You are given that mass of 3^7Li = 7.0160 u, mass of 2^4He = 4.0026 u and mass of 1^1H = 1.0079 u. W

Q: Why do we subtract the mass of products from reactants?

In an exothermic nuclear reaction, a tiny amount of mass is converted into energy. The reactants have slightly more mass than the products, and this difference (mass defect) is what becomes the released energy according to $E=mc^2$.

Q: Why did we multiply by 931?

The atomic mass unit (u) is a very small unit of mass. When 1 u of mass is completely converted into energy, it yields approximately 931 MeV. This is a standard conversion factor used in nuclear physics.

Q: How did we find the number of atoms in 20g of Lithium?

We used the mole concept. The number of moles is the given mass (20g) divided by the molar mass of Lithium-7 (7g/mol). Multiplying the number of moles by Avogadro's number ($6.023 \times 10^{23}$) gives the total number of atoms.

Q: Why convert MeV to Joules first?

Kilowatt-hour (kWh) is a macroscopic unit of energy based on Joules ($1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$). MeV is a microscopic unit. To bridge the gap, we must first convert MeV to the SI unit (Joules) using $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.

Q: What does 'proton capture' mean?

Proton capture is a type of nuclear reaction where a target nucleus absorbs a proton. Here, Lithium-7 absorbs a proton (Hydrogen-1 nucleus) to briefly form Beryllium-8, which immediately splits into two stable Helium-4 nuclei.

Question

You are given that mass of ${}_{3}^{7}	ext{Li} = 7.0160	ext{ u}$, mass of ${}_{2}^{4}	ext{He} = 4.0026	ext{ u}$ and mass of ${}_{1}^{1}	ext{H} = 1.0079	ext{ u}$. When $20	ext{ g}$ of ${}_{3}^{7}	ext{Li}$ is converted into ${}_{2}^{4}	ext{He}$ by proton capture, the energy liberated (in $	ext{kWh}$), is
[Mass of nucleon $= 1	ext{ GeV/c}^{2}$]

Options:

(A) $4.5 	imes 10^{5}$

(B) $8 	imes 10^{6}$

(C) $6.82 	imes 10^{5}$

(D) $1.33 	imes 10^{6}$

Sigma Percentile · Accepted Answer

### The Power of the Nucleus

Imagine holding just $20	ext{ g}$ of Lithium in your hand. It feels light, almost insignificant. But hidden within its atomic structure is a staggering amount of energy waiting to be unlocked. When a Lithium-7 nucleus captures a stray proton (a Hydrogen-1 nucleus), it undergoes a violent transformation. It briefly becomes highly unstable and immediately splits into two stable Helium-4 nuclei (alpha particles).

This process is a classic example of an exothermic nuclear reaction. The equation is elegantly simple:

$${}_{3}^{7}	ext{Li} + {}_{1}^{1}	ext{H} ightarrow 2 {}_{2}^{4}	ext{He} + Q$$

But where does this mysterious energy $Q$ come from? It comes from the very fabric of mass itself.

### Calculating the Mass Defect

In the quantum realm, mass and energy are two sides of the same coin, bound by Einstein's iconic $E=mc^2$. If we carefully weigh the ingredients before and after the reaction, we notice something profound: the products are slightly lighter than the reactants. This missing mass hasn't vanished; it has transformed into pure energy. We call this the **mass defect** ($\Delta m$).

Let's calculate it by subtracting the mass of the products from the reactants:

$$\Delta m = m({}_{3}^{7}	ext{Li}) + m({}_{1}^{1}	ext{H}) - 2m({}_{2}^{4}	ext{He})$$

Substituting the given atomic masses:

$$\Delta m = 7.0160 + 1.0079 - 2(4.0026)$$

$$\Delta m = 8.0239 - 8.0052 = 0.0187 	ext{ u}$$

### The Energy of a Single Reaction

Now that we know exactly how much mass is converted per reaction, we can find the energy. In nuclear physics, a standard conversion factor is used: $1	ext{ u}$ of mass yields approximately $931	ext{ MeV}$ of energy.

$$Q = \Delta m 	imes 931 	ext{ MeV}$$

$$Q = 0.0187 	imes 931 = 17.41 	ext{ MeV}$$

So, every single time a Lithium atom captures a proton, $17.41	ext{ MeV}$ of energy bursts forth.

### Scaling Up: The Mole Concept

But we don't just have one atom; we have $20	ext{ g}$ of Lithium. To find out how many atoms are participating in this grand energy release, we turn to the mole concept. The number of atoms $N$ is the given mass divided by the molar mass, multiplied by Avogadro's number ($N_A$).

$$N = \frac{20}{7} 	imes 6.023 	imes 10^{23}$$

$$N = 1.72 	imes 10^{24} 	ext{ atoms}$$

This is an unfathomably large number of reactions happening simultaneously!

### The Final Conversion: MeV to kWh

To find the total energy released, we multiply the energy of one reaction by the total number of atoms:

$$E_{	ext{total}} = (1.72 	imes 10^{24}) 	imes (17.41 	ext{ MeV})$$

However, the question asks for the answer in macroscopic, everyday units: **kilowatt-hours (kWh)**. To get there, we must first convert MeV to the standard SI unit of energy, Joules. We know that $1	ext{ MeV} = 10^6 	imes 1.6 	imes 10^{-19}	ext{ J}$.

$$E_{	ext{total}} = 1.72 	imes 10^{24} 	imes 17.41 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J}$$

$$E_{	ext{total}} = 4.79 	imes 10^{12} 	ext{ J}$$

Finally, we bridge the gap to kilowatt-hours. Since $1	ext{ kWh} = 3.6 	imes 10^6	ext{ J}$, we divide our total Joules by this factor:

$$E_{	ext{kWh}} = \frac{4.79 	imes 10^{12}}{3.6 	imes 10^6}$$

$$E_{	ext{kWh}} = 1.33 	imes 10^6 	ext{ kWh}$$

Just pause and think about that. A mere $20	ext{ g}$ of Lithium can produce over a million kilowatt-hours of energy—enough to power a small town! This is the breathtaking power of nuclear physics.

Select Answer:

Visualized Solution

\text{Nuclear Reaction}

\text{Mass Defect } (\Delta m)

\text{Substitution}

\text{Q-value Calculation}

\text{Number of Atoms } (N)

\text{Total Atoms}

\text{Total Energy Setup}

\text{Conversion to Joules}

\text{Conversion to kWh}

The Sigma Insight: Nucleus and Nuclear Reaction

The Power of the Nucleus

Calculating the Mass Defect

The Energy of a Single Reaction

Scaling Up

The Final Conversion

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