Find the binding energy of the nucleus of lighium isotope `._(3)Li^(7)` and hence find the binding energy per nucleon in it.

Explanation

To find the binding energy of the lithium isotope \( _{3}^{7}\text{Li} \) and the binding energy per nucleon, we will follow these steps:

### Step 1: Identify the Masses
We need to know the mass of the lithium nucleus, the mass of a proton, and the mass of a neutron. The given values are:
- Mass of lithium \( _{3}^{7}\text{Li} = 7.01454353 \, \text{amu} \)
- Mass of proton \( m_p = 1.007826 \, \text{amu} \)
- Mass of neutron \( m_n = 1.00867 \, \text{amu} \)

### Step 2: Calculate the Number of Protons and Neutrons
For the lithium isotope \( _{3}^{7}\text{Li} \):
- Number of protons \( Z = 3 \)
- Number of neutrons \( N = A - Z = 7 - 3 = 4 \)

### Step 3: Calculate the Mass Defect (\( \Delta m \))
The mass defect is calculated using the formula:
\[ \Delta m = (Z \cdot m_p + N \cdot m_n) - m_{\text{Li}} \]
Substituting the values:
\[ \Delta m = (3 \cdot 1.007826 + 4 \cdot 1.00867) - 7.01454353 \]
Calculating the values:
\[ \Delta m = (3.023478 + 4.03468) - 7.01454353 \]
\[ \Delta m = 7.058158 - 7.01454353 = 0.04361447 \, \text{amu} \]

### Step 4: Calculate the Binding Energy (\( BE \))
The binding energy can be calculated using the mass defect:
\[ BE = \Delta m \times 931.5 \, \text{MeV/amu} \]
Substituting the mass defect:
\[ BE = 0.04361447 \, \text{amu} \times 931.5 \, \text{MeV/amu} \]
Calculating the binding energy:
\[ BE \approx 40.6 \, \text{MeV} \]

### Step 5: Calculate the Binding Energy per Nucleon
The binding energy per nucleon is given by:
\[ \text{Binding Energy per nucleon} = \frac{BE}{A} \]
Where \( A \) is the mass number (7 for \( _{3}^{7}\text{Li} \)):
\[ \text{Binding Energy per nucleon} = \frac{40.6 \, \text{MeV}}{7} \approx 5.8 \, \text{MeV/nucleon} \]

### Final Answers
- Binding Energy of \( _{3}^{7}\text{Li} \) = \( 40.6 \, \text{MeV} \)
- Binding Energy per nucleon = \( 5.8 \, \text{MeV/nucleon} \)