The mass of `._3^7Li` is 0.042 amu less than the sum of masses of its constituents. The binding energy per nucleon is

Options

A. 5.586 MeV
B. 10.522 MeV
C. 2.433 MeV
D. 3.739 MeV

Explanation

To find the binding energy per nucleon for the lithium-7 nucleus (Li-7), we will follow these steps:

### Step 1: Understand the mass defect
The mass defect (ΔM) is given as 0.042 amu. This is the difference between the sum of the masses of the individual nucleons (protons and neutrons) and the actual mass of the nucleus.

### Step 2: Convert mass defect to energy
The binding energy (BE) can be calculated using the mass defect and Einstein's mass-energy equivalence principle. The formula to convert mass defect in amu to energy in MeV is:

\[ BE = \Delta M \times 931.5 \text{ MeV/amu} \]

Substituting the given mass defect:

\[ BE = 0.042 \text{ amu} \times 931.5 \text{ MeV/amu} \]

### Step 3: Calculate the total binding energy
Now, we perform the multiplication:

\[ BE = 0.042 \times 931.5 \approx 39.1 \text{ MeV} \]

### Step 4: Determine the binding energy per nucleon
The binding energy per nucleon is calculated by dividing the total binding energy by the mass number (A) of the nucleus. For lithium-7, the mass number \( A = 7 \).

\[ BE_{\text{per nucleon}} = \frac{BE}{A} = \frac{39.1 \text{ MeV}}{7} \]

### Step 5: Perform the final calculation
Now we calculate the binding energy per nucleon:

\[ BE_{\text{per nucleon}} \approx \frac{39.1}{7} \approx 5.586 \text{ MeV} \]

### Conclusion
The binding energy per nucleon for lithium-7 is approximately **5.586 MeV**.

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