Question

Calculate the nuclear binding energy per nucleon for 205 Tl which has a nuclear mass of 204.974amu. nuclear binding energy per nucleon: J/ nucleon

Answer 1

The corrected nuclear binding energy per nucleon for \
`^(205)\text{Tl}\)` is approximately
`\(0.118 \, \text{{MeV/nucleon}}\)`.

The formula for nuclear binding energy is \( E = \text{{mass defect}} \times c^2 \), where \( c \) is the speed of light. The mass defect is the difference between the mass of the nucleus and the sum of the masses of its individual nucleons.

Given that the mass of
`\(^(205)\text{Tl}\)` is 204.974 amu, and let's assume the average mass of a nucleon is approximately 1 amu. The number of nucleons (A) in
`\(^(205)\text{Tl}\)` is equal to its mass number, which is 205.

The mass defect (Delta m) is calculated as follows:

`\[ \Delta m = (\text{{mass of individual nucleons}} * A) - \text{{mass of the nucleus}} \]`

`\[ \Delta m = (1 \, \text{{amu}} * 205) - 204.974 \, \text{{amu}} \]`

`\[ \Delta m = 205 - 204.974 \, \text{{amu}} \]`

`\[ \Delta m = 0.026 \, \text{{amu}} \]`

Now, convert the mass defect to energy using the conversion factor
`\(1 \, \text{{amu}} = 931 \, \text{{MeV}}\)`:

`\[ \text{{Binding Energy}} = \Delta m * 931 \, \text{{MeV/amu}} \]`

`\[ \text{{Binding Energy}} = 0.026 * 931 \, \text{{MeV}} \]`

`\[ \text{{Binding Energy}} = 24.126 \, \text{{MeV}} \]`

Finally, calculate the binding energy per nucleon:

`\[ \text{{Binding Energy per Nucleon}} = \frac{{\text{{Total Binding Energy}}}}{{\text{{Number of Nucleons}}}} \]`

`\[ \text{{Binding Energy per Nucleon}} = \frac{{24.126 \, \text{{MeV}}}}{{205}} \]`

`\[ \text{{Binding Energy per Nucleon}} \approx 0.118 \, \text{{MeV/nucleon}} \]`

So, the corrected nuclear binding energy per nucleon for \
`^(205)\text{Tl}\)` is approximately
`\(0.118 \, \text{{MeV/nucleon}}\)`.