Calculate the binding energy per nucleon (B.E./nucleon) in the nuclei of `._(26)Fe^(56)`. Given :
`mp[._(26)Fe^(56)]=55.934939 " amu",m[._(0)n^(1)]=1.00865 " amu" , m[._(1)"II"^(1)]=1.00782 " amu"`

Options

A. 477.45 MeV
B. 8.52 MeV
C. 577 MeV
D. 10.52 MeV

Explanation

To calculate the binding energy per nucleon (B.E./nucleon) in the nucleus of \( _{26}^{56}Fe \), we will follow these steps:

### Step 1: Determine the number of protons and neutrons
The nucleus of \( _{26}^{56}Fe \) has:
- Number of protons (Z) = 26
- Mass number (A) = 56

To find the number of neutrons (N):
\[ N = A - Z = 56 - 26 = 30 \]

### Step 2: Calculate the total mass of protons and neutrons
The mass of a proton (\( m_p \)) is given as \( 1.00782 \, \text{amu} \) and the mass of a neutron (\( m_n \)) is given as \( 1.00865 \, \text{amu} \).

Total mass of protons:
\[ \text{Total mass of protons} = Z \times m_p = 26 \times 1.00782 \, \text{amu} = 26.20332 \, \text{amu} \]

Total mass of neutrons:
\[ \text{Total mass of neutrons} = N \times m_n = 30 \times 1.00865 \, \text{amu} = 30.2595 \, \text{amu} \]

### Step 3: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the total mass of the individual nucleons and the actual mass of the nucleus.

Given the mass of \( _{26}^{56}Fe \) is \( 55.934939 \, \text{amu} \):
\[ \Delta m = (\text{Total mass of protons} + \text{Total mass of neutrons}) - \text{mass of } _{26}^{56}Fe \]
\[ \Delta m = (26.20332 + 30.2595) - 55.934939 \]
\[ \Delta m = 56.46282 - 55.934939 = 0.527881 \, \text{amu} \]

### Step 4: Convert mass defect to energy
Using Einstein's equation \( E = \Delta m c^2 \) and knowing that \( 1 \, \text{amu} \) corresponds to \( 931.5 \, \text{MeV} \):
\[ E = 0.527881 \, \text{amu} \times 931.5 \, \text{MeV/amu} = 491.70 \, \text{MeV} \]

### Step 5: Calculate binding energy per nucleon
The binding energy per nucleon is given by:
\[ \text{B.E./nucleon} = \frac{E}{A} = \frac{491.70 \, \text{MeV}}{56} \]
\[ \text{B.E./nucleon} \approx 8.78 \, \text{MeV/nucleon} \]

### Final Answer
The binding energy per nucleon in the nucleus of \( _{26}^{56}Fe \) is approximately \( 8.78 \, \text{MeV/nucleon} \).

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