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

To solve for the energy released, we start by analyzing the nuclear conversion of Lithium-7 into Helium-4 nuclei. The process involves one Lithium nucleus capturing one proton to split into two Helium nuclei.

1. First, we determine the energy released per reaction event using Einstein's mass-energy equivalence principle ($E = \Delta m c^2$).
The mass of the reactants (${}^7_3\text{Li} + {}^1\text{p}$) is $7.0183 + 1.008 = 8.0263 \text{ u}$.
The mass of the products ($2 \times {}^4_2\text{He}$) is $2 \times 4.004 = 8.008 \text{ u}$.
The difference, or mass defect ($\Delta m$), is $8.0263 - 8.008 = 0.0183 \text{ u}$.
The energy released per nucleus of Li is $0.0183 \times 931 \text{ MeV} = 17.0373 \text{ MeV}$.

2. Next, we find how many nuclei of Lithium are present in $\frac{7}{17.13} \text{ kg}$.
Mass of sample $= \frac{7000}{17.13} \text{ grams}$.
Number of atoms $= \text{moles} \times N_A = \left( \frac{7000 / 17.13}{7} \right) \times 6 \times 10^{23} = \frac{1000 \times 6 \times 10^{23}}{17.13} = \frac{6 \times 10^{26}}{17.13}$.

3. Finally, we calculate total energy in eV.
Total energy $= \text{Number of atoms} \times \text{Energy per atom}$
Total energy $= \frac{6 \times 10^{26}}{17.13} \times 17.0373 \times 10^6 \text{ eV}$
Total energy $= \frac{17.0373}{17.13} \times 6 \times 10^{32} \text{ eV}$
Since $\frac{17.0373}{17.13}$ is very close to $1$ (approx $0.9946$), the expression simplifies to:
Total energy $\approx 0.9946 \times 6 \times 10^{32} \text{ eV} \approx 5.967 \times 10^{32} \text{ eV}$.

Comparing with $\alpha \times 10^{32}$, the value of $\alpha$ is $6$.