I E Irodov Atomic And Nuclear Physics (Nuclear Reactions) Q. 6.270 Solution

The I E Irodov Atomic and Nuclear Physics chapter is one of the most important chapters of Advanced Physics for students preparing for JEE Advanced, Physics Olympiads, KVPY level exams and other competitive entrance exams.

Students learn how nuclear particles interact, how energy is released in nuclear reactions, how the laws of conservation of energy and momentum are applied to solve advanced numerical problems.

I E Irodov Atomic and Nuclear Physics (Nuclear Reactions) Question 6.270 on this page, concept, formula and relation used in the solution are also given. Detailed step by step solution for.

This is an advanced level numerical problem which involves multiple ideas of Modern Physics and students appearing for JEE Advanced are advised to understand the concepts involved before solving this problem.

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I E Irodov Nuclear Reactions Question 6.270

Question

Protons with kinetic energy

T = 1.0 MeV

strike a lithium target and induce the nuclear reaction

p + Li⁷ → 2He⁴

Find:

• The kinetic energy of each alpha particle.
• The angle of divergence between the two alpha particles, provided their directions of motion are symmetrical with respect to the incoming proton.

Concepts Used in I E Irodov Q. 6.270

This problem is based on advanced Nuclear Reactions.

Important concepts used include:

The two produced alpha particles are identical and move symmetrically. Thus, the two particles have same kinetic energy and same magnitude of momentum..

Easy Solution of I E Irodov Nuclear Reactions Q. 6.270

Given

Initial proton kinetic energy

Tp = 1.0 MeV

Reaction

p + Li⁷ → 2He⁴

Initially,

Lithium nucleus is at rest.

Let

• Kα = Kinetic Energy of each Alpha Particle
• pp = Momentum of Incident Proton
• pα = Momentum of Alpha Particle
• θ = Angle made by each alpha particle with proton direction
• φ = Angle between both alpha particles

We need to determine

• Kinetic energy of each alpha particle
• Divergence angle between the two alpha particles

Step 1: Calculate the Q-Value of the Nuclear Reaction

The energy released during the nuclear reaction is

Q = [Mass of Reactants − Mass of Products] c²

Using the atomic masses,

Hydrogen

1.007825 u

Lithium-7

7.016003 u

Helium-4

4.002603 u

Therefore,

Q ≈ 17.35 MeV

This means the nuclear reaction releases approximately 17.35 MeV of energy.

This format is clean, easy to copy into WordPress, and has the same structure as your Relativistic Mechanics Q. 1.388 page. In Part 2 I’ll continue with Step 2 via the FAQs in the same copy-friendly format.

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Step 2: Apply Conservation of Energy

The total kinetic energy following the nuclear reaction is the sum of the initial kinetic energy of the incident proton and the energy released in the nuclear reaction.

Therefore,

Total Final Kinetic Energy = Initial Proton Energy + Q-value

Substituting the values,

Total Final Kinetic Energy = 1.0 MeV + 17.35 MeV

Total Final Kinetic Energy = 18.35 MeV

Since two identical alpha particles are produced and move symmetrically, they share the total kinetic energy equally.

Therefore,

2Kα = 18.35 MeV

Hence,

Kα = 18.35 / 2

Kα ≈ 9.17 MeV

Thus, the kinetic energy of each alpha particle is approximately 9.17 MeV.

Step 3: Apply Conservation of Momentum

The lithium nucleus is initially at rest.

Therefore, the initial momentum of the system is only the momentum of the incoming proton.

Since both alpha particles move symmetrically, their momentum components perpendicular to the proton direction cancel each other.

Only the components along the original proton direction remain.

Hence,

Proton Momentum = 2 × Alpha Particle Momentum × cos θ

Using the momentum relation,

p = √(2mK)

we get

cos θ = pp / 2pα

Substituting

• Proton kinetic energy = 1.0 MeV
• Alpha particle kinetic energy = 9.17 MeV
• Alpha particle mass ≈ 4 × Proton Mass

we obtain

cos θ ≈ 0.0828

Therefore,

θ ≈ 85.25°

Step 4: Calculate the Divergence Angle

Each alpha particle moves at an angle θ with respect to the original proton direction.

Hence, the angle between the two alpha particles is

φ = 2θ

Therefore,

φ = 2 × 85.25°

φ ≈ 170.5°

Hence, the angle of divergence between the two alpha particles is

approximately 170.5° (≈171°).

Final Answer

Kinetic Energy of Each Alpha Particle

9.17 MeV

Angle of Divergence Between the Alpha Particles

170.5° (Approximately 171°)

Download JEE Advanced Revision Notes

After solving this question, students should revise the important concepts of Atomic and Nuclear Physics and practice more advanced numerical problems.

Simple Explanation of the Answer

The proton initially carries 1 MeV of kinetic energy.

In the nuclear reaction when it hits the lithium nucleus, another 17.35 MeV is released because a small amount of mass is converted to energy by Einstein’s mass-energy relation.

The total available energy therefore becomes

18.35 MeV

Since two identical alpha particles are produced and the reaction is perfectly symmetric, each alpha particle gets exactly half of the available kinetic energy.

The proton carries momentum , so the alpha particles can not be exactly opposite , but they are traveling in nearly opposite directions . The two particles move at a small angle to the original direction of the proton in an effort to conserve momentum. This gives a divergence angle of ~170.5°.

Key Takeaways

Common Mistakes Students Should Avoid

Students frequently make the following mistakes while solving this question:

• Ignoring the Q-value released during the nuclear reaction.
• Using only the proton’s kinetic energy to calculate the alpha particle energies.
• Forgetting that the lithium nucleus is initially at rest.
• Assuming the alpha particles move exactly opposite to each other.
• Using incorrect atomic masses while calculating the Q-value.
• Applying the momentum equation incorrectly.
• Forgetting that the divergence angle is 2θ, not θ.
• Mixing momentum formulas with energy equations.

Avoiding these mistakes helps students solve similar Nuclear Physics problems more accurately.

part3

Why Practice I E Irodov Problems for JEE Advanced?

I E Irodov problems help students develop a deeper understanding of advanced Physics concepts.

Regular practice can improve:

• Conceptual clarity
• Mathematical application
• Analytical thinking
• Multi-step problem solving
• Confidence in advanced numerical questions
• Preparation for JEE Advanced and Physics Olympiads

Students should attempt selected I E Irodov questions only after doing NCERT, class notes, coaching modules and previous year JEE questions.

Is I E Irodov Q. 6.270 a JEE Main Question?

No.

I E Irodov Q. 6.270 is more advanced than a typical JEE Main question because it combines:

• Nuclear reaction Q-value
• Conservation of energy
• Conservation of momentum
• Symmetrical particle emission
• Angular analysis of reaction products

This question is better suited for:

• JEE Advanced aspirants
• Physics Olympiad students
• Advanced Nuclear Physics practice
• Students strengthening Modern Physics concepts

Students for JEE Main should focus on NCERT, official syllabus topics, previous year questions and mock tests first.

Is Nuclear Physics Important for JEE?

Yes.

Nuclear Physics is an important part of Modern Physics and can be a scoring area when the concepts are clear.

Students should prepare the following topics carefully:

• Atomic nucleus
• Mass defect
• Binding energy
• Radioactivity
• Nuclear reactions
• Q-value
• Nuclear fission
• Nuclear fusion
• Alpha, beta and gamma decay
• Conservation laws in nuclear processes

Questions in Nuclear Physics are often formula-based, but more advanced questions may require careful use of conservation of energy and momentum.

Important Topics in Atomic and Nuclear Physics

How Should Students Use I E Irodov for JEE Advanced?

A recommended preparation order is:

Students should not try to solve every I E Irodov problem without understanding the underlying theory.

I E Irodov for NRI JEE Students

Selected I E Irodov questions can be useful for advanced Physics practice for NRI students preparing from:

• USA
• UAE
• Saudi Arabia
• Qatar
• Oman
• Kuwait
• Bahrain

Students of AP, IB, American or any other International boards should first match their school syllabus with JEE syllabus.

A suitable preparation approach is:

• Complete the JEE Physics syllabus
• Solve chapter-wise JEE questions
• Practice previous year papers
• Attempt JEE Advanced problems
• Use I E Irodov for selected advanced topics
• Revise errors regularly

Online JEE coaching can help NRI students in choosing the right I E Irodov questions instead of wasting time on problems out of their preparation needs.

Related I E Irodov Nuclear Reactions Questions

Students can also practice:

• I E Irodov Nuclear Reactions Q. 6.254
• I E Irodov Nuclear Reactions Q. 6.258
• I E Irodov Nuclear Reactions Q. 6.264
• I E Irodov Nuclear Reactions Q. 6.287

Link each question to its corresponding TestprepKart solution page.

Practice Next

After completing I E Irodov Q. 6.270, students can continue with:

• JEE Advanced Nuclear Physics revision notes
• Nuclear reaction practice questions
• JEE previous year question papers
• Atomic Physics revision
• Radioactivity problems
• Binding energy questions
• JEE Advanced mock tests

Useful JEE and NRI Resources

FAQs on I E Irodov Nuclear Reactions Q. 6.270

9.17 MeV

170.5°

17.35 MeV