The energy released when $$\frac{7}{17.13}$$ kg of $$^7_3$$Li is converted into $$^4_2$$He by proton bombardment is $$\alpha \times 10^{32}$$ eV. The value of $$\alpha$$ is _______. (Nearest integer)
(Mass of $$^7_3$$Li = 7.0183 u, mass of $$^4_2$$He = 4.004 u, mass of proton = 1.008 u and 1 u = 931 MeV/c$$^2$$ and Avogadro number = $$6.0 \times 10^{23}$$)

The nuclear reaction for the proton bombardment of Lithium-7 is:

$$ ^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow 2 ^{4}_{2}\text{He} $$

$$ \Delta m = m(^{7}_{3}\text{Li}) + m(\text{proton}) - 2 \times m(^{4}_{2}\text{He}) $$

$$ \Delta m = 7.0183 \text{ u} + 1.008 \text{ u} - 2(4.004 \text{ u}) $$

$$ \Delta m = 8.0263 \text{ u} - 8.0080 \text{ u} $$

$$ \Delta m = 0.0183 \text{ u} $$

The energy released per reaction ($$ Q $$) is calculated using the given conversion factor $$ 1 \text{ u} = 931 \text{ MeV/c}^2 $$:

$$ Q = 0.0183 \times 931 \text{ MeV} $$

$$ Q = 17.0373 \text{ MeV} $$

$$ Q = 17.0373 \times 10^6 \text{ eV} $$

Next, calculate the number of Lithium atoms ($$ N $$) in the given sample mass.

The given mass is $$ m = \frac{7}{17.13} \text{ kg} = \frac{7000}{17.13} \text{ g} $$

The molar mass of $$ ^{7}_{3}\text{Li} $$ is approximately $$ 7 \text{ g/mol} $$.

Calculate the number of moles ($$ n $$):

$$ n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{7000 / 17.13}{7} $$

$$ n = \frac{1000}{17.13} \text{ mol} $$

Calculate the total number of atoms ($$ N $$) using Avogadro's number ($$ N_A = 6.0 \times 10^{23} $$):

$$ N = n \times N_A $$

$$ N = \left( \frac{1000}{17.13} \right) \times 6.0 \times 10^{23} $$

$$ N = \frac{6.0 \times 10^{26}}{17.13} \text{ atoms} $$

The total energy released ($$ E $$) is the total number of atoms multiplied by the energy released per reaction:

$$ E = N \times Q $$

$$ E = \left( \frac{6.0 \times 10^{26}}{17.13} \right) \times (17.0373 \times 10^6 \text{ eV}) $$

Rearranging the terms for simpler calculation:

$$ E = 6.0 \times \left( \frac{17.0373}{17.13} \right) \times 10^{32} \text{ eV} $$

Notice that the fraction is very close to 1 ($$ \frac{17.0373}{17.13} \approx 0.9946 $$):

$$ E \approx 6.0 \times 0.9946 \times 10^{32} \text{ eV} $$

$$ E \approx 5.967 \times 10^{32} \text{ eV} $$

Rounding to the nearest integer, we get:

$$ E = 6 \times 10^{32} \text{ eV} $$

Comparing this with the given format $$ \alpha \times 10^{32} \text{ eV} $$, we find the value of $$ \alpha $$:

$$ \alpha = 6 $$