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Flashcards

Nuclear Physics

Kerala Board · Class 12 · Physics

Flashcards for Nuclear Physics — Kerala Board Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.

45 questions24 flashcards5 concepts
24 Flashcards
Card 1Atomic Nucleus

What is the atomic number (Z) and mass number (A) of a nucleus? How are they related to the number of protons and neutrons?

Answer

Atomic number (Z) = Number of protons in the nucleus = Number of electrons in a neutral atom Mass number (A) = Total number of nucleons = Number of protons + Number of neutrons Number of neutrons (N)

Card 2Atomic Nucleus

Define isotopes, isobars, and isotones with examples.

Answer

Isotopes: Same Z, different A (same element, different neutrons) Example: ¹²₆C and ¹⁴₆C Isobars: Same A, different Z (different elements, same mass number) Example: ⁴⁰₁₈Ar and ⁴⁰₂₀Ca Isotones: Same

Card 3Nuclear Size

What is the formula for nuclear radius and calculate the radius of ¹⁶₈O nucleus?

Answer

Nuclear radius formula: R = r₀A¹/³ where r₀ = 1.2 × 10⁻¹⁵ m (unit nuclear radius) For ¹⁶₈O: A = 16 R = 1.2 × 10⁻¹⁵ × (16)¹/³ R = 1.2 × 10⁻¹⁵ × 2.52 R = 3.02 × 10⁻¹⁵ m = 3.02 fm

Card 4Nuclear Mass

What is unified atomic mass unit (u)? Express the masses of proton and neutron in u.

Answer

1u = (1/12) × mass of ¹²₆C atom = 1.66 × 10⁻²⁷ kg = 931.3 MeV/c² Mass of proton (mₚ) = 1.00727 u = 938.3 MeV/c² Mass of neutron (mₙ) = 1.00865 u = 939.6 MeV/c² Mass of electron (mₑ) = 0.000549 u = 0.

Card 5Mass-Energy Relations

Define mass defect and binding energy. How are they related?

Answer

Mass defect (Δm): Difference between sum of masses of individual nucleons and actual nuclear mass Δm = [Z·mₚ + (A-Z)·mₙ] - M Binding Energy (BE): Energy equivalent of mass defect BE = Δm × c² = Δm ×

Card 6Mass-Energy Relations

Calculate the binding energy per nucleon for ⁷₃Li nucleus. Given: mass of ⁷₃Li = 6.015 u, mₚ = 1.00727 u, mₙ = 1.00865 u

Answer

Given: M(⁷₃Li) = 6.015 u, Z = 3, A = 7, N = 4 Mass of constituents: 3 protons = 3 × 1.00727 = 3.02181 u 4 neutrons = 4 × 1.00865 = 4.0346 u Total = 7.05641 u Mass defect = 7.05641 - 6.015 = 1.04141

Card 7Nuclear Forces

What are the characteristic properties of nuclear forces?

Answer

1. Short range: Effective only up to ~10⁻¹⁵ m 2. Charge independent: Same for p-p, p-n, and n-n interactions 3. Strongly attractive: Much stronger than electromagnetic forces 4. Saturated: Each nucleo

Card 8Radioactivity

What is radioactivity? Who discovered it and how?

Answer

Radioactivity: Spontaneous disintegration of unstable nuclei with emission of α, β, or γ radiations to achieve stability. Discovered by: A.H. Becquerel (1896) Discovery: Accidentally found that urani

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Frequently Asked Questions

What are the important topics in Nuclear Physics for Kerala Board Class 12 Physics?

Nuclear Physics covers several key topics that are frequently asked in Kerala Board Class 12 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.

Start by understanding all key concepts. Practise previous year questions from this chapter. Revise formulas and definitions regularly. Use flashcards for quick revision before the exam.

There are 24 flashcards for Nuclear Physics covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.