Nuclear Energy Calculation: Lithium-7 to Helium-4

This problem involves calculating the energy released during a nuclear reaction, specifically the bombardment of Lithium-7 (${}^7_3\text{Li}$) with a proton (${}^1_1\text{p}$), leading to the formation of Helium-4 (${}^4_2\text{He}$). The energy released is given in electron volts (eV).

1. Determine the Nuclear Reaction

Proton bombardment of Lithium-7 (${}^7_3\text{Li}$) typically results in the formation of Helium-4 (${}^4_2\text{He}$). To balance the reaction based on the provided particles, the most likely reaction is:

$ {}^7_3\text{Li} + {}^1_1\text{p} \rightarrow 2 \times {}^4_2\text{He} $

2. Calculate the Mass Defect ($\Delta m$)

The mass defect is the difference between the total mass of the reactants and the total mass of the products.

• Mass of Reactants = Mass(${}^7_3\text{Li}$) + Mass(${}^1_1\text{p}$) = $7.0183 \text{ u} + 1.008 \text{ u} = 8.0263 \text{ u}$
• Mass of Products = $2 \times$ Mass(${}^4_2\text{He}$) = $2 \times 4.004 \text{ u} = 8.008 \text{ u}$
• Mass Defect ($\Delta m$) = Mass of Reactants - Mass of Products = $8.0263 \text{ u} - 8.008 \text{ u} = 0.0183 \text{ u}$

3. Calculate Energy Released per Reaction

Use Einstein's mass-energy equivalence, $E = \Delta m c^2$. The conversion factor $1 \text{ u} = 931 \text{ MeV/}c^2$ is given.

• Energy per reaction ($E_{\text{reaction}}$) = $0.0183 \text{ u} \times 931 \text{ MeV/u}$
• $E_{\text{reaction}} = 17.0373 \text{ MeV}$
• Convert MeV to eV: $1 \text{ MeV} = 10^6 \text{ eV}$.
• $E_{\text{reaction}} = 17.0373 \times 10^6 \text{ eV}$

4. Calculate the Number of Lithium-7 Nuclei

The mass of ${}^7_3\text{Li}$ is given as $M = \frac{7}{17.13} \text{ kg}$. We need to find the number of ${}^7_3\text{Li}$ nuclei in this mass.

• Molar mass of ${}^7_3\text{Li}$ $\approx 7.0183 \text{ g/mol}$.
• Avogadro number ($N_A$) $= 6.0 \times 10^{23} \text{ mol}^{-1}$.
• Convert mass to grams: $M = \frac{7}{17.13} \text{ kg} = \frac{7000}{17.13} \text{ g}$.
• Number of moles = $\frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} = \frac{7000 / 17.13}{7.0183}$ mol.
• Number of nuclei ($N$) = moles $\times N_A$
• $N = \left( \frac{7000}{17.13 \times 7.0183} \right) \times (6.0 \times 10^{23})$
• $N \approx \left( \frac{7000}{120.17} \right) \times (6.0 \times 10^{23}) \approx 58.25 \times 6.0 \times 10^{23}$
• $N \approx 349.5 \times 10^{23} = 3.495 \times 10^{25}$ nuclei.

5. Calculate Total Energy Released

Total Energy = Number of nuclei $\times$ Energy per reaction

• Total Energy = $(3.495 \times 10^{25}) \times (17.0373 \times 10^6 \text{ eV})$
• Total Energy $\approx 59.54 \times 10^{31} \text{ eV}$
• Total Energy $\approx 5.954 \times 10^{32} \text{ eV}$

6. Determine the Value of $\alpha$

The total energy is given in the form $\alpha \times 10^{32} \text{ eV}$. By comparing our calculated value:

• $5.954 \times 10^{32} \text{ eV} = \alpha \times 10^{32} \text{ eV}$
• Therefore, $\alpha = 5.954$.

The question asks for the nearest integer value.

• The nearest integer to $5.954$ is $6$.