Obtain the binding energy of the nuclei `._26Fe^(56)` and `._83Bi^(209)` in units of MeV form the following data: `m(._26Fe^(56))=55.934939a.m.u.` , `m=(._83Bi^(209))=208.980388 a m u`. Which nucleus has greater binding energy per nucleon? Take `1a.m.u 931.5MeV`

Explanation

To find the binding energy of the nuclei \( _{26}^{56}\text{Fe} \) and \( _{83}^{209}\text{Bi} \), we will follow these steps:

### Step 1: Determine the number of protons and neutrons
For \( _{26}^{56}\text{Fe} \):
- Number of protons (Z) = 26
- Number of neutrons (N) = A - Z = 56 - 26 = 30

For \( _{83}^{209}\text{Bi} \):
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