Given `({:(56),(26):}Fe)= 55. 934939u and m`
`({:(209),(83):}Bi)= 208. 980388u8`
`m _("proton") =1.007825u, m _("nutron")=1.008665u`
Then, BE per nucleon of Fe and Bi are respectively

Options

A. `9. 790 MeV, 7.848MeV`
B. `7.75 MeV, 6.84 MeV`
C. `7.5 MeV, 6.5 MeV`
D. data insufficient

Answer

To find the binding energy per nucleon for Iron (Fe) and Bismuth (Bi), we will follow these steps:

### Step 1: Calculate the Number of Protons and Neutrons
For Iron (Fe):
- Atomic number (Z) = 26 (number of protons)
- Atomic mass (A) = 56
- Number of neutrons (N) = A - Z = 56 - 26 = 30

For Bismuth (Bi):
- Atomic number (Z) = 83 (number of protons)
- Atomic mass (A) = 209
- Number of neutrons (N) = A - Z = 209 - 83 = 126

### Step 2: Calculate the Total Mass of Protons and Neutrons
For Iron (Fe):
- Mass of protons = Number of protons × Mass of one proton
= 26 × 1.007825 u = 26.20345 u
- Mass of neutrons = Number of neutrons × Mass of one neutron
= 30 × 1.008665 u = 30.25995 u
- Total mass = Mass of protons + Mass of neutrons
= 26.20345 u + 30.25995 u = 56.4634 u

For Bismuth (Bi):
- Mass of protons = Number of protons × Mass of one proton
= 83 × 1.007825 u = 83.64 u
- Mass of neutrons = Number of neutrons × Mass of one neutron
= 126 × 1.008665 u = 127.09 u
- Total mass = Mass of protons + Mass of neutrons
= 83.64 u + 127.09 u = 210.73179 u

### Step 3: Calculate the Mass Defect
For Iron (Fe):
- Mass defect (Δm) = Total mass - Actual mass
= 56.4634 u - 55.934939 u = 0.528461 u

For Bismuth (Bi):
- Mass defect (Δm) = Total mass - Actual mass
= 210.73179 u - 208.980388 u = 1.75179 u

### Step 4: Calculate the Binding Energy
The binding energy (BE) can be calculated using the formula:
\[ \text{BE} = \Delta m \times 931.5 \text{ MeV} \]

For Iron (Fe):
\[ \text{BE} = 0.528461 \times 931.5 \text{ MeV} = 492.26 \text{ MeV} \]

For Bismuth (Bi):
\[ \text{BE} = 1.75179 \times 931.5 \text{ MeV} = 1631.7 \text{ MeV} \]

### Step 5: Calculate the Binding Energy per Nucleon
For Iron (Fe):
\[ \text{BE per nucleon} = \frac{\text{BE}}{A} = \frac{492.26 \text{ MeV}}{56} \approx 8.79 \text{ MeV} \]

For Bismuth (Bi):
\[ \text{BE per nucleon} = \frac{\text{BE}}{A} = \frac{1631.7 \text{ MeV}}{209} \approx 7.8 \text{ MeV} \]

### Final Answer
The binding energy per nucleon for Iron (Fe) is approximately **8.79 MeV** and for Bismuth (Bi) is approximately **7.8 MeV**.

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