What is the binding energy per nucleon for Helium in MeV if the mass defect for the nucleus of Helium is 0.0303 amu.

Options

A. 28
B. 7
C. 4
D. 1

Explanation

To find the binding energy per nucleon for Helium given the mass defect, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the mass defect**: The mass defect for Helium is given as \( \Delta m = 0.0303 \, \text{amu} \).

2. **Convert mass defect to energy**: We use the equivalence of mass and energy given by Einstein’s equation \( E = mc^2 \). In nuclear physics, we use the conversion factor that \( 1 \, \text{amu} \) corresponds to \( 931.5 \, \text{MeV} \). Therefore, the binding energy \( E_b \) can be calculated as:
\[ E_b = \Delta m \times 931.5 \, \text{MeV} \]
Substituting the values:
\[ E_b = 0.0303 \, \text{amu} \times 931.5 \, \text{MeV/amu} = 28.22 \, \text{MeV} \]

3. **Determine the number of nucleons in Helium**: Helium has 2 protons and 2 neutrons, giving a total of:
\[ \text{Number of nucleons} = 2 + 2 = 4 \]

4. **Calculate binding energy per nucleon**: To find the binding energy per nucleon, we divide the total binding energy by the number of nucleons:
\[ \text{Binding energy per nucleon} = \frac{E_b}{\text{Number of nucleons}} = \frac{28.22 \, \text{MeV}}{4} \]
Performing the calculation:
\[ \text{Binding energy per nucleon} = 7.055 \, \text{MeV} \]

5. **Final result**: Rounding the result, we find that the binding energy per nucleon for Helium is approximately:
\[ \text{Binding energy per nucleon} \approx 7.11 \, \text{MeV} \]

### Summary of the Solution:
The binding energy per nucleon for Helium, given the mass defect of \( 0.0303 \, \text{amu} \), is approximately \( 7.11 \, \text{MeV} \).