Binding energy of deuterium is 2.23MeV. Mass defect in amu is

Options

A. 0.0012
B. 0.0024
C. 0.0036
D. `-0.0012`

Answer

To find the mass defect of deuterium given its binding energy, we can use the relationship between binding energy and mass defect derived from Einstein's equation \(E = mc^2\).

### Step-by-Step Solution:

1. **Understand the relationship**: The binding energy (BE) of a nucleus is related to its mass defect (\(\Delta m\)) by the equation:
\[ BE = \Delta m \cdot c^2 \]
where \(c\) is the speed of light.

2. **Rearrange the equation**: To find the mass defect, we can rearrange the equation:
\[ \Delta m = \frac{BE}{c^2} \]

3. **Convert binding energy to joules**: The binding energy of deuterium is given as 2.23 MeV. We need to convert this energy into joules. The conversion factor is:
\[ 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J} \]
Therefore,
\[ BE = 2.23 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} = 3.568 \times 10^{-13} \text{ J} \]

4. **Use the speed of light**: The speed of light \(c\) is approximately \(3 \times 10^8 \text{ m/s}\). We need to calculate \(c^2\):
\[ c^2 = (3 \times 10^8 \text{ m/s})^2 = 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]

5. **Calculate the mass defect**: Now substitute the values into the rearranged equation:
\[ \Delta m = \frac{3.568 \times 10^{-13} \text{ J}}{9 \times 10^{16} \text{ m}^2/\text{s}^2} \]
\[ \Delta m = 3.964 \times 10^{-30} \text{ kg} \]

6. **Convert mass defect to amu**: To convert the mass defect from kilograms to atomic mass units (amu), we use the conversion factor:
\[ 1 \text{ amu} = 1.66 \times 10^{-27} \text{ kg} \]
Therefore,
\[ \Delta m = \frac{3.964 \times 10^{-30} \text{ kg}}{1.66 \times 10^{-27} \text{ kg/amu}} \approx 0.00239 \text{ amu} \]

7. **Final result**: The mass defect of deuterium is approximately:
\[ \Delta m \approx 0.00239 \text{ amu} \]

### Summary:
The mass defect of deuterium, given its binding energy of 2.23 MeV, is approximately 0.00239 amu.