Determine the binding energy of `""_(3)Li^(7)` if its mass is 7.00 amu
Use 1 amu = 931 `MeVc^(2) (m_(p) = 1.007825 u, m_(n) = 1.008665 u)`.

Answer

To determine the binding energy of \( _{3}^{7}\text{Li} \), we will follow these steps:

### Step 1: Identify the number of protons and neutrons
The atomic number of Lithium (Li) is 3, which means it has 3 protons. The mass number (A) is 7, which means the total number of nucleons (protons + neutrons) is 7.

To find the number of neutrons (N):
\[ N = A - Z = 7 - 3 = 4 \]
So, Lithium-7 has 3 protons and 4 neutrons.

### Step 2: Calculate the total mass of the nucleons
The mass of the protons and neutrons can be calculated using their respective masses:
- Mass of one proton, \( m_p = 1.007825 \, \text{amu} \)
- Mass of one neutron, \( m_n = 1.008665 \, \text{amu} \)

Now, calculate the total mass of the nucleons:
\[ \text{Total mass of nucleons} = (3 \times m_p) + (4 \times m_n) \]
Substituting the values:
\[ \text{Total mass of nucleons} = (3 \times 1.007825) + (4 \times 1.008665) \]
\[ = 3.023475 + 4.03466 = 7.058135 \, \text{amu} \]

### Step 3: Calculate the mass defect (\( \Delta m \))
The mass defect is the difference between the total mass of the nucleons and the actual mass of the nucleus:
\[ \Delta m = \text{Total mass of nucleons} - \text{Mass of nucleus} \]
Given that the mass of the nucleus is 7.00 amu:
\[ \Delta m = 7.058135 - 7.00 = 0.058135 \, \text{amu} \]

### Step 4: Convert the mass defect to binding energy
The binding energy (BE) can be calculated using the formula:
\[ \text{BE} = \Delta m \times c^2 \]
We are given that \( 1 \, \text{amu} = 931 \, \text{MeV/c}^2 \). Therefore:
\[ \text{BE} = 0.058135 \, \text{amu} \times 931 \, \text{MeV/amu} \]
Calculating this:
\[ \text{BE} = 0.058135 \times 931 \approx 54.12 \, \text{MeV} \]

### Final Answer
The binding energy of \( _{3}^{7}\text{Li} \) is approximately \( 54.12 \, \text{MeV} \).

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