The mass defect for the nucleus of helium is `0.0303` a.m.u. What is the binding energy per nucleon for helium in `MeV` ?

Options

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

Answer

To find the binding energy per nucleon for the helium nucleus given a mass defect of `0.0303` a.m.u, we can follow these steps:

### Step 1: Understand the concept of binding energy
The binding energy (BE) of a nucleus can be calculated using the mass defect (Δm) and the equation:
\[ BE = Δm \cdot c^2 \]
where \( c \) is the speed of light.

### Step 2: Use the conversion factor
We know that:
\[ c^2 = 931.5 \text{ MeV/a.m.u} \]
This means that for every atomic mass unit (a.m.u) of mass defect, the binding energy is equivalent to 931.5 MeV.

### Step 3: Calculate the total binding energy
Substituting the values into the binding energy formula:
\[ BE = Δm \cdot c^2 = 0.0303 \text{ a.m.u} \cdot 931.5 \text{ MeV/a.m.u} \]

### Step 4: Perform the multiplication
Calculating the binding energy:
\[ BE = 0.0303 \cdot 931.5 = 28.2 \text{ MeV} \]

### Step 5: Determine the number of nucleons in helium
Helium has a mass number of 4, which means it has 4 nucleons (2 protons and 2 neutrons).

### Step 6: Calculate the binding energy per nucleon
The binding energy per nucleon (B.E. per nucleon) is given by:
\[ \text{B.E. per nucleon} = \frac{BE}{\text{Number of nucleons}} = \frac{28.2 \text{ MeV}}{4} \]

### Step 7: Perform the division
Calculating the binding energy per nucleon:
\[ \text{B.E. per nucleon} = \frac{28.2}{4} = 7.05 \text{ MeV} \]

### Final Answer
The binding energy per nucleon for helium is approximately \( 7 \text{ MeV} \).

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