Problem 25 - A Complete Resource Book in Chemistry for JEE Main

Question

In the nuclear fusion \({ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \longrightarrow{ }_{2} \mathrm{He}^{4}\), the masses of \({ }_{1} \mathrm{H}^{2}\) and \({ }_{2} \mathrm{H}^{4}\) are \(2.014 \mathrm{~m}_{\mathrm{a}}\) and \(4.003 \mathrm{~m}_{\mathrm{u}}\) respectively. The energy released per atom of helium formed is ........ MeV (a) \(4.31\) (b) \(13.26\) (c) \(16.76\) (d) \(23.275\)

Step-by-Step Solution

Answer

Option (d) 23.275 MeV is correct.

Step 1: Understanding the Reaction

In the given nuclear fusion reaction, two Deuterium nuclei (\( {}_{1} \mathrm{H}^{2} \)) are fusing to form one Helium nucleus (\( {}_{2} \mathrm{He}^{4} \)). We need to calculate the energy released per atom of Helium produced.

Step 2: Calculating Total Initial Mass

The initial reactants are two Deuterium atoms. The mass of each \( {}_{1} \mathrm{H}^{2} \) is \( 2.014 \mathrm{~u} \), so the total mass of the reactants is: \[2 imes 2.014 = 4.028 \mathrm{~u}\]

Step 3: Identifying Final Mass

After the fusion, the product is one Helium nucleus, \( {}_{2} \mathrm{He}^{4} \), which has a mass of \( 4.003 \mathrm{~u} \).

Step 4: Calculating Mass Defect

The mass defect (\(\Delta m\)) is the difference between the initial total mass and the final total mass:\[\Delta m = 4.028 \mathrm{~u} - 4.003 \mathrm{~u} = 0.025 \mathrm{~u}\]

Step 5: Converting Mass Defect to Energy

The energy released is given by Einstein’s mass-energy equivalence principle: \[E = \Delta m \cdot c^2\]where the mass defect \(\Delta m\) is in atomic mass units and needs to be converted to energy using the conversion factor \(931.5 \mathrm{~MeV/u}\): \[E = 0.025 \mathrm{~u} \times 931.5 \mathrm{~MeV/u} = 23.2875 \mathrm{~MeV}\]

Step 6: Choosing the Closest Answer

The calculated energy release is approximately \(23.2875 \mathrm{~MeV}\). Comparing this with the given options:(a) \(4.31\), (b) \(13.26\), (c) \(16.76\), (d) \(23.275\)The closest is option (d) \(23.275 \mathrm{~MeV}\).

Key Concepts

Mass Defect

In nuclear fusion, the term "mass defect" explains a fascinating aspect of atomic interactions. When Deuterium nuclei combine to form a Helium nucleus, the total mass of the resulting Helium is less than the sum of the original Deuterium masses. This difference in mass is what we call mass defect.
When calculating the mass defect, it's about finding the difference between the initial total mass and the resultant mass of the new nucleus. In our example:

• Initial mass: Two Deuterium atoms at 2.014 u each, totaling 4.028 u.
• Final mass: One Helium nucleus at 4.003 u.
• Mass defect: 4.028 u - 4.003 u = 0.025 u.

Understanding mass defect helps us measure the mass converted into energy during fusion.

Einstein’s Mass-Energy Equivalence

Einstein's mass-energy equivalence principle is succinctly captured in the famous equation \[E = mc^2\]. This states that mass can be converted into energy, and vice versa. In the world of nuclear reactions, this principle is key to understanding how mass loss (or mass defect) translates to energy release.
When dealing with nuclear fusion, such as the fusion of Deuterium into Helium, the lost mass is converted into a significant amount of energy. This is calculated using the mass defect and converting it from atomic mass units into energy (MeV):

• Mass defect: 0.025 u
• Energy released: 0.025 u multiplied by conversion factor (931.5 MeV/u)
• Resulting in energy output of approximately 23.2875 MeV.

This energy is what powers stars and has the potential for clean energy on Earth.

Deuterium Nuclei

Deuterium is an isotope of hydrogen consisting of one proton and one neutron, hence sometimes referred to as heavy hydrogen. Its symbol is \({ }_{1} ext{H}^{2}\), denoting its atomic number (1 for hydrogen) and mass number (2 for the total number of protons and neutrons).
In nuclear fusion processes, Deuterium nuclei are an essential ingredient. They are relatively abundant, as hydrogen's most significant isotopic mixture in nature includes Deuterium. When two Deuterium nuclei fuse, they form a Helium nucleus.
Deuterium plays a critical role in nuclear fusion research due to:

• Its abundance in nature as it can be extracted from water.
• Its role in fusion reactions at achievable temperatures and pressures.

Understanding Deuterium's properties is essential for grasping the mechanics of nuclear fusion.

Helium Nucleus Formation

The formation of a Helium nucleus from two Deuterium nuclei during fusion is a process that releases an enormous amount of energy. This transformation involves an internal rearrangement of nuclear particles, where two protons and two neutrons come together to form a stable Helium nucleus, noted as \({ }_{2} ext{He}^{4}\).
During this fusion:

• The two separate nuclei (Deuterium) combine under high energy conditions found in stars.
• They form a Helium nucleus, which is more stable due to the strong nuclear force.
• This reaction releases energy due to the mass defect—the lighter Helium nucleus compares to its original ingredients (Deuterium), unleashing energy by Einstein’s equation \(E=mc^2\).

Helium nucleus formation is fundamental in processes fueling stars and serves as an ideal reaction model for potential future clean energy sources.