Energy Released in Proton Bombardment of Lithium

The reaction for the conversion of $_3^7\text{Li}$ into $_2^4\text{He}$ by proton bombardment is given by: $_1^1\text{H} + _3^7\text{Li} \rightarrow _2^4\text{He} + _2^4\text{He}$

Step 1: Calculate the mass defect ($\Delta m$) for a single reaction. $\Delta m = \left[ m(_1^1\text{H}) + m(_3^7\text{Li}) \right] - 2 \cdot m(_2^4\text{He})$

Substitute the given values of masses: $\Delta m = (1.008 + 7.0183) - 2 \times 4.004$ $\Delta m = 8.0263 - 8.008 = 0.0183\text{ u}$

Step 2: Calculate the energy released ($Q$) per reaction. $Q = \Delta m \times 931\text{ MeV}$ $Q = 0.0183 \times 931\text{ MeV} = 17.0373\text{ MeV} = 17.0373 \times 10^6\text{ eV}$

Step 3: Calculate the total number of $_3^7\text{Li}$ nuclei ($N$). The mass of Lithium provided is $m = \frac{7}{17.13}\text{ kg} = \frac{7000}{17.13}\text{ g}$. Since the molar mass of $_3^7\text{Li}$ is $7\text{ g/mol}$:

$N = \frac{m}{M} \times N_A$ $N = \frac{\frac{7000}{17.13}}{7} \times 6.0 \times 10^{23}$ $N = \frac{1000}{17.13} \times 6.0 \times 10^{23} = \frac{6.0 \times 10^{26}}{17.13}\text{ nuclei}$

Step 4: Calculate the total energy released ($E$). $E = N \times Q$ $E = \left(\frac{6.0 \times 10^{26}}{17.13}\right) \times \left(17.0373 \times 10^6\text{ eV}\right)$ $E = \frac{6.0 \times 17.0373}{17.13} \times 10^{32}\text{ eV}$ $E = \frac{102.2238}{17.13} \times 10^{32}\text{ eV} \approx 5.9675 \times 10^{32}\text{ eV}$

Comparing with $E = \alpha \times 10^{32}\text{ eV}$: $\alpha \approx 5.9675$

Rounding off to the nearest integer, we get: $\alpha = 6$