If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxyge

Q: Why do we need 4 Helium nuclei?

To satisfy the conservation of mass number and atomic number. Oxygen has a mass number of 16 and an atomic number of 8. Since Helium has a mass number of 4 and an atomic number of 2, exactly 4 Helium nuclei are required ($4 \times 4 = 16$ and $4 \times 2 = 8$).

Q: What is mass defect?

Mass defect is the difference between the sum of the masses of the individual reactants and the actual mass of the product nucleus. This "missing" mass is converted into binding energy.

Q: Why do we multiply by 931.5?

The value $931.5 \text{ MeV}$ is the energy equivalent of $1 \text{ atomic mass unit (amu)}$. It is derived from Einstein's equation $E = mc^2$, converting mass in amu to energy in MeV.

Q: Is this reaction possible on Earth?

Currently, we cannot sustain this specific fusion reaction on Earth. It requires the extreme temperatures and pressures found only in the cores of stars to overcome the electrostatic repulsion between the nuclei.

Q: Why is energy released in this process?

Energy is released because the binding energy per nucleon of Oxygen-16 is higher than that of Helium-4. The system moves to a more stable, lower-energy state, releasing the excess energy.

Question

If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is :
(Mass of the helium nucleus is $4.0026	ext{ amu}$ and mass of oxygen nucleus is $15.9994	ext{ amu}$)

Options:

(A) $7.6	ext{ MeV}$

(B) $56.12	ext{ MeV}$

(C) $10.24	ext{ MeV}$

(D) $23.4	ext{ MeV}$

Sigma Percentile · Accepted Answer

### The Stellar Forge: Fusing Helium into Oxygen

Imagine you are deep inside the core of an aging star. The hydrogen fuel that once powered it has been exhausted, and the core has contracted, driving temperatures and pressures to unimaginable extremes. In this stellar forge, a new process begins: the fusion of heavier elements. Our problem asks us to calculate the energy released when a star converts helium nuclei entirely into oxygen nuclei.

This is not just a math problem; it is the very mechanism that creates the elements necessary for life. Let's break down the physics behind this cosmic event.

### Analyzing the Setup

The problem states that helium nuclei (${}_2^4	ext{He}$) are fusing to form oxygen nuclei (${}_8^{16}	ext{O}$). To balance the nucleons, we can easily see that it takes exactly four helium nuclei to create one oxygen nucleus:

$$ 4(_{2}^{4}	ext{He}) ightarrow \ _{8}^{16}	ext{O} + \Delta E $$

Here, $\Delta E$ represents the energy released in the process. But where does this energy come from? It comes from a tiny loss of mass during the fusion process, a phenomenon known as the **mass defect**.

### The Master Equation: Mass Defect

According to Einstein's mass-energy equivalence principle, $E = mc^2$, mass and energy are interchangeable. In a nuclear exothermic reaction, the total mass of the products is slightly less than the total mass of the reactants. This "missing" mass ($\Delta m$) is converted into the energy released ($\Delta E$).

The mass defect is calculated as:

$$ \Delta m = 	ext{Mass of Reactants} - 	ext{Mass of Products} $$

For our specific reaction, the reactants are four helium nuclei, and the product is one oxygen nucleus.

$$ \Delta m = 4 	imes m_{	ext{He}} - m_{	ext{O}} $$

### Final Calculation

Now, let's substitute the given values into our equation. The mass of a helium nucleus is $4.0026 	ext{ amu}$, and the mass of an oxygen nucleus is $15.9994 	ext{ amu}$.

$$ \Delta m = 4(4.0026 	ext{ amu}) - 15.9994 	ext{ amu} $$

Calculating the total mass of the reactants:

$$ 4 	imes 4.0026 = 16.0104 	ext{ amu} $$

Now, subtract the mass of the product:

$$ \Delta m = 16.0104 	ext{ amu} - 15.9994 	ext{ amu} = 0.011 	ext{ amu} $$

We have found the mass defect, but we need the energy in Mega electron-volts (MeV). We know that $1 	ext{ amu}$ of mass is equivalent to approximately $931.5 	ext{ MeV}$ of energy.

$$ \Delta E = \Delta m 	imes 931.5 	ext{ MeV/amu} $$
$$ \Delta E = 0.011 	imes 931.5 	ext{ MeV} $$
$$ \Delta E \approx 10.24 	ext{ MeV} $$

Thus, the energy released per oxygen nucleus formed is **$10.24 	ext{ MeV}$**. This immense release of energy is what counteracts the crushing force of gravity, keeping the star stable and shining brightly in the night sky.

Animated Solution for Physics - Atoms and Nuclei: If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is : (Mass of the helium nucleus is $4.0026 amu$ and mass of oxygen nucleus is $15.9994 amu$ )

Select Answer:

Visualized Solution

$Nuclear Fusion$

$Mass Defect$

$Substitution$

$Calculating Mass Defect$

$Energy Released$

$Food for Thought$

The Sigma Insight: Nucleus and Nuclear Reaction

The Stellar Forge

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

The binding energy per nucleon of deuteron ( $_{1}^{2} H$ ) and helium nucleus ( $_{2}^{4} He$ ) is $1.1 MeV$ and $7 MeV$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron $= 1.6725 \times 1 0^{- 27}$ kg, mass of proton $= 1.6725 \times 1 0^{- 27}$ kg, mass of electron $= 9 \times 1 0^{- 31}$ kg)

The binding energies per nucleon for deuteron ( $_{1} H^{2}$ ) and helium ( $_{2} He^{4}$ ) are $1.1 MeV$ and $7.0 MeV$ respectively. The energy released when two deuterons fuse to form a helium nucleus ( $_{2} He^{4}$ ) is ......... .

Consider the nuclear fission $^{20} Ne ⟶ 2^{4} He +^{12} C$ Given that the binding energy/nucleon of $^{20} Ne$ , $^{4} He$ and $^{12} C$ are respectively, $8.03 MeV$ , $7.07 MeV$ and $7.86 MeV$ , identify the correct statement.

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

If the binding energy per nucleon in $_{3}^{7} Li$ and $_{2}^{4} He$ nuclei are $5.60 MeV$ and $7.06 MeV$ respectively, then in the reaction $p +_{3}^{7} Li \to 2_{2}^{4} He$ energy of proton must be

Consider the reaction : $_{1}^{2} H +_{1}^{2} H =_{2}^{4} He + Q$ . Mass of the deuterium atom = 2.0141 u. Mass of helium atom = 4.0024 u. This is a nuclear ........ reaction in which the energy $Q$ released is ...... MeV.

In a reactor, $2 kg$ of $_{92} U^{235}$ fuel is fully used up in $30 days$ . The energy released per fission is $200 MeV$ . Given that, the Avogadro number, $N = 6.023 \times 1 0^{26} kmol^{- 1}$ and $1 eV = 1.6 \times 1 0^{- 19} J$ . The power output of the reactor is close to

If $M_{o}$ is the mass of an oxygen isotope $_{8} O^{17}$ , $M_{p}$ and $M_{n}$ are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nuclear Fusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass Defect

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Mass Defect

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Released

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought