Question

Compute the total binding energy and the binding energy per nucleon for Lithium-7.

Answer 1

Answer on Question #84474 Physics / Other

Compute the total binding energy and the binding energy per nucleon for Lithium-7.

Solution:

The total binding energy by definition

$$BE = \Delta m \cdot c^2 = \Delta m(u) \cdot 931.494 \text{ MeV}$$

where $\Delta m$ is a mass defect, that is given by relationship

$$\Delta m = Z m_p + (A - Z) m_n - M_{\text{nuclide}}$$

Since

$$m_p = 1.007825 \text{ u}, \quad m_n = 1.008665 \text{ u}$$

and for Lithium-7 ($A/Z = 3$ Li)

$$Z = 3, \quad A = 7, \quad M_{\text{nuclide}} = 7.016004 \text{ u}$$

we obtain

$$\begin{array}{l} \Delta m = 3 \times (1.007825 \text{ u}) + 4 \times (1.008665 \text{ u}) - 7.016004 \text{ u} \\ = 0.042131 \text{ u} \end{array}$$

So, the total nuclear binding energy for Lithium-7

$$BE = (0.042131 \text{ u}) \cdot 931.494 \text{ MeV} = 39.245 \text{ MeV}$$

The total nuclear binding energy per nucleon for Lithium-7

$$\frac{BE}{A} = \frac{39.245 \text{ MeV}}{7} = 5.606 \text{ MeV}$$

Answer: 39.245 MeV, 5.606 MeV

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