Question

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

Hint:

Calculate the mass defect per reaction, find the energy released per reaction in MeV, then determine the total number of atoms in the given mass of Lithium to find the cumulative energy released.

Answer 1

Correct Answer: 6

The nuclear reaction for the proton bombardment of Lithium-7 to produce Helium-4 can be written as:
$${}^7_3\text{Li} + {}^1_1\text{H} \rightarrow 2 \times {}^4_2\text{He}$$
In this process, a proton (${}^1_1\text{H}$) strikes a lithium nucleus, resulting in the creation of two alpha particles (helium nuclei).

Step 1: Calculate the mass defect ($\Delta m$) of a single reaction.
The mass defect is the difference between the total mass of the reactants and the total mass of the products.
Mass of reactants $= \text{Mass of } {}^7_3\text{Li} + \text{Mass of proton} = 7.0183 \text{ u} + 1.008 \text{ u} = 8.0263 \text{ u}$.
Mass of products $= 2 \times \text{Mass of } {}^4_2\text{He} = 2 \times 4.004 \text{ u} = 8.008 \text{ u}$.
$$\Delta m = 8.0263 \text{ u} - 8.008 \text{ u} = 0.0183 \text{ u}$$

Step 2: Calculate the energy released per single reaction.
Using the conversion $1 \text{ u} = 931 \text{ MeV}$, the energy released ($E_{single}$) is:
$$E_{single} = 0.0183 \times 931 \text{ MeV} \approx 17.0373 \text{ MeV}$$

Step 3: Find the total number of Lithium atoms in the given mass.
The mass of the sample is $m = \frac{7}{17.13} \text{ kg} = \frac{7000}{17.13} \text{ g}$.
Molar mass of ${}^7_3\text{Li} = 7 \text{ g/mol}$.
Number of nuclei ($N$) $= \frac{\text{Mass}}{\text{Molar mass}} \times N_A = \frac{7000 / 17.13}{7} \times 6.0 \times 10^{23}$
$$N = \frac{1000}{17.13} \times 6.0 \times 10^{23} = \frac{6 \times 10^{26}}{17.13}$$

Step 4: Calculate the total energy released.
Total energy $E_{total} = N \times E_{single}$
$$E_{total} = \left( \frac{6 \times 10^{26}}{17.13} \right) \times 17.0373 \text{ MeV}$$
$$E_{total} \approx 6 \times 10^{26} \times \left( \frac{17.0373}{17.13} \right) \text{ MeV} \approx 5.967 \times 10^{26} \text{ MeV}$$
Convert to electron-volts (eV) where $1 \text{ MeV} = 10^6 \text{ eV}$:
$$E_{total} \approx 5.967 \times 10^{26} \times 10^6 \text{ eV} = 5.967 \times 10^{32} \text{ eV}$$
Comparing this with $\alpha \times 10^{32} \text{ eV}$, we get $\alpha \approx 5.967$.
Rounding to the nearest integer, $\alpha = 6$.