What is the binding energy per nucleon of `""^(262)Bh` ? The mass of the atom is `262.1231 u`.

Answer

To find the binding energy per nucleon of the isotope \(^{262}\text{Bh}\), we will follow these steps:

### Step 1: Identify the number of protons and neutrons
The atomic number of \(^{262}\text{Bh}\) is 107, which means it has 107 protons. The mass number is 262, which means the total number of nucleons (protons + neutrons) is 262.

To find the number of neutrons:
\[ \text{Number of neutrons} = \text{Mass number} - \text{Atomic number} = 262 - 107 = 155 \]

### Step 2: Calculate the total mass of protons and neutrons
We will use the following masses:
- Mass of a proton \(m_p = 1.007825 \, u\)
- Mass of a neutron \(m_n = 1.008665 \, u\)

Now, calculate the total mass of protons and neutrons:
\[ \text{Total mass} = (\text{Number of protons} \times m_p) + (\text{Number of neutrons} \times m_n) \]
\[ \text{Total mass} = (107 \times 1.007825 \, u) + (155 \times 1.008665 \, u) \]
Calculating this gives:
\[ \text{Total mass} = 107.836775 \, u + 156.344075 \, u = 264.180850 \, u \]

### Step 3: Calculate the mass defect
The mass defect \(\Delta m\) is given by:
\[ \Delta m = \text{Total mass of nucleons} - \text{Mass of the nucleus} \]
Given that the mass of the atom (nucleus) is \(262.1231 \, u\):
\[ \Delta m = 264.180850 \, u - 262.1231 \, u = 2.057750 \, u \]

### Step 4: Calculate the binding energy
The binding energy \(E_b\) can be calculated using the mass defect:
\[ E_b = \Delta m \cdot c^2 \]
Using the conversion factor \(1 \, u = 931.494 \, \text{MeV}/c^2\):
\[ E_b = 2.057750 \, u \cdot 931.494 \, \text{MeV}/c^2 \]
Calculating this gives:
\[ E_b \approx 1916.316 \, \text{MeV} \]

### Step 5: Calculate the binding energy per nucleon
The binding energy per nucleon \(E_b/n\) is given by:
\[ E_b/n = \frac{E_b}{\text{Number of nucleons}} = \frac{1916.316 \, \text{MeV}}{262} \]
Calculating this gives:
\[ E_b/n \approx 7.314 \, \text{MeV} \]

### Final Answer
The binding energy per nucleon of \(^{262}\text{Bh}\) is approximately \(7.314 \, \text{MeV}\).

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