Calculate the binding energy per nucleon of `._(20)^(40)Ca`. Given that mass of `._(20)^(40)Ca` nucleus `= 39.962589 u`, mass of proton `= 1.007825 u`. Mass of Neutron `= 1.008665 u` and `1 u` is equivalent to `931 MeV`.

Options

A. 18.32 MeV
B. 8.55 MeV
C. 9.94 MeV
D. 14.72 MeV

Answer

To calculate the binding energy per nucleon of \( _{20}^{40}\text{Ca} \), we will follow these steps:

### Step 1: Determine the number of protons and neutrons
For \( _{20}^{40}\text{Ca} \):
- The atomic number (Z) is 20, which means there are 20 protons.
- The mass number (A) is 40, which means the total number of nucleons (protons + neutrons) is 40.
- Therefore, the number of neutrons (N) can be calculated as:
\[ N = A - Z = 40 - 20 = 20 \]

### Step 2: Calculate the total mass of protons and neutrons
Using the given masses:
- Mass of one proton = \( 1.007825 \, u \)
- Mass of one neutron = \( 1.008665 \, u \)

The total mass of the protons and neutrons is:
\[ \text{Total mass} = (20 \times \text{mass of proton}) + (20 \times \text{mass of neutron}) \]
\[ = (20 \times 1.007825 \, u) + (20 \times 1.008665 \, u) \]
Calculating this:
\[ = 20 \times 1.007825 + 20 \times 1.008665 = 20.1565 \, u + 20.1733 \, u = 40.3298 \, u \]

### Step 3: Calculate the mass defect (\( \Delta m \))
The mass defect is given by:
\[ \Delta m = \text{Total mass of protons and neutrons} - \text{mass of nucleus} \]
\[ = 40.3298 \, u - 39.962589 \, u = 0.367211 \, u \]

### Step 4: Convert the mass defect to energy (binding energy)
Using the conversion factor \( 1 \, u = 931 \, \text{MeV} \):
\[ \text{Binding Energy} = \Delta m \times 931 \, \text{MeV} \]
\[ = 0.367211 \, u \times 931 \, \text{MeV/u} = 342.045 \, \text{MeV} \]

### Step 5: Calculate the binding energy per nucleon
The binding energy per nucleon is calculated by dividing the total binding energy by the number of nucleons:
\[ \text{Binding Energy per Nucleon} = \frac{\text{Total Binding Energy}}{\text{Number of Nucleons}} = \frac{342.045 \, \text{MeV}}{40} \]
\[ = 8.551125 \, \text{MeV} \]

### Final Answer
The binding energy per nucleon of \( _{20}^{40}\text{Ca} \) is approximately \( 8.55 \, \text{MeV} \).

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