Calculate the binding energy per nucleon of `._17^35Cl` nucleus. Given that mass of `._17^35Cl` nucleus = 34.98000 u, mass of proton = 1.007825 u, mass of neutron = 1.008665 u and 1 u is equivalent to 931 Mev.

Options

A. 6.2 MeV
B. 7.2 Mev
C. 4.2 Mev
D. 8.2 Mev

Answer

To calculate the binding energy per nucleon of the chlorine-35 nucleus (denoted as \( _{17}^{35}\text{Cl} \)), we will follow these steps:

### Step 1: Identify the Masses
We are given the following data:
- Mass of \( _{17}^{35}\text{Cl} \) nucleus, \( M = 34.98000 \, \text{u} \)
- Mass of proton, \( m_p = 1.007825 \, \text{u} \)
- Mass of neutron, \( m_n = 1.008665 \, \text{u} \)

### Step 2: Determine the Number of Protons and Neutrons
For the chlorine-35 nucleus:
- Atomic number \( Z = 17 \) (number of protons)
- Mass number \( A = 35 \) (total number of nucleons)

The number of neutrons \( N \) can be calculated as:
\[ N = A - Z = 35 - 17 = 18 \]

### Step 3: Calculate the Theoretical Mass of the Nucleus
The theoretical mass of the nucleus can be calculated using the masses of protons and neutrons:
\[ \text{Theoretical mass} = Z \cdot m_p + N \cdot m_n \]
Substituting the values:
\[ \text{Theoretical mass} = 17 \cdot 1.007825 + 18 \cdot 1.008665 \]
Calculating this:
\[ \text{Theoretical mass} = 17.132025 + 18.15597 = 35.287995 \, \text{u} \]

### Step 4: Calculate the Mass Defect
The mass defect \( \Delta m \) is given by:
\[ \Delta m = \text{Theoretical mass} - \text{Actual mass} \]
Substituting the values:
\[ \Delta m = 35.287995 \, \text{u} - 34.98000 \, \text{u} = 0.307995 \, \text{u} \]

### Step 5: Convert Mass Defect to Binding Energy
To find the binding energy \( BE \), we use the conversion factor \( 1 \, \text{u} = 931 \, \text{MeV} \):
\[ BE = \Delta m \cdot 931 \, \text{MeV} \]
Substituting the mass defect:
\[ BE = 0.307995 \, \text{u} \cdot 931 \, \text{MeV} \approx 287.67 \, \text{MeV} \]

### Step 6: Calculate Binding Energy per Nucleon
The binding energy per nucleon \( BE/A \) is calculated as:
\[ BE/A = \frac{BE}{A} \]
Substituting the values:
\[ BE/A = \frac{287.67 \, \text{MeV}}{35} \approx 8.2 \, \text{MeV} \]

### Final Answer
The binding energy per nucleon of the \( _{17}^{35}\text{Cl} \) nucleus is approximately \( 8.2 \, \text{MeV} \).

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