Question

If the binding energy Per nucleon in \({ }_{3}^{7} \mathrm{Li}\) and \({ }^{4}{ }_{2}\) He nuclear is \(5.6 \mathrm{MeV}\) and \(7.06 \mathrm{MeV}\) respectively, then in the reaction \(\mathrm{P}+{ }_{3} \mathrm{Li} \rightarrow 2\left({ }_{2}^{4} \mathrm{He}\right)\) (P here retrent Proton) energy of Proton must be (A) \(1.46 \mathrm{MeV}\) (B) \(39.2 \mathrm{MeV}\) (C) \(17.28 \mathrm{MeV}\) (D) \(28.24 \mathrm{MeV}\)

Short answer

The energy of the proton must be \(17.28 \mathrm{MeV}\). Thus, option (C) is the correct answer.

Step-by-step solution

Understand Conservation of Energy at a Nuclear Level

In a nuclear reaction, energy is conserved. This means that the total energy before the reaction should be equal to the total energy after the reaction. The binding energy given here is the energy required to disassemble a nucleus into individual nucleons. Let's call the binding energy per nucleon for lithium and helium as \(E_{Li}\) and \(E_{He}\) respectively and number of nucleons in helium as \(n_{He}\).

Determine Energy Before the Reaction

The energy before the reaction would be the sum of the total input energy and the total binding energy for lithium. The proton is our input energy and it's our unknown, so we will call that 'P'. The total binding energy for lithium is the product of binding energy per nucleon and the total number of nucleons. As Lithium-7 has 7 nucleons, this gives us a total initial energy of \( P + 7*E_{Li}\).

Identify Energy After the Reaction

Similarly, the total energy after the reaction is the total binding energy for both the helium nuclei, each helium having 4 nucleons, we get \( 2*4*E_{He}\).

Use Conservation of Energy to Determine the Proton's Energy

Let's equate the energy before and after the reaction. This gives: \( P + 7*E_{Li} = 2*4*E_{He}\). From this, we can isolate the unknown, P, our proton's energy: \( P = 2*4*E_{He} - 7*E_{Li}\).

Substitute the Binding Energies and Solve for P

Let's substitute \(E_{Li} = 5.6 \mathrm{MeV}\) and \(E_{He} = 7.06 \mathrm{MeV}\) into our equation to solve for P: \( P = 2*4*7.06 - 7*5.6 = 56.48 - 39.2 = 17.28 \mathrm{MeV}\). So, the energy of proton must be \(17.28 \mathrm{MeV}\). Thus, option (C) is the correct answer.

Key concepts

Binding Energy

The concept of binding energy is pivotal in understanding nuclear reactions. Binding energy is the energy required to completely separate all the nucleons (protons and neutrons) in a nucleus. It effectively measures the stability of the nucleus. The greater the binding energy, the more stable the nucleus. In nuclear physics, binding energy per nucleon is often used, which gives an idea of the average energy holding each nucleon in the nucleus.

For example,

• In the isotope \(_3^7 \text{Li}\), the binding energy per nucleon is 5.6 MeV.
• For \(^4_2 \text{He}\), it is 7.06 MeV.

These values indicate that helium is more stable than lithium. When performing calculations related to nuclear reactions, it is crucial to take binding energies into account to understand the energy changes involved.

Conservation of Energy

In nuclear reactions, much like in other physical systems, the principle of conservation of energy is a key concept. This principle states that energy cannot be created or destroyed in an isolated system, only transformed from one form to another. In the context of nuclear reactions, the total energy of the system before the reaction must equal the total energy after.

To apply conservation of energy in the reaction \((P + _3^7 \text{Li} \rightarrow 2(_2^4 \text{He}))\):

• Calculate the total energy before the reaction, which is the sum of the proton's energy and lithium's total binding energy.
• Calculate the total energy after the reaction, which is the binding energy of both helium nuclei.

By equating these two energies, we can solve for unknowns such as the energy of the proton required for the reaction.

Nucleons

Nucleons are the building blocks of the atomic nucleus, consisting of protons and neutrons. Understanding nucleons is fundamental when calculating and comprehending nuclear reactions. Each nucleon has a certain amount of binding energy, which contributes to the total stability of the nucleus.

For practical purposes:

• The nucleus of \(_3^7 \text{Li}\) consists of 7 nucleons.
• The nucleus of \(_2^4 \text{He}\) consists of 4 nucleons.

In nuclear physics problems, it’s essential to determine the number of nucleons involved, as this information helps in calculating the total binding energies and understanding more about the reactions and their outcomes.

Nuclear Physics

Nuclear physics is the field of physics studying atomic nuclei and their interactions. This field covers concepts like nuclear reactions, the forces at play within a nucleus, and the energy dynamics involved.

Key areas in nuclear physics include:

• Studying different isotopes and their nuclear properties.
• Understanding nuclear reactions like fission, fusion, and decay.
• Analyzing the forces that bind nucleons together within the nucleus.

Nuclear physics also explores how various nuclear processes can result in energy release or absorption, often leveraging binding energy calculations. This knowledge is not only critical for solving textbook problems but also for practical applications such as nuclear power generation.