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Nuclei

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Chapter Notes

Overview
This chapter explores the structure and properties of the atomic nucleus, including its composition, size, mass, and the forces that bind it together. It explains the phenomenon of radioactivity, the mass-energy equivalence principle, and the concepts of nuclear binding energy, fission, and fusion. The one big idea a student must walk away with is that the nucleus is a dense, positively charged core of protons and neutrons held together by a strong short-range nuclear force, and that transformations of nuclei (radioactive decay, fission, fusion) involve enormous energy changes governed by Einstein's E=mc².
Key Concepts (16)
Atomic mass unit (u)
1/12th of the mass of the carbon (¹²C) atom.
e.g. The mass of a carbon-12 atom is exactly 12 u, so 1 u = 1.660539 × 10⁻²⁷ kg.
Isotopes
Atomic species of the same element having the same atomic number (Z) but different mass numbers (A), i.e., same number of protons but different number of neutrons.
e.g. Hydrogen has three isotopes: protium (¹H), deuterium (²H), and tritium (³H).
Isobars
Nuclides with the same mass number (A) but different atomic number (Z).
e.g. ³₁H and ³₂He are isobars.
Isotones
Nuclides with the same neutron number (N) but different atomic number (Z).
e.g. ¹⁹⁸₈₀Hg and ¹⁹⁷₇₉Au are isotones.
Nucleon
A collective term for a proton or a neutron present in the nucleus.
e.g. The number of nucleons in an atom is its mass number A.
Mass defect (ΔM)
The difference between the total mass of the individual nucleons (protons and neutrons) and the actual mass of the nucleus.
e.g. For ¹⁶₈O, ΔM = 0.13691 u.
Binding energy (Eb)
The energy required to separate a nucleus into its individual protons and neutrons, given by Eb = ΔM c².
e.g. The binding energy of ¹⁶₈O is 127.5 MeV.
Binding energy per nucleon (Ebn)
The average energy per nucleon needed to separate a nucleus into its individual nucleons, calculated as Eb / A.
e.g. For iron (A=56), Ebn is about 8.75 MeV, the maximum on the binding energy curve.
Radioactivity
The phenomenon in which an unstable nucleus undergoes spontaneous decay by emitting α, β, or γ radiation.
e.g. Uranium-238 decays by α-emission to thorium-234.
α-decay
A type of radioactive decay in which a helium nucleus (⁴₂He) is emitted from the parent nucleus.
e.g. ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He.
β-decay
A type of radioactive decay in which an electron or a positron is emitted from the nucleus.
e.g. ¹⁴₆C → ¹⁴₇N + e⁻ + ν̄ₑ.
γ-decay
A type of radioactive decay in which high-energy photons (gamma rays) are emitted from an excited nucleus.
e.g. After α or β decay, the daughter nucleus may be in an excited state and emit a γ-ray photon to reach ground state.
Nuclear fission
A nuclear reaction in which a heavy nucleus (e.g., ²³⁵U) splits into two or more lighter nuclei, releasing a large amount of energy.
e.g. ¹n + ²³⁵₉₂U → ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + 3¹n + 200 MeV.
Nuclear fusion
A nuclear reaction in which two light nuclei combine to form a heavier nucleus, releasing energy.
e.g. ²₁H + ²₁H → ³₂He + n + 3.27 MeV.
Thermonuclear fusion
Fusion achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the Coulomb repulsion.
e.g. The proton-proton cycle in the sun is a thermonuclear fusion process.
Q-value
The energy released or absorbed in a nuclear reaction, given by Q = (sum of initial masses – sum of final masses) c².
e.g. For fission of ²³⁵U, Q ≈ 200 MeV.
Formulas & Equations (11)
1 u = 1/12 × mass of one ¹²C atom = 1.660539 × 10⁻²⁷ kg
Reads as: One atomic mass unit equals one-twelfth the mass of a carbon-12 atom, which is 1.660539 times ten to the minus twenty-seven kilograms.
Variables: u = atomic mass unit (dimensionless in u, but in kg: 1.660539 × 10⁻²⁷ kg)
Use when: Converting atomic masses from u to kg or to energy equivalent.
⚠ Students forget that 1 u = 931.5 MeV/c², not 931.5 MeV directly.
R = R₀ A¹/³
Reads as: The radius of a nucleus equals R₀ times the cube root of the mass number A.
Variables: R = nuclear radius (m); R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm; A = mass number (dimensionless)
Use when: Calculating nuclear radius or comparing sizes of different nuclei.
⚠ Students use R₀ = 1.2 × 10⁻¹⁵ m but forget to cube it when finding volume.
E = mc²
Reads as: Energy equals mass times the speed of light squared.
Variables: E = energy (J); m = mass (kg); c = speed of light in vacuum ≈ 3 × 10⁸ m/s
Use when: Converting mass defect into binding energy or calculating energy released in nuclear reactions.
⚠ Students forget to square c, or use c = 3 × 10⁸ but then get units wrong.
ΔM = [Z mₚ + (A – Z) mₙ] – M
Reads as: Mass defect equals the sum of the masses of Z protons and (A minus Z) neutrons minus the actual nuclear mass.
Variables: ΔM = mass defect (u or kg); Z = atomic number; mₚ = mass of proton (1.00727 u); A = mass number; mₙ = mass of neutron (1.00866 u); M = mass of nucleus (u or kg)
Use when: Calculating the mass defect of a nucleus.
⚠ Students forget to subtract the mass of electrons when using atomic masses instead of nuclear masses.
E_b = ΔM c²
Reads as: Binding energy equals mass defect times the speed of light squared.
Variables: E_b = binding energy (J or MeV); ΔM = mass defect (kg or u); c = speed of light (3 × 10⁸ m/s)
Use when: Finding the energy required to break a nucleus into its nucleons.
⚠ Students use ΔM in u and forget to multiply by 931.5 to get MeV.
E_bn = E_b / A
Reads as: Binding energy per nucleon equals total binding energy divided by mass number.
Variables: E_bn = binding energy per nucleon (MeV); E_b = total binding energy (MeV); A = mass number
Use when: Comparing the stability of different nuclei; higher E_bn means more stable.
⚠ Students confuse E_bn with E_b and think a larger nucleus always has more binding energy per nucleon.
1 u = 931.5 MeV/c²
Reads as: One atomic mass unit is equivalent to 931.5 megaelectronvolts divided by the speed of light squared.
Variables: u = atomic mass unit; MeV = megaelectronvolt; c = speed of light
Use when: Converting mass defect in u directly to energy in MeV.
⚠ Students write 1 u = 931.5 MeV, forgetting the /c², which is dimensionally incorrect.
Q = (sum of initial masses – sum of final masses) c²
Reads as: The Q-value of a nuclear reaction equals the difference between the total initial mass and total final mass, multiplied by c squared.
Variables: Q = energy released (MeV); masses in u or kg; c = speed of light
Use when: Determining if a nuclear reaction is exothermic (Q > 0) or endothermic (Q < 0).
⚠ Students forget to include the mass of neutrons or electrons when balancing the reaction.
¹n + ²³⁵₉₂U → ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + 3¹n
Reads as: A neutron plus uranium-235 yields barium-144, krypton-89, and three neutrons.
Variables: n = neutron; U = uranium; Ba = barium; Kr = krypton
Use when: Writing a typical fission reaction of uranium-235.
⚠ Students forget to balance the number of neutrons on both sides.
²₁H + ²₁H → ³₂He + n + 3.27 MeV
Reads as: Two deuterons fuse to form helium-3, a neutron, and release 3.27 MeV of energy.
Variables: ²₁H = deuteron; ³₂He = helium-3; n = neutron; MeV = megaelectronvolt
Use when: Writing a typical fusion reaction.
⚠ Students confuse this with the proton-proton cycle and forget the energy released.
4¹₁H → ⁴₂He + 2e⁺ + 2νₑ + 6γ + 26.7 MeV
Reads as: Four hydrogen atoms fuse to form helium-4, two positrons, two neutrinos, six gamma photons, and release 26.7 MeV.
Variables: ¹₁H = hydrogen nucleus (proton); ⁴₂He = helium-4; e⁺ = positron; νₑ = electron neutrino; γ = gamma photon
Use when: Describing the net result of the proton-proton cycle in the sun.
⚠ Students forget the positrons and neutrinos, thinking only helium and energy are produced.
How It Works
Rutherford's α-particle scattering experiment for nuclear size determination
  1. Step 1: α-particles of known kinetic energy (e.g., 5.5 MeV) are directed at a thin gold foil.
  2. Step 2: The distance of closest approach to the gold nucleus is calculated from Coulomb repulsion, found to be about 4.0 × 10⁻¹⁴ m.
  3. Step 3: Since scattering follows pure Coulomb repulsion, the actual nuclear radius must be less than this distance.
  4. Step 4: Using higher energy α-particles or electron scattering, the nuclear radius formula R = R₀ A¹/³ is established.
💡 The distance of closest approach gives an upper limit for nuclear size, not the exact radius.
Mass defect and binding energy calculation
  1. Step 1: Determine the number of protons (Z) and neutrons (N = A – Z) in the nucleus.
  2. Step 2: Calculate the total mass of individual nucleons: Z × mₚ + N × mₙ.
  3. Step 3: Subtract the actual nuclear mass M from this total to get the mass defect ΔM.
  4. Step 4: Convert ΔM to energy using E = ΔM c², or multiply ΔM (in u) by 931.5 to get binding energy in MeV.
  5. Step 5: Divide by A to get binding energy per nucleon.
💡 Always use nuclear mass, not atomic mass, or subtract the mass of electrons if using atomic mass.
Nuclear fission chain reaction
  1. Step 1: A slow neutron is absorbed by a ²³⁵U nucleus, forming ²³⁶U in an excited state.
  2. Step 2: The excited ²³⁶U splits into two intermediate-mass fragments (e.g., ¹⁴⁴Ba and ⁸⁹Kr) and releases 2-3 neutrons.
  3. Step 3: The released neutrons can then induce fission in other ²³⁵U nuclei, creating a chain reaction.
  4. Step 4: Each fission releases about 200 MeV of energy, which appears as kinetic energy of fragments and neutrons, eventually becoming heat.
💡 The chain reaction requires a critical mass of fissile material; otherwise, neutrons escape without causing further fission.
Proton-proton cycle in the sun
  1. Step 1: Two protons fuse to form a deuteron, a positron, and a neutrino, releasing 0.42 MeV.
  2. Step 2: The positron annihilates with an electron, producing two gamma photons (1.02 MeV).
  3. Step 3: A deuteron fuses with another proton to form ³₂He and a gamma photon (5.49 MeV).
  4. Step 4: Two ³₂He nuclei fuse to form ⁴₂He, two protons, and release 12.86 MeV.
  5. Step 5: Net effect: 4 protons combine to form ⁴₂He, 2 positrons, 2 neutrinos, and 26.7 MeV of energy.
💡 The cycle requires very high temperatures (≈1.5 × 10⁷ K) to overcome Coulomb repulsion between protons.
Exam Traps — Don't Make These Mistakes
⚠ Confusing atomic mass with nuclear mass
✗ Using the atomic mass of oxygen (15.99493 u) directly as the nuclear mass in mass defect calculation.
→ Atomic mass includes the mass of electrons. For ¹⁶₈O, the atomic mass includes 8 electrons. To get nuclear mass, subtract the mass of 8 electrons (8 × 0.00055 u).
✓ Nuclear mass = atomic mass – Z × mₑ. For ¹⁶₈O: 15.99493 u – 8 × 0.00055 u = 15.99053 u.
⚠ Forgetting the /c² in 1 u = 931.5 MeV/c²
✗ 1 u = 931.5 MeV
→ Mass and energy are different physical quantities. The correct equivalence is 1 u c² = 931.5 MeV, so 1 u = 931.5 MeV/c². Writing 1 u = 931.5 MeV is dimensionally incorrect.
✓ 1 u = 931.5 MeV/c², meaning the energy equivalent of 1 u is 931.5 MeV.
⚠ Mixing up isotopes, isobars, and isotones
✗ Isobars have the same number of neutrons.
→ Isobars have the same mass number A (total nucleons), not the same neutron number. Isotones have the same neutron number N.
✓ Isotopes: same Z, different N. Isobars: same A, different Z. Isotones: same N, different Z.
⚠ Assuming binding energy per nucleon is constant for all nuclei
✗ Binding energy per nucleon is about 8 MeV for all nuclei.
→ The binding energy per nucleon is nearly constant (≈8 MeV) only for nuclei in the mass range 30 < A < 170. It is lower for very light (A < 30) and very heavy (A > 170) nuclei.
✓ Binding energy per nucleon is maximum around A = 56 (≈8.75 MeV), decreases for lighter and heavier nuclei.
⚠ Thinking fission and fusion are the same process
✗ Both fission and fusion involve splitting of nuclei.
→ Fission is the splitting of a heavy nucleus into lighter ones, while fusion is the combining of light nuclei into a heavier one. Both release energy because the products have higher binding energy per nucleon.
✓ Fission: heavy nucleus splits into lighter fragments (e.g., ²³⁵U). Fusion: light nuclei combine to form a heavier nucleus (e.g., ²H + ²H → ³He).
⚠ Neglecting the role of neutrons in fission chain reaction
✗ Fission of uranium produces energy and two fragments.
→ Fission also releases 2-3 neutrons per event, which are essential for sustaining a chain reaction. Without mentioning neutrons, the explanation is incomplete.
✓ Fission of ²³⁵U produces two intermediate-mass fragments and 2-3 neutrons, which can induce further fission, leading to a chain reaction.
Compare & Contrast
Definition based on atomic number (Z) and mass number (A)
Isotopes
Isobars
Isotopes have same Z but different A; isobars have same A but different Z.
Type of nuclear reaction
Nuclear fission
Nuclear fusion
Fission splits a heavy nucleus into lighter ones; fusion combines light nuclei into a heavier one. Both release energy because the products have higher binding energy per nucleon.
Particle emitted
α-decay
β-decay
α-decay emits a helium nucleus (⁴₂He); β-decay emits an electron (or positron) and a neutrino (or antineutrino).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total 11 marks
Most asked: Mass defect and binding energy calculations · Nuclear radius formula R = R₀ A¹/³ · Q-value of nuclear reactions · Fission and fusion reactions with examples · Properties of α, β, γ radiation
Don't skip: Numerical problems on binding energy per nucleon using mass defect; Writing and balancing nuclear fission/fusion reactions; Explanation of the binding energy curve and its implications
One-Line Revision (read aloud before the exam)
  1. The nucleus contains protons and neutrons, collectively called nucleons.
  2. Atomic mass unit (u) is 1/12th the mass of a ¹²C atom, equal to 1.660539 × 10⁻²⁷ kg.
  3. Isotopes have same Z, different N; isobars have same A, different Z; isotones have same N, different Z.
  4. Nuclear radius R = R₀ A¹/³, with R₀ = 1.2 fm.
  5. Nuclear density is constant at about 2.3 × 10¹⁷ kg/m³, independent of A.
  6. Mass defect ΔM = [Z mₚ + (A – Z) mₙ] – M.
  7. Binding energy E_b = ΔM c², and 1 u = 931.5 MeV/c².
  8. Binding energy per nucleon is maximum for iron (A=56) at 8.75 MeV.
  9. Nuclear force is short-range, charge-independent, and much stronger than Coulomb force.
  10. Radioactivity includes α-decay (⁴₂He), β-decay (electron/positron), and γ-decay (high-energy photon).
  11. Nuclear fission splits a heavy nucleus (e.g., ²³⁵U) into lighter fragments, releasing ~200 MeV.
  12. Nuclear fusion combines light nuclei (e.g., ²H + ²H) into heavier ones, releasing energy.
  13. The proton-proton cycle in the sun converts 4 protons into ⁴₂He, releasing 26.7 MeV.
  14. Thermonuclear fusion requires temperatures ~10⁸ K to overcome Coulomb barrier.
  15. Q-value = (initial mass – final mass) c²; positive Q means exothermic reaction.

MCQs — 10 Questions with Answers

Q1. What is the atomic mass unit (u) defined as, and what is its value in kilograms?

  • A. 1/12 of the mass of ¹²C atom = 1.66054 × 10⁻²⁷ kg ✓
  • B. 1/16 of the mass of ¹⁶O atom = 1.66054 × 10⁻²⁷ kg
  • C. Mass of a proton = 1.67262 × 10⁻²⁷ kg
  • D. 1/2 of the mass of a neutron = 5.24 × 10⁻²⁸ kg

Answer: A — The atomic mass unit is defined as exactly 1/12 of the ¹²C atom mass, which equals 1.66054 × 10⁻²⁷ kg by international convention.

Q2. The nuclei of deuterium (²₁H) and tritium (³₁H) are isotopes of hydrogen. What conclusion can be drawn about their internal structure?

  • A. Both contain one proton each; deuterium has one neutron and tritium has two neutrons ✓
  • B. Both contain different numbers of protons; deuterium has 1 and tritium has 2
  • C. Both contain only protons and no neutrons; their mass difference is due to electrons
  • D. Both contain equal numbers of protons and neutrons

Answer: A — Isotopes of the same element have identical Z (protons); mass differences arise from different N (neutrons). Since ¹H, ²H, ³H are isotopes, they all have Z=1, so ²H has N=1 and ³H has N=2.

Q3. If the atomic mass of chlorine is 35.46 u and it consists of two isotopes ³⁵Cl (34.98 u) and ³⁷Cl (36.98 u), what is the approximate relative abundance of ³⁵Cl?

  • A. 75.4% ✓
  • B. 50%
  • C. 24.6%
  • D. 60%

Answer: A — Using weighted average: 35.46 = (x/100) × 34.98 + ((100-x)/100) × 36.98, solving gives x ≈ 75.4%, which matches the study material value.

Q4. Which statement about the nucleus is INCORRECT?

  • A. The nucleus contains more than 99.9% of the atom's mass
  • B. The nuclear radius is approximately 10⁴ times smaller than the atomic radius
  • C. A free neutron is stable and does not decay ✓
  • D. The nucleus contains protons and neutrons bound by the nuclear force

Answer: C — A free neutron is unstable with a half-life of about 1000 s; it decays into a proton, electron, and antineutrino, though it remains stable inside the nucleus.

Q5. In the notation ¹⁹⁷₇₉Au, what do the superscript 197 and subscript 79 represent, and how many neutrons are in this nucleus?

  • A. 197 is mass number A, 79 is atomic number Z; neutron number N = 197 − 79 = 118 ✓
  • B. 197 is proton number, 79 is neutron number; total neutrons = 197 + 79 = 276
  • C. 197 is atomic number Z, 79 is mass number A; neutrons cannot be calculated
  • D. 197 is electron count, 79 is proton count; neutrons = 79

Answer: A — In ᴬₖX notation, A (superscript) is mass number = protons + neutrons, Z (subscript) is atomic number = protons; therefore N = A − Z = 197 − 79 = 118 neutrons.

Q6. Chadwick discovered the neutron by bombarding beryllium with alpha-particles and observing neutral radiation. What experimental evidence ruled out photons as the neutral radiation?

  • A. Photon energy required from conservation of momentum and energy was much higher than available from the bombardment ✓
  • B. Photons cannot knock protons out of nuclei due to lack of charge
  • C. The neutral radiation had mass, while photons are massless
  • D. Beryllium nuclei do not emit photons when bombarded with alpha-particles

Answer: A — The study material states that applying conservation of energy and momentum showed photons would need far higher energy than available from the ¹⁰Be + α reaction, proving the radiation was massive neutrons instead.

Q7. The mass of a proton is 1.00727 u and the mass of a ¹H atom is 1.00783 u. What does this mass difference imply?

  • A. The mass difference equals the electron mass (≈0.00055 u), showing the atom includes one electron outside the nucleus ✓
  • B. There is a mass defect due to binding energy in the nucleus
  • C. The proton contains an electron bound inside it
  • D. The difference is measurement error and should be ignored

Answer: A — Atomic mass = nuclear mass + electron mass; the difference 1.00783 − 1.00727 = 0.00056 u equals the electron mass, confirming one electron orbits the ¹H nucleus.

Q8. Two nuclides ³⁵Cl and ³⁷Cl are examples of _____, while ³H and ³₂He are examples of _____.

  • A. isotopes; isobars ✓
  • B. isobars; isotopes
  • C. isotones; isobars
  • D. isotopes; isotones

Answer: A — Isotopes have same Z but different A (³⁵Cl and ³⁷Cl both Z=17, different N); isobars have same A but different Z (³H has Z=1, ³₂He has Z=2, both A=3).

Q9. If a nucleus contains 26 protons and 30 neutrons, what is its mass number A and atomic number Z, and identify the element.

  • A. A = 56, Z = 26; the element is Iron (Fe) ✓
  • B. A = 56, Z = 30; the element is Zinc (Zn)
  • C. A = 30, Z = 26; the element is Iron (Fe)
  • D. A = 26, Z = 56; the element is Barium (Ba)

Answer: A — A = Z + N = 26 + 30 = 56 (mass number); Z = 26 (atomic number = proton count), which corresponds to Iron (Fe), notated as ⁵⁶₂₆Fe.

Q10. Assertion (A): A neutron is stable when inside a nucleus but decays when free. Reason (R): The nuclear force binds neutrons and protons together inside the nucleus, stabilizing the neutron. Select the correct option:

  • A. Both A and R are true, and R is the correct explanation of A ✓
  • B. Both A and R are true, but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false and R is false

Answer: A — Assertion is correct: free neutrons decay in ~1000 s into p + e⁻ + ν̄ₑ, but remain stable in nuclei. Reason is correct: the strong nuclear force stabilizes neutrons by binding them to the nucleus, making the decay energetically unfavorable.

Flashcards

What is an atomic mass unit (u) and what is its value in kilograms?

1 u is 1/12th of the mass of ¹²C atom, equal to 1.66054 × 10⁻²⁷ kg.

Define isotopes with one example from the study material.

Isotopes are atoms of the same element with same atomic number Z but different mass numbers A due to different neutron numbers; example: ¹H and ²H (deuterium).

What is the mass number A in nuclear notation and how is it calculated?

A is the total number of nucleons (protons + neutrons) in a nucleus, calculated as A = Z + N.

How did Chadwick's discovery of the neutron solve the mass discrepancy in hydrogen isotopes?

Chadwick proved deuterium and tritium contain protons plus neutral particles (neutrons) equal in mass to protons, explaining the 1:2:3 mass ratio.

State the relationship between nucleus radius and atom radius from the study material.

Nuclear radius is smaller than atomic radius by a factor of about 10⁴, making the nucleus about 10⁻¹² times the volume of the atom.

What do the symbols Z, N, and A represent in nuclear notation ᴬₖX?

Z is atomic number (proton count), N is neutron number, and A is mass number (total nucleons); X is the chemical symbol.

Why is a free neutron unstable but stable inside the nucleus?

A free neutron decays into a proton, electron, and antineutrino with a half-life of ~1000 s, but the nuclear force inside the nucleus keeps it stable.

Calculate the number of neutrons in ¹⁹⁷₇₉Au using the study material symbols.

N = A − Z = 197 − 79 = 118 neutrons in the gold nucleus.

What is the mass of a proton in atomic mass units and in kilograms?

Proton mass is 1.00727 u or 1.67262 × 10⁻²⁷ kg.

Distinguish between isobars and isotones with examples.

Isobars have the same mass number A but different Z (e.g., ³₁H and ³₂He); isotones have the same neutron number N but different Z (e.g., ¹⁹⁸₈₀Hg and ¹⁹⁷₇₉Au).

Important Board Questions

Define atomic mass unit (u) and write the expression relating it to kilogram. What is the mass of one proton in u? [2 marks]

State that 1 u = 1/12 of ¹²C mass, give the SI value (1.66054 × 10⁻²⁷ kg), and recall proton mass = 1.00727 u from the textbook definition and values provided.

Explain with an example how the concept of isotopes accounts for the non-integral atomic masses of elements. Show the calculation for chlorine (masses 34.98 u and 36.98 u with abundances 75.4% and 24.6%) to arrive at 35.46 u. [5 marks]

Define isotopes as atoms with same Z but different A (and N). Apply the weighted average formula: average mass = (fraction₁ × mass₁) + (fraction₂ × mass₂). Calculate: (0.754 × 34.98) + (0.246 × 36.98) = 35.46 u to show why atomic mass is not integral.

Derive the relationship between mass number A, atomic number Z, and neutron number N for any nucleus. Using this relationship, explain how Chadwick's discovery of the neutron solved the puzzle of why deuterium and tritium have masses in the ratio 1:2:3 despite being isotopes of hydrogen. Show all working. [6 marks]

State A = Z + N, so N = A − N. For hydrogen isotopes: ¹H (A=1, Z=1, N=0), ²H (A=2, Z=1, N=1), ³H (A=3, Z=1, N=2). Explain that before neutron discovery, the mass discrepancy was unexplained; the neutron's existence (mₙ ≈ mₚ) accounts for why N increases by 1 unit for each isotope, giving the observed 1:2:3 mass ratio. Chadwick proved this experimentally and won the Nobel Prize.

Assertion-Reason Questions for Nuclei

Q1. Assertion (A):

Isotopes of chlorine (³⁴Cl and ³⁶Cl) exhibit identical chemical properties despite different masses.

Reason (R):

Chemical properties depend on the number of electrons, which is determined by atomic number Z, and both isotopes have the same Z value.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why isotopes have identical chemical properties. The electronic structure (determined by Z) is identical for isotopes, so they occupy the same position in the periodic table and show identical chemical behaviour.

Q2. Assertion (A):

A free neutron decays spontaneously into a proton, electron, and antineutrino with a mean life of about 1000 seconds.

Reason (R):

Neutrons are composed of quarks and are less stable particles than protons in a free state.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—free neutrons do decay with mean life ~1000 s. The reason is also factually true (quarks, stability), but it does not explain WHY free neutrons decay. The reason provides particle composition but doesn't address the energy or decay mechanism. This tests the distinction between 'what happens' and 'why it happens.'

Q3. Assertion (A):

The nucleus of gold-197 (¹⁹⁷₇₉Au) contains 118 neutrons.

Reason (R):

The mass number equals the sum of protons and neutrons, so N = A - Z = 197 - 79 = 118.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are correct, and the reason provides the exact calculation showing why 118 neutrons are present. The mass number formula A = Z + N is fundamental and correctly applied here. Students must practice this calculation repeatedly.

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Case-Based Questions (CBQ) for Nuclei

Case Study:

A nuclear physics research team in India is studying isotopic abundance patterns of chlorine samples collected from different geological regions. Sample A from a coastal region shows isotopic composition: ³⁴Cl at 75.4% abundance with mass 34.98 u, and ³⁶Cl at 24.6% abundance with mass 36.98 u. Sample B from an inland region shows slight variations: ³⁴Cl at 74.8% abundance and ³⁶Cl at 25.2% abundance. The team aims to understand how geographical factors influence isotopic distribution and predict variations in average atomic mass between regions.

Q1. What is the average atomic mass of chlorine in Sample A?

  • A. 35.46 u ✓
  • B. 35.50 u
  • C. 35.98 u
  • D. 35.82 u

Answer: A

Q2. If Sample B's average atomic mass is calculated as (0.748 × 34.98) + (0.252 × 36.98), what is the result approximately?

  • A. 35.42 u
  • B. 35.48 u
  • C. 35.54 u ✓
  • D. 35.90 u

Answer: C

Q3. Both samples contain the same element (chlorine) with the same atomic number Z=17. Why do they show different average atomic masses?

  • A. Different numbers of protons in each sample
  • B. Different relative abundances of the two isotopes ✓
  • C. Different numbers of electrons in each sample
  • D. Geographical factors alter the nuclear charge

Answer: B

Q4. Which of the following is NOT true regarding these chlorine samples?

  • A. Both samples contain isotopes with the same atomic number
  • B. Both samples show identical chemical properties in reactions
  • C. The heavier isotope ³⁶Cl has more neutrons than ³⁴Cl
  • D. The average atomic mass can be used to predict isotopic composition ✓

Answer: D

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Previous Year Questions (PYQ) for Nuclei

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 55/1/1 · 1 mark

The ‘distance of closest approach’ of an alpha-particle is ‘d’ when it moves with a velocity v head-on towards the target nucleus. If the velocity of alpha particle is halved, the new ‘distance of closest approach’ will be –

  • d/2
  • 2d
  • d/4
  • 4d

CBSE 2026 · Set 55/1/1 · 1 mark

Assertion (A): The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R): Energy is absorbed when the nucleons are bound together to form a nucleus.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Both Assertion (A) and Reason (R) are false.

CBSE 2026 · Set 55/1/1 · 2 marks

Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

CBSE 2026 · Set 55/1/2 · 1 mark

The distance-of-closest-approach for an alpha particle is 'd' when it moves head-on with speed v towards a target nucleus. If the alpha particle is replaced by a proton moving with the same speed, the new distance-of-closest-approach will be

  • 2d
  • d
  • d/2
  • d/√2

CBSE 2026 · Set 55/1/2 · 2 marks

Draw a graph showing variation of binding energy per nucleon as a function of mass number A. The binding energy per nucleon for heavy nuclei (A > 170) decreases with increase in mass number. Explain its significance.

CBSE 2026 · Set 55/1/3 · 2 marks

Suppose a nucleus with mass number A = 240 and B.E./A = 7.6 MeV, breaks into two nuclei, each of mass number A = 120 with B.E./A = 8.5 MeV. Calculate the energy released in the process.

CBSE 2026 · Set 55/2/1 · 3 marks

(a) Consider the following nuclides: ¹²₆C, ¹⁹⁸₈₀Hg, ¹⁴₆C, ¹⁹⁷₇₉Au. Group them into isotopes and isotones. (b) How does the size of a nucleus depend on its mass number A? Hence prove that the density of nucleus is a constant, independent of A, for all nuclei.

CBSE 2026 · Set 55/2/3 · 3 marks

(a) Consider the following nuclides: ¹²₆C, ¹⁹⁸₈₀Hg, ¹⁴₆C, ¹⁹⁷₇₉Au. Group them into isotopes and isotones.

CBSE 2026 · Set 55(B)_Physics (Theory) · 1 mark

Which of the following particles do not exist in ₉₂U²³⁸ nucleus ?

  • 92 protons
  • 92 electrons
  • 146 neutrons
  • 238 nucleons

CBSE 2026 · Set 55(B)_Physics (Theory) · 1 mark

The main source for energy generation in the stars is –

  • fusion of lighter nuclei
  • fission of heavy nuclei
  • chemical reaction
  • fusion of heavy nuclei

CBSE 2026 · Set 55/3/1 · 1 mark

If r₁ and r₂ are the radii of atomic nuclei of mass numbers 64 and 27 respectively, then the value of (r₁/r₂) is:

  • 1
  • 4/3
  • 3/4
  • 27/64

CBSE 2026 · Set 55/3/1 · 1 mark

Assertion (A): Nuclear forces are always attractive. Reason (R): The nuclear force between protons and neutrons in a nucleus is a weak force.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Both Assertion (A) and Reason (R) are false.

See all 82 board PYQs for this chapter — open PYQ tab →

Class 12 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Physics Previous Year Questions, Chapter-wise

All 962 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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Class 12 Physics Derivations

All 68 derivations, chapter-wise, step by step with diagrams and marked with the CBSE board years (2022–2026) each was asked in. Free PDF download.

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