Calculate the binding energy per nucleon of `Li` isotope, which has the isotopic mass of `7.016 am u`. The individual masses of neutron and proton are `1.008665 am u` and `1.007277 am u`, respectively and the mass of electron `= 0.000548 am u`.

Explanation

To calculate the binding energy per nucleon of the lithium isotope with an isotopic mass of 7.016 amu, we can follow these steps:

### Step 1: Identify the number of protons and neutrons in the lithium isotope.
Lithium (Li) has an atomic number of 3, which means it has 3 protons. The isotope in question has a mass number of 7, so the number of neutrons can be calculated as follows:
\[ \text{Number of neutrons} = \text{Mass number} - \text{Number of protons} = 7 - 3 = 4 \]

### Step 2: Calculate the total mass of the individual nucleons (protons and neutrons).
- Mass of 3 protons:
\[ \text{Mass of protons} = 3 \times 1.007277 \, \text{amu} = 3.021831 \, \text{amu} \]

- Mass of 4 neutrons:
\[ \text{Mass of neutrons} = 4 \times 1.008665 \, \text{amu} = 4.03466 \, \text{amu} \]

- Mass of 3 electrons (not needed for binding energy calculation, but included for completeness):
\[ \text{Mass of electrons} = 3 \times 0.000548 \, \text{amu} = 0.001644 \, \text{amu} \]

### Step 3: Calculate the total mass of the lithium nucleus.
The total mass of the lithium nucleus (without considering electrons) is:
\[ \text{Total mass of nucleons} = \text{Mass of protons} + \text{Mass of neutrons} \]
\[ \text{Total mass of nucleons} = 3.021831 \, \text{amu} + 4.03466 \, \text{amu} = 7.056491 \, \text{amu} \]

### Step 4: Calculate the mass defect.
The mass defect (\( \Delta m \)) is the difference between the total mass of the nucleons and the actual mass of the lithium isotope:
\[ \Delta m = \text{Total mass of nucleons} - \text{Isotopic mass} \]
\[ \Delta m = 7.056491 \, \text{amu} - 7.016 \, \text{amu} = 0.040491 \, \text{amu} \]

### Step 5: Convert the mass defect to energy using Einstein's equation.
Using the conversion factor \( 1 \, \text{amu} \approx 931.478 \, \text{MeV} \):
\[ \Delta E = \Delta m \times 931.478 \, \text{MeV/amu} \]
\[ \Delta E = 0.040491 \, \text{amu} \times 931.478 \, \text{MeV/amu} \approx 37.7 \, \text{MeV} \]

### Step 6: Calculate the binding energy per nucleon.
The binding energy per nucleon is given by:
\[ \text{Binding Energy per Nucleon} = \frac{\Delta E}{\text{Number of nucleons}} \]
\[ \text{Binding Energy per Nucleon} = \frac{37.7 \, \text{MeV}}{7} \approx 5.39 \, \text{MeV} \]

### Final Answer:
The binding energy per nucleon of the lithium isotope is approximately **5.39 MeV**.