If the binding energy per nucleon of deuterium is 1.115 MeV, its mass defect in atomic mass unit is

Options

A. 0.0048
B. 0.0024
C. 0.0012
D. 0.0006

Explanation

To find the mass defect of deuterium in atomic mass units (amu) given its binding energy per nucleon, we can follow these steps:

### Step 1: Identify the given values
- Binding energy per nucleon of deuterium (²H) = 1.115 MeV
- Mass number of deuterium = 2 (since it consists of one proton and one neutron)

### Step 2: Calculate the total binding energy
The total binding energy (BE) of deuterium can be calculated by multiplying the binding energy per nucleon by the mass number:
\[ \text{Total Binding Energy} = \text{Binding Energy per Nucleon} \times \text{Mass Number} \]
\[ \text{Total Binding Energy} = 1.115 \, \text{MeV} \times 2 = 2.230 \, \text{MeV} \]

### Step 3: Relate binding energy to mass defect
The mass defect (Δm) can be calculated using the formula:
\[ \text{Binding Energy} = \Delta m \times 931.5 \, \text{MeV/amu} \]
Rearranging this gives:
\[ \Delta m = \frac{\text{Binding Energy}}{931.5 \, \text{MeV/amu}} \]

### Step 4: Substitute the total binding energy into the equation
Now, substitute the total binding energy we calculated:
\[ \Delta m = \frac{2.230 \, \text{MeV}}{931.5 \, \text{MeV/amu}} \]

### Step 5: Calculate the mass defect
Now perform the calculation:
\[ \Delta m \approx \frac{2.230}{931.5} \approx 0.0024 \, \text{amu} \]

### Final Answer
The mass defect of deuterium is approximately \(0.0024 \, \text{amu}\).

---