 3: Detailed Explanation:
A Helium-4 nucleus (\(^4_2He\)) consists of \(Z = 2\) protons and \(A - Z = 4 - 2 = 2\) neutrons.
Find sum of individual masses:
Mass of 2 protons = \(2 \times 1.0073\text{u} = 2.0146\text{u}\)
Mass of 2 neutrons = \(2 \times 1.0087\text{u} = 2.0174\text{u}\)
Total constituent mass = \(2.0146 + 2.0174 = 4.0320\text{u}\)
Calculate mass defect \(\Delta m\):
Given actual mass = 4.0015u.
\[ \Delta m = 4.0320 - 4.0015 = 0.0305\text{u} \] Calculate binding energy:
\[ \text{BE} = 0.0305\text{u} \times 931.5 \text{ MeV/u} \] \[ \text{BE} \approx 28.41 \text{ MeV} \] This is approximately 28.4 MeV.
Step 4: Final Answer:
The binding energy is approximately 28.4 MeV.