Question

Calculate the binding energy per nucleon (in MeV) of a helium nucleus \(\left( ^4_2\text{He} \right)\). Given: \[ \begin{aligned} m\left(^4_2\text{He}\right) &= 4.002603\,\text{u} \\ m_n &= 1.008665\,\text{u} \\ m_H &= 1.007825\,\text{u} \\ 1\,\text{u} &= 931.5\,\text{MeV}/c^2 \end{aligned} \]

Hint:

To find binding energy per nucleon: 1. Add the masses of free protons and neutrons. 2. Subtract the actual nucleus mass to get mass defect. 3. Multiply by \( 931.5 \,\text{MeV/u} \), then divide by number of nucleons.

Answer 1

The helium nucleus has: - 2 protons (use \( m_H \) since proton mass is included in hydrogen atom mass) - 2 neutrons Step 1: Calculate mass of individual nucleons \[ \text{Mass of 2 protons} = 2 \times 1.007825 = 2.015650\,\text{u} \] \[ \text{Mass of 2 neutrons} = 2 \times 1.008665 = 2.017330\,\text{u} \] \[ \text{Total mass of free nucleons} = 2.015650 + 2.017330 = 4.032980\,\text{u} \] Step 2: Calculate mass defect \[ \Delta m = \text{mass of nucleons} - \text{mass of nucleus} = 4.032980 - 4.002603 = 0.030377\,\text{u} \] Step 3: Calculate binding energy \[ \text{Total B.E.} = \Delta m \times 931.5 = 0.030377 \times 931.5 \approx 28.30\,\text{MeV} \] Step 4: Binding energy per nucleon \[ \text{B.E./nucleon} = \frac{28.30}{4} = \boxed{7.075\,\text{MeV}} \]