Nuclear Composition and Properties | Kumar Physics Classes

🔬 Searching for a Physics Tutor? Don't understand Nuclear Composition, Atomic Number, Mass Number, Isotopes, Isobars, Isotones, Nuclear Radius, Nuclear Density or Nuclear Force?

Contact Kumar Sir for one-to-one online Physics classes  |  📞 +91-9958461445  |  ✉️ kumarsirphysics@gmail.com  |  🌐 kumarphysicsclasses.com

N
Nuclei  •  Chapter Page 01

Nuclear
Composition
And Properties

Study composition of nucleus, proton, neutron, atomic number, mass number, isotopes, isobars, isotones, nuclear radius, nuclear density and nuclear force.

CBSE NEET JEE Main JEE Advanced IB Physics IGCSE ICSE A-Level
1Composition of Nucleus

Understanding what the nucleus is made of — the foundation of nuclear physics.

What is the Nucleus?

The nucleus is the tiny, dense, positively charged core at the centre of an atom. It was discovered by Ernest Rutherford through his famous gold foil experiment (1911).


Key facts about the nucleus:

  • The nucleus is composed of two types of particles: protons and neutrons.
  • Protons carry a positive electric charge (+e).
  • Neutrons are electrically neutral (charge = 0).
  • Together, protons and neutrons are called nucleons.
  • Nearly the entire mass of the atom is concentrated in the nucleus (>99.9%).
  • The nucleus occupies an extremely small fraction of the total atomic volume (radius ≈ 10⁻¹⁵ m vs atomic radius ≈ 10⁻¹⁰ m).
  • Electrons revolve around the nucleus in orbits (shells).
  • The nucleus is held together by the very strong nuclear force, which overcomes the electrostatic repulsion between protons.
Structure of an Atom p⁺ p⁺ n n e⁻ e⁻ e⁻ e⁻ e⁻ e⁻ e⁻ Nucleus Proton (p⁺) Neutron (n) Electron (e⁻) K-shell (1st) L-shell (2nd) M-shell (3rd) Proton Neutron Electron

Fig 1.1 — Structure of a typical atom showing nucleus (protons + neutrons) and electrons revolving in shells.

Key Comparison: Nucleus vs Atom
  • Atomic radius ≈ 10⁻¹⁰ m  |  Nuclear radius ≈ 10⁻¹⁵ m
  • Nucleus is about 10⁵ times smaller than the atom.
  • If the atom were enlarged to the size of a football stadium, the nucleus would be like a tiny pea at the centre.
2The Proton

The positively charged particle inside the nucleus — it defines the identity of an element.

Properties of the Proton

  • The proton is a positively charged particle found inside the nucleus.
  • It was discovered by Ernest Rutherford in 1919.
  • The number of protons in a nucleus is called the atomic number (Z).
  • Each element has a unique number of protons — changing the proton count changes the element.
  • Protons contribute to both the mass and the charge of the nucleus.
  • Protons experience electrostatic repulsion with each other, which is overcome by the nuclear force.
  • Symbol: p or 11H (hydrogen nucleus)
Charge of proton = +e = +1.6 × 10⁻¹⁹ C
Mass of proton (mₚ) ≈ 1.6726 × 10⁻²⁷ kg
mₚ ≈ 1.007276 u  (atomic mass units)
mₚ ≈ 938.3 MeV/c²
3The Neutron

The electrically neutral particle inside the nucleus — essential for nuclear stability.

Properties of the Neutron

  • The neutron is an electrically neutral particle found inside the nucleus.
  • It was discovered by James Chadwick in 1932.
  • Neutrons contribute to the mass number (A) but not to the atomic number (Z).
  • Neutrons help stabilise the nucleus by adding mass without increasing electrostatic repulsion.
  • Free neutrons are unstable and undergo beta decay with a half-life of about 10 minutes.
  • Inside a stable nucleus, neutrons remain stable.
  • Symbol: n or 10n
Charge of neutron = 0
Mass of neutron (mₙ) ≈ 1.6749 × 10⁻²⁷ kg
mₙ ≈ 1.008665 u
mₙ ≈ 939.6 MeV/c²
Note: The neutron is very slightly heavier than the proton.
mₙ − mₚ ≈ 2.3 × 10⁻³⁰ kg ≈ 1.29 MeV/c²
4Atomic Number (Z) and Mass Number (A)

The two fundamental numbers that describe every nucleus.

Nuclear Notation: ᴬ_Z X X A Z Mass Number Atomic Number Chemical Symbol A = Z + N Z = no. of protons N = no. of neutrons N = A − Z

Fig 4.1 — Standard nuclear notation showing mass number (A) and atomic number (Z).

Z = Atomic Number = Number of Protons
A = Mass Number = Number of Protons + Number of Neutrons
N = Number of Neutrons = A − Z
A = Z + N

Examples

NucleusSymbolZ (Protons)A (Mass No.)N = A−Z (Neutrons)Nucleons
Carbon-12126C61212−6 = 612
Oxygen-16168O81616−8 = 816
Uranium-23523592U92235235−92 = 143235
5Isotopes

Same element, different masses — isotopes share the same atomic number but differ in neutron count.

Definition

Isotopes are atoms of the same element that have the same atomic number (Z) but different mass numbers (A). They differ only in the number of neutrons in their nuclei.


Same: Z (atomic number), number of protons, chemical properties, position in periodic table
Different: A (mass number), N (neutrons), nuclear properties, radioactive behaviour

Examples of Isotopes

NucleusNameZAN = A−ZType
11HProtium110Isotopes of H
21HDeuterium121
31HTritium132
126CCarbon-126126Isotopes of C
136CCarbon-136137
146CCarbon-146148
23592UUranium-23592235143Isotopes of U
23892UUranium-23892238146
📌 Key Point: Isotopes have identical chemical properties because chemical behaviour depends on the number of electrons (= Z), which is the same for all isotopes of an element. However, their nuclear properties (stability, radioactivity) may differ significantly.
6Isobars

Same mass number, different elements — isobars are nuclei of different elements with equal A.

Definition

Isobars are atoms of different elements that have the same mass number (A) but different atomic numbers (Z).


Same: A (mass number), total number of nucleons
Different: Z (atomic number), N (neutrons), chemical element, chemical properties

Examples of Isobars

NucleusElementZAN = A−ZNote
4018ArArgon184022A = 40 (Isobars)
4020CaCalcium204020
146CCarbon6148A = 14 (Isobars)
147NNitrogen7147
31HTritium132A = 3 (Isobars)
32HeHelium-3231
7Isotones

Same neutron number, different elements — isotones share the same N = A − Z.

Definition

Isotones are nuclei that have the same number of neutrons (N) but different atomic numbers (Z) and different mass numbers (A). The condition is: N = A − Z = constant.

Examples of Isotones

NucleusAZN = A − ZConclusion
146C14614 − 6 = 8Isotones (N=8)
157N15715 − 7 = 8
3014Si301430 − 14 = 16Isotones (N=16)
3115P311531 − 15 = 16
3919K391939 − 19 = 20Isotones (N=20)
4020Ca402040 − 20 = 20
136C13613 − 6 = 7Isotones (N=7)
147N14714 − 7 = 7
📌 Quick Summary — Iso family: Isotopes → same Z | Isobars → same A | Isotones → same N
8Nuclear Radius

How big is the nucleus? The empirical formula relates nuclear radius to mass number.

Empirical Formula for Nuclear Radius

Experiments show that the nuclear radius depends on mass number A as:

R = R₀ A^(1/3)
where R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm (femtometre)
1 fm = 1 femtometre = 10⁻¹⁵ m

Key Implications

  • R ∝ A^(1/3) → nuclear radius is proportional to the cube root of mass number.
  • Larger nuclei (higher A) have larger radius, but the growth is slow (cube root).
  • Volume V = (4/3)πR³ = (4/3)πR₀³ A  →  V ∝ A (volume is directly proportional to A).
  • R vs A^(1/3) graph is a straight line through origin with slope R₀.

Solved Numericals — Nuclear Radius

📘 Example R-1: Find the radius of Aluminium nucleus (2713Al)
Find the nuclear radius of 2713Al. Given R₀ = 1.2 × 10⁻¹⁵ m.
▶ View Solution

Given: A = 27, R₀ = 1.2 × 10⁻¹⁵ m

Formula: R = R₀ A^(1/3)

R = 1.2 × 10⁻¹⁵ × (27)^(1/3)
(27)^(1/3) = ∛27 = 3
R = 1.2 × 10⁻¹⁵ × 3 = 3.6 × 10⁻¹⁵ m

R = 3.6 × 10⁻¹⁵ m = 3.6 fm

💡 Exam Tip: Remember cube roots of perfect cubes: ∛8=2, ∛27=3, ∛64=4, ∛125=5, ∛216=6.
📘 Example R-2: Find the radius of Tin nucleus (12550Sn)
Find the nuclear radius of 125Sn. Given R₀ = 1.2 fm.
▶ View Solution

Given: A = 125, R₀ = 1.2 × 10⁻¹⁵ m

R = 1.2 × 10⁻¹⁵ × (125)^(1/3)
(125)^(1/3) = ∛125 = 5
R = 1.2 × 10⁻¹⁵ × 5 = 6.0 × 10⁻¹⁵ m

R = 6.0 fm

💡 Exam Tip: ∛125 = 5. Always check if A is a perfect cube for quick calculation.
📘 Example R-3: Compare radii for A = 8 and A = 64
Compare the nuclear radii of two nuclei with A₁ = 8 and A₂ = 64.
▶ View Solution

Formula: R₁/R₂ = (A₁/A₂)^(1/3)

R₁/R₂ = (8/64)^(1/3) = (1/8)^(1/3) = 1/2
∴ R₂ = 2 R₁

The nucleus with A=64 has twice the radius of the nucleus with A=8.

Ratio of Volumes: V₁/V₂ = (R₁/R₂)³ = (1/2)³ = 1/8  →  V₂ = 8 V₁

💡 Exam Tip: When A ratio is 8:1, radius ratio is 2:1 and volume ratio is 8:1.
9Nuclear Density

One of the most remarkable results in nuclear physics — nuclear density is the same for ALL nuclei.

Derivation of Nuclear Density

Let mₙ = mass of a nucleon (approximately same for proton and neutron)


Step 1: Mass of nucleus

Mass of nucleus (M) ≈ A × mₙ

Step 2: Volume of nucleus

R = R₀ A^(1/3)
V = (4/3) π R³ = (4/3) π (R₀ A^(1/3))³ = (4/3) π R₀³ A

Step 3: Nuclear density

ρ = M / V = A mₙ / [(4/3) π R₀³ A]
The factor A cancels out!
ρ = 3 mₙ / (4π R₀³)
ρ = 3mₙ / (4πR₀³)
ρ ≈ 2.3 × 10¹⁷ kg m⁻³
ρ is INDEPENDENT of mass number A
🔑 Most Important Conclusion: Nuclear density is approximately constant for all nuclei regardless of their mass number A. This is because both the mass and the volume of the nucleus are proportional to A — so A cancels in the density formula. Nuclear density (≈ 2.3 × 10¹⁷ kg/m³) is about 10¹⁴ times greater than the density of ordinary matter!
Comparison: Density of water = 10³ kg/m³  |  Density of iron ≈ 7.8 × 10³ kg/m³  |  Nuclear density ≈ 2.3 × 10¹⁷ kg/m³
If a teaspoon of nuclear matter were brought to Earth, it would weigh about one billion tonnes.
10Radius Formula Applications

NEET/JEE style shortcuts and solved problems using R = R₀A^(1/3).

R = R₀ A^(1/3)   →   R ∝ A^(1/3)
Ratio of radii: R₁/R₂ = (A₁/A₂)^(1/3)
Ratio of volumes: V₁/V₂ = (R₁/R₂)³ = A₁/A₂
Ratio of surface areas: S₁/S₂ = (A₁/A₂)^(2/3)
Nuclear density: ρ = constant (independent of A)
📘 App-1: Ratio of nuclear radii of 64Cu and 27Al
Find the ratio of nuclear radii of copper (64Cu) and aluminium (27Al).
▶ View Solution
R_Cu / R_Al = (A_Cu / A_Al)^(1/3) = (64/27)^(1/3)
(64/27)^(1/3) = ∛64 / ∛27 = 4/3

R_Cu : R_Al = 4 : 3

💡 Shortcut: Always try to express the ratio as ratio of perfect cubes for quick calculation.
📘 App-2: Ratio of volumes of 125Te and 27Al
Find the ratio of volumes of nuclei 125Te (A=125) and 27Al (A=27).
▶ View Solution
V₁/V₂ = A₁/A₂ = 125/27

V(Te) : V(Al) = 125 : 27

Note: Volume ∝ A (direct proportion), not A^(1/3)!

💡 Volume ratio = mass number ratio (directly, not cube root).
📘 App-3: Compare nuclear size with atomic size
For a carbon atom (A=12), find the ratio of nuclear radius to atomic radius. Atomic radius of carbon ≈ 77 pm.
▶ View Solution
R_nucleus = 1.2 × 10⁻¹⁵ × (12)^(1/3) m
(12)^(1/3) ≈ 2.289
R_nucleus ≈ 1.2 × 2.289 × 10⁻¹⁵ ≈ 2.75 × 10⁻¹⁵ m
R_atom = 77 pm = 77 × 10⁻¹² m = 7.7 × 10⁻¹¹ m
Ratio = R_nucleus / R_atom = 2.75 × 10⁻¹⁵ / 7.7 × 10⁻¹¹ ≈ 3.6 × 10⁻⁵

Nuclear radius ≈ 1/28000 of atomic radius

💡 The nucleus is roughly 10⁴ to 10⁵ times smaller than the atom — this is why atoms are mostly empty space.
11Nuclear Force

The strongest fundamental force — holds the nucleus together against electrostatic repulsion.

What is Nuclear Force?

The nuclear force (also called the strong nuclear force) is the fundamental force that binds protons and neutrons together inside the nucleus. It acts between all nucleons (proton–proton, neutron–neutron, and proton–neutron pairs).


  • It is an extremely strong attractive force at internucleon distances of about 1–2 fm.
  • It has a very short range — effective only up to about 1–2 femtometres (fm). Beyond ~3 fm it becomes negligible.
  • At very small separations (< 0.5 fm), it becomes strongly repulsive — this prevents the nucleus from collapsing.
  • It is charge-independent — the nuclear force between p–p, n–n, and p–n pairs is approximately the same.
  • It has saturation property — a nucleon interacts only with its nearest neighbours, not with all nucleons.
  • It is spin-dependent and non-central in nature (unlike gravity or electromagnetic force).
Nuclear Force vs Internucleon Distance Internucleon Distance r (fm) Nuclear Force F 0 0.5 1.0 1.5 2.0 2.5 3.0 Repulsive (+) Attractive (−) r₀≈1fm Very Repulsive Force → 0 (beyond ~3 fm) Max Attractive force at ~1 fm

Fig 11.1 — Nuclear force (F) vs internucleon distance (r): strongly repulsive at very small r, attractive near 1 fm, negligible beyond 3 fm.

Range of Nuclear Force: The nuclear force is effective only up to 1 fm to 2 fm. At distances greater than about 3 fm, the nuclear force becomes completely negligible. This is why nuclear forces only affect nearest neighbours in a nucleus (saturation property).
12Properties of Nuclear Force

Eight key properties every physics student must know.

1. Strongest Force

Nuclear force is the strongest of all fundamental forces at short ranges. It is about 100 times stronger than the electromagnetic force and 10³⁸ times stronger than gravity at nuclear distances.

2. Short Range Force

It acts only within a very small range of about 1–2 fm. Beyond 3 fm it becomes effectively zero. This is why it does not affect electrons in their orbits.

3. Attractive at Normal Separation

At typical nucleon separations (~1 fm), the nuclear force is strongly attractive, holding the nucleus together against electrostatic repulsion between protons.

4. Repulsive at Very Small Separation

At distances less than about 0.5 fm, the nuclear force becomes strongly repulsive. This prevents the nucleus from collapsing under the attractive force.

5. Charge Independent

The nuclear force between p–p, n–n, and p–n pairs is approximately the same. It does not depend on the electric charge of the nucleons.

6. Saturation Property

Each nucleon interacts with only a limited number of nearest neighbours, not with all nucleons. This is why binding energy per nucleon remains roughly constant for medium and heavy nuclei.

7. Non-Central Nature

The nuclear force is not purely central. It depends on the orientation of nucleon spins relative to the line joining them, unlike gravitational or Coulomb forces.

8. Responsible for Stability

Nuclear force is responsible for the stability of the nucleus. Without it, the nucleus would fly apart due to electrostatic repulsion between protons.

Comparison: Nuclear Force vs Electrostatic Force vs Gravitational Force

PropertyNuclear ForceElectrostatic ForceGravitational Force
RangeShort (~2 fm)Long range (∞)Long range (∞)
StrengthStrongest (1)~1/100 of nuclearWeakest (~10⁻³⁸)
NatureAttractive + RepulsiveAttractive + RepulsiveAlways Attractive
Charge dependenceCharge-independentDepends on chargeIndependent of charge
Acts betweenNucleons onlyCharged particlesAll massive bodies
Carrier particlePions (π mesons)Photons (γ)Gravitons (theoretical)
SaturationYesNoNo
Non-central?YesNo (central)No (central)
13Important Graphs

These five graphs appear directly in CBSE, NEET and JEE exams — study each carefully.

Graph 1: R vs A^(1/3) A^(1/3) → R → slope = R₀ = 1.2 fm O

Straight line through origin. Slope = R₀ = 1.2 fm

Graph 2: Volume vs A Mass Number A → Volume V → V ∝ A (direct proportion) O

Straight line through origin. V = (4/3)πR₀³ × A

Graph 3: Density vs A Mass Number A → Density ρ → ρ = constant ≈ 2.3 × 10¹⁷ kg/m³ O

Constant horizontal line — nuclear density is independent of A.

Graph 4: Potential Energy vs r r (distance) → PE → 0 r₀ Repulsive (PE large +) Min PE PE→0

Nuclear potential energy vs distance — potential well at r₀ ≈ 1 fm.

Graph 5: Nuclear Force vs Distance — already shown in detail in Section 11 (Fig 11.1). Key takeaway: Repulsive at r < 0.5 fm, attractive maximum near 1 fm, negligible beyond 3 fm.
14Important Numericals — 30 Solved Problems

Each problem includes: Question | Given | Formula | Solution | Final Answer | Exam Tip

📘 Q1. Find Z, A, N for Chlorine-35
For 3517Cl, find the atomic number, mass number, and neutron number.
▶ Show Solution
Given: Z = 17, A = 35
Formula: N = A − Z
Z = 17 (protons)
A = 35 (nucleons)
N = 35 − 17 = 18 (neutrons)
Z = 17, A = 35, N = 18
💡 Always subtract Z from A to get N. Never reverse.
📘 Q2. Find nuclear radius of 64Zn
Calculate the nuclear radius of 64Zn. (R₀ = 1.2 fm)
▶ Show Solution
Given: A = 64, R₀ = 1.2 × 10⁻¹⁵ m
R = R₀ A^(1/3) = 1.2 × 10⁻¹⁵ × (64)^(1/3)
(64)^(1/3) = 4
R = 1.2 × 4 × 10⁻¹⁵ = 4.8 × 10⁻¹⁵ m
R = 4.8 fm
💡 ∛64 = 4. Memorise cube roots: 8→2, 27→3, 64→4, 125→5, 216→6.
📘 Q3. Ratio of radii of 8Be and 1H
Find the ratio of nuclear radii of 8Be (A=8) and 1H (A=1).
▶ Show Solution
R_Be / R_H = (8/1)^(1/3) = ∛8 = 2
R_Be : R_H = 2 : 1
💡 The nucleus of Beryllium-8 is twice the radius of the proton.
📘 Q4. Which are isotopes — identify the pair
Which of the following pairs are isotopes? (a) 14N and 14C   (b) 12C and 13C   (c) 39K and 40Ca
▶ Show Solution
(a) 14N: Z=7, A=14   14C: Z=6, A=14 → Same A, different Z → Isobars
(b) 12C: Z=6, A=12   13C: Z=6, A=13 → Same Z, different A → Isotopes ✓
(c) 39K: Z=19, N=20   40Ca: Z=20, N=20 → Same N → Isotones
Pair (b) are isotopes.
💡 Check Z first: same Z = isotopes. Same A = isobars. Same N = isotones.
📘 Q5. Nuclear density calculation
Calculate nuclear density given mₙ = 1.67 × 10⁻²⁷ kg and R₀ = 1.2 × 10⁻¹⁵ m.
▶ Show Solution
Formula: ρ = 3mₙ / (4πR₀³)
R₀³ = (1.2 × 10⁻¹⁵)³ = 1.728 × 10⁻⁴⁵ m³
4πR₀³ = 4 × 3.14159 × 1.728 × 10⁻⁴⁵ = 2.171 × 10⁻⁴⁴ m³
ρ = (3 × 1.67 × 10⁻²⁷) / (2.171 × 10⁻⁴⁴)
ρ = 5.01 × 10⁻²⁷ / 2.171 × 10⁻⁴⁴ = 2.31 × 10¹⁷ kg/m³
ρ ≈ 2.3 × 10¹⁷ kg/m³
💡 Nuclear density ≈ 2.3 × 10¹⁷ kg/m³ — memorise this value directly for MCQs.
📘 Q6. Identify isotones
From the following, identify isotones: 146C, 157N, 168O, 179F
▶ Show Solution
146C: N = 14−6 = 8
157N: N = 15−7 = 8
168O: N = 16−8 = 8
179F: N = 17−9 = 8
All four are isotones (N = 8 for each)!
💡 Multiple nuclei can be isotones simultaneously if they all have the same N.
📘 Q7. Ratio of nuclear volumes
The ratio of nuclear radii of two nuclei is 3:1. Find the ratio of their mass numbers.
▶ Show Solution
R₁/R₂ = (A₁/A₂)^(1/3) = 3/1
A₁/A₂ = (3)³ = 27
A₁ : A₂ = 27 : 1
💡 If radius ratio = k, then mass number ratio = k³.
📘 Q8. Neutron number from notation
A nucleus has A = 197 and Z = 79. Find the number of neutrons and identify the element.
▶ Show Solution
N = A − Z = 197 − 79 = 118
Z = 79 → Element is Gold (Au)
N = 118 neutrons. Element = Gold (19779Au)
💡 Gold-197 is the only stable isotope of gold. Common in nuclear physics problems.
📘 Q9. Radius of nucleus with A = 216
Find the nuclear radius of a nucleus with mass number 216. (R₀ = 1.2 fm)
▶ Show Solution
R = 1.2 × (216)^(1/3) = 1.2 × 6 = 7.2 × 10⁻¹⁵ m
∛216 = 6
R = 7.2 fm
💡 ∛216 = 6. Perfect cube — common in exam questions.
📘 Q10. Find nucleus from Z and N
A nucleus has 29 protons and 34 neutrons. Write its complete symbol.
▶ Show Solution
Z = 29 → Element is Copper (Cu)
A = Z + N = 29 + 34 = 63
6329Cu — Copper-63
💡 Always add Z + N to get A, then look up the element by Z.
📘 Q11. Density comparison of two nuclei
Compare the nuclear densities of 12C and 208Pb.
▶ Show Solution
ρ = 3mₙ / (4πR₀³) — independent of A
Therefore ρ(C) = ρ(Pb)
Nuclear densities are equal. ρ(¹²C) = ρ(²⁰⁸Pb) ≈ 2.3 × 10¹⁷ kg/m³
💡 Nuclear density is the same for ALL nuclei — this is one of the most important facts in nuclear physics.
📘 Q12. Surface area ratio of nuclei
Find the ratio of surface areas of nuclei with A₁ = 27 and A₂ = 125.
▶ Show Solution
Surface area S = 4πR² ∝ R² ∝ A^(2/3)
S₁/S₂ = (A₁/A₂)^(2/3) = (27/125)^(2/3)
(27/125)^(1/3) = 3/5, so (27/125)^(2/3) = (3/5)² = 9/25
S₁ : S₂ = 9 : 25
💡 Surface area ∝ A^(2/3). This is between R∝A^(1/3) and V∝A.
📘 Q13. How many isotopes of hydrogen exist?
List all isotopes of hydrogen and give their composition.
▶ Show Solution
Protium (1H): Z=1, N=0, A=1 → 1 proton, 0 neutrons (most abundant)
Deuterium (2H or D): Z=1, N=1, A=2 → 1 proton, 1 neutron (stable)
Tritium (3H or T): Z=1, N=2, A=3 → 1 proton, 2 neutrons (radioactive, t₁/₂ = 12.3 yr)
3 known isotopes of hydrogen: ¹H, ²H, ³H
💡 Hydrogen isotopes are so important they have individual names: Protium, Deuterium, Tritium.
📘 Q14. Nucleus volume if radius is doubled
If the nuclear radius is doubled, by what factor does the volume change?
▶ Show Solution
V = (4/3)πR³
If R → 2R, then V_new = (4/3)π(2R)³ = (4/3)π × 8R³ = 8V
Volume increases by a factor of 8
💡 Volume scales as R³. Double the radius → 8× the volume (cube relationship).
📘 Q15. Isobar identification
From the following, identify pairs of isobars: (i) 4018Ar, 4020Ca   (ii) 136C, 147N   (iii) 11H, 21H
▶ Show Solution
(i) A=40 for both, Z differs (18 vs 20) → Isobars ✓
(ii) A=13 vs A=14 → different A → Not isobars
(iii) A=1 vs A=2 → different A, same Z → Isotopes
Pair (i) are isobars.
💡 Isobars → same A but different Z (different elements).
📘 Q16. A nucleus with A = 1000, find radius
Estimate the radius of a nucleus with A = 1000. (R₀ = 1.2 fm)
▶ Show Solution
R = 1.2 × (1000)^(1/3) fm
(1000)^(1/3) = 10
R = 1.2 × 10 = 12 fm
R = 12 × 10⁻¹⁵ m = 12 fm
💡 ∛1000 = 10. Easy calculation! A=1000 gives R=10R₀.
📘 Q17. Nuclear force range
Two nucleons are 5 fm apart. Is there a significant nuclear force between them?
▶ Show Solution
Range of nuclear force ≈ 1–2 fm
At 5 fm distance, nuclear force becomes negligible (effectively zero)
At 5 fm, only electromagnetic force acts (if both are protons)
No. Nuclear force is negligible at 5 fm. Only electromagnetic (Coulomb) force acts.
💡 Beyond 3 fm ≈ nuclear force ≈ 0. This is a conceptual MCQ favourite.
📘 Q18. Number of nucleons in 1g of carbon
How many nucleons are present in 1 g of 12C? (Avogadro number Nₐ = 6.022 × 10²³ mol⁻¹)
▶ Show Solution
Molar mass of 12C = 12 g/mol
No. of atoms in 1g = (1/12) × 6.022 × 10²³ = 5.018 × 10²²
Each carbon nucleus has A = 12 nucleons
Total nucleons = 12 × 5.018 × 10²² = 6.022 × 10²³
Number of nucleons = 6.022 × 10²³ ≈ Nₐ
💡 For any element, nucleons in 1 g = Nₐ × A / (molar mass) = Nₐ (since molar mass ≈ A grams).
📘 Q19. Ratio of radii — NEET style
The ratio of volumes of two nuclei is 8:125. Find the ratio of their nuclear radii.
▶ Show Solution
V ∝ R³, so V₁/V₂ = R₁³/R₂³ = 8/125
R₁/R₂ = (8/125)^(1/3) = 2/5
R₁ : R₂ = 2 : 5
💡 Volume ratio = cube of radius ratio. Take cube root to get radius ratio.
📘 Q20. Nuclear force between p-p vs n-n
Is the nuclear force between two protons equal to the nuclear force between two neutrons at the same separation?
▶ Show Solution
Nuclear force is charge-independent
F(p-p) = F(n-n) = F(p-n) at the same separation
(Electrostatic repulsion between protons is additional and separate from nuclear force)
Yes. Nuclear force between p-p equals that between n-n at same distance.
💡 Nuclear force is charge-independent — a key property. The electrostatic repulsion between protons is SEPARATE from the nuclear force.
📘 Q21. Find A when radius is given
The nuclear radius of a nucleus is 3.6 fm. Find its mass number. (R₀ = 1.2 fm)
▶ Show Solution
R = R₀ A^(1/3) → A^(1/3) = R/R₀ = 3.6/1.2 = 3
A = 3³ = 27
A = 27 (this is Aluminium-27)
💡 A = (R/R₀)³. Cube the ratio to get A. Reverse of the normal formula.
📘 Q22. Isotones of 13C
Find two isotones of 136C.
▶ Show Solution
N for 13C = 13 − 6 = 7
Need nuclei with N = 7 but different Z
Z=7: A = Z+N = 7+7 = 14 → 147N
Z=8: A = 8+7 = 15 → 158O
Isotones of 13C: 147N and 158O
💡 To find isotones, keep N fixed and change Z (and A accordingly).
📘 Q23. Nuclear force vs electromagnetic force strength
At a distance of 1 fm, compare the order of magnitude of nuclear force vs electrostatic force between two protons.
▶ Show Solution
Nuclear force at 1 fm ≈ 10⁴ N (approximately)
Electrostatic force = kq²/r² = (9×10⁹ × (1.6×10⁻¹⁹)²) / (10⁻¹⁵)²
= (9×10⁹ × 2.56×10⁻³⁸) / 10⁻³⁰ = 230 N ≈ 10² N
Ratio = nuclear/electromagnetic ≈ 10⁴/10² = 10²
Nuclear force is ~100 times stronger than electrostatic force at 1 fm.
💡 Nuclear force strength / Electromagnetic force strength ≈ 100 at nuclear distances.
📘 Q24. 238U nucleus — find all quantities
For 23892U find: (a) number of protons (b) neutrons (c) nucleons (d) nuclear radius
▶ Show Solution
(a) Protons = Z = 92
(b) Neutrons = N = A−Z = 238−92 = 146
(c) Nucleons = A = 238
(d) R = 1.2 × (238)^(1/3) = 1.2 × 6.20 ≈ 7.44 fm
[(238)^(1/3) ≈ 6.20 using calculator]
Z=92, N=146, A=238, R≈7.44 fm
💡 Uranium-238 is the most abundant isotope of uranium (99.27% natural abundance).
📘 Q25. Mass of nucleus in terms of u
Write the approximate mass of a nucleus with A = 56 in atomic mass units (u). (mₙ ≈ 1.008 u)
▶ Show Solution
Mass ≈ A × mₙ ≈ 56 × 1.008 u = 56.448 u
(This is approximate — actual mass differs due to binding energy)
M ≈ 56.45 u (approximately 56 u)
💡 1 u = 1.66 × 10⁻²⁷ kg. Nuclear mass ≈ A × 1 u is a useful approximation.
📘 Q26. Ratio of densities of nucleus and water
Calculate the ratio of nuclear density to the density of water.
▶ Show Solution
ρ_nuclear ≈ 2.3 × 10¹⁷ kg/m³
ρ_water = 1000 kg/m³ = 10³ kg/m³
Ratio = 2.3 × 10¹⁷ / 10³ = 2.3 × 10¹⁴
Nuclear density is about 2.3 × 10¹⁴ times denser than water
💡 Nuclear matter is unimaginably dense — a sugar-cube-sized piece would weigh ~1 billion tons.
📘 Q27. Find isotopes from given data
Which of these are isotopes? A(Z=8,A=16), B(Z=8,A=17), C(Z=9,A=18), D(Z=8,A=18)
▶ Show Solution
A: Z=8 (Oxygen-16)   B: Z=8 (Oxygen-17)   C: Z=9 (Fluorine-18)   D: Z=8 (Oxygen-18)
Isotopes = same Z: A, B, D all have Z=8
A and C have same A=16? No. B and C have same A? No. C,D have same A=18 but different Z → Isobars
A, B, and D are isotopes (all are oxygen isotopes: ¹⁶O, ¹⁷O, ¹⁸O)
💡 Remember: Isotopes → same element (same Z), different A.
📘 Q28. Mass number from protons and neutrons
A nucleus has 50 protons and 69 neutrons. Identify the element and write its full symbol.
▶ Show Solution
Z = 50 → Element = Tin (Sn)
A = Z + N = 50 + 69 = 119
11950Sn — Tin-119
💡 Z=50 is Tin (Sn) — one of the magic numbers in nuclear physics!
📘 Q29. Two nuclei have same number of neutrons — which type?
Nuclei P (2311Na) and Q (2412Mg) — what is their relationship?
▶ Show Solution
N(Na) = 23 − 11 = 12
N(Mg) = 24 − 12 = 12
Same N=12, different Z (11 vs 12), different A (23 vs 24)
P and Q are Isotones (same N = 12)
💡 When both Z and A differ but N is same → Isotones.
📘 Q30. At what separation is nuclear force zero?
At what internucleon separation is nuclear force (a) maximum attractive (b) zero (changing from attractive to repulsive)?
▶ Show Solution
(a) Nuclear force is maximum attractive at r ≈ 1 fm
(b) The force changes from attractive to repulsive at r ≈ 0.8 fm
For r < 0.8 fm → repulsive; for r > 0.8 fm (up to ~3 fm) → attractive
For r > 3 fm → negligible
(a) Max attractive at ~1 fm   (b) Zero (equilibrium) at ~0.8 fm
💡 r ≈ 0.8 fm is the equilibrium distance where nuclear force = 0 (changes sign).
15Previous Year Questions (PYQs)

Exam-wise PYQs with complete solutions. Select your exam board below.

📚 CBSE Board Questions

CBSE 2023 | 1 MarkQ1. Define atomic number and mass number of a nucleus.
▶ Solution
Atomic number (Z): The number of protons in a nucleus. It determines the chemical identity of the element.
Mass number (A): The total number of nucleons (protons + neutrons) in a nucleus. A = Z + N.
CBSE 2023 | 2 MarksQ2. Define isotopes and give two examples with their neutron numbers.
▶ Solution
Isotopes are atoms of the same element having same Z but different A.
Examples: 1H (N=0), 2H (N=1), 3H (N=2) — all have Z=1.
12C (N=6), 14C (N=8) — both have Z=6.
CBSE 2022 | 2 MarksQ3. What are isobars? Give one example.
▶ Solution
Isobars are nuclei of different elements with same mass number A but different atomic number Z.
Example: 4018Ar and 4020Ca — both have A=40, but Z=18 and Z=20 respectively.
CBSE 2022 | 3 MarksQ4. State the formula for nuclear radius. Show that nuclear density is independent of mass number.
▶ Solution
R = R₀A^(1/3) where R₀ = 1.2 × 10⁻¹⁵ m.
Mass M ≈ Amₙ; Volume V = (4/3)πR³ = (4/3)πR₀³A
ρ = M/V = Amₙ/[(4/3)πR₀³A] = 3mₙ/(4πR₀³)
Since A cancels, ρ is independent of mass number A.
CBSE 2021 | 1 MarkQ5. Write the approximate value of nuclear density.
▶ Solution
Nuclear density ≈ 2.3 × 10¹⁷ kg m⁻³
CBSE 2021 | 2 MarksQ6. State two properties of nuclear force that distinguish it from the gravitational force.
▶ Solution
1. Range: Nuclear force is short range (~1–2 fm); gravitational force is long-range (infinite).
2. Nature: Nuclear force can be repulsive at very small separations; gravity is always attractive.
3. Strength: Nuclear force is far stronger at nuclear distances; gravity is the weakest force.
CBSE 2020 | 3 MarksQ7. The ratio of the nuclear radii of two nuclei is 2:3. Find the ratio of their (i) mass numbers (ii) nuclear densities.
▶ Solution
(i) R₁/R₂ = (A₁/A₂)^(1/3) = 2/3 → A₁/A₂ = (2/3)³ = 8/27
(ii) Nuclear density is independent of A, so ρ₁ = ρ₂ (ratio = 1:1)
CBSE 2020 | 1 MarkQ8. Name the force responsible for holding nucleons together in a nucleus.
▶ Solution
Nuclear force (also called the strong nuclear force).
CBSE 2019 | 2 MarksQ9. Distinguish between isotopes and isotones with one example each.
▶ Solution
Isotopes: Same Z, different A. Example: 12C and 14C (both Z=6).
Isotones: Same N, different Z and A. Example: 14C (N=8) and 15N (N=8).
CBSE 2019 | 3 MarksQ10. Calculate the radius of 27Al nucleus. Compare it with the radius of 64Cu. (R₀ = 1.2 fm)
▶ Solution
R(Al) = 1.2 × (27)^(1/3) = 1.2 × 3 = 3.6 fm
R(Cu) = 1.2 × (64)^(1/3) = 1.2 × 4 = 4.8 fm
R(Al)/R(Cu) = 3.6/4.8 = 3/4. So R(Cu) is 4/3 times R(Al).
CBSE 2018 | 2 MarksQ11. Why is the nuclear density of all elements nearly the same?
▶ Solution
Because nuclear volume is directly proportional to mass number A (V ∝ A), and nuclear mass is also ∝ A. So ρ = M/V = Amₙ/(kA) = mₙ/k = constant. Both mass and volume scale equally with A, so their ratio (density) is constant.
CBSE 2018 | 1 MarkQ12. What is the charge of a neutron?
▶ Solution
The charge of a neutron is zero (0 coulombs). It is electrically neutral.
CBSE 2017 | 2 MarksQ13. A nucleus ZAX has Z protons and N neutrons. Express N in terms of A and Z. If Z=11 and A=23, find N.
▶ Solution
N = A − Z
For Z=11, A=23: N = 23 − 11 = 12 neutrons (this is Sodium-23)
CBSE 2017 | 2 MarksQ14. State the saturation property of nuclear force.
▶ Solution
The saturation property states that each nucleon in a nucleus interacts with only a limited number of its nearest neighbours, not with all other nucleons. This is why binding energy per nucleon remains approximately constant for medium-mass nuclei, and why nuclear force does not grow with the total number of nucleons.
CBSE 2016 | 3 MarksQ15. Draw a plot of potential energy of a pair of nucleons as a function of their separation. Mark the equilibrium separation and explain the graph.
▶ Solution
The potential energy vs separation graph shows:
• For r < r₀ (≈0.8 fm): PE is positive and increases steeply → repulsive region
• At r = r₀ (≈0.8 fm): PE = 0 (equilibrium, force changes sign)
• For r slightly > r₀: PE becomes negative (minimum around 1 fm) → maximum attractive force
• For r > 3 fm: PE → 0 (force negligible)
The minimum of PE corresponds to the equilibrium separation of nucleons.
CBSE 2016 | 1 MarkQ16. What is the mass number of a nucleus with 8 protons and 8 neutrons?
▶ Solution
A = Z + N = 8 + 8 = 16. This is Oxygen-16 (16O).
CBSE 2015 | 2 MarksQ17. Two nuclei have mass numbers in the ratio 1:8. Find the ratio of their nuclear radii.
▶ Solution
R₁/R₂ = (A₁/A₂)^(1/3) = (1/8)^(1/3) = 1/2
∴ R₁ : R₂ = 1 : 2
CBSE 2015 | 2 MarksQ18. Give two differences between nuclear force and electrostatic force.
▶ Solution
1. Nuclear force is short range (~2 fm); electrostatic force is long range (infinite).
2. Nuclear force is charge-independent; electrostatic force depends on charges of particles.
3. Nuclear force is stronger at nuclear distances; electrostatic can be repulsive or attractive.
CBSE 2014 | 3 MarksQ19. What are isotones? Find two isotones of 146C.
▶ Solution
Isotones: nuclei with same N but different Z and A.
For 14C: N = 14−6 = 8
Isotones: 157N (N=8) and 168O (N=8)
CBSE 2014 | 2 MarksQ20. The nuclear radius of 8Be is 1.2 × 2^(1/3) fm. Using R = R₀A^(1/3), verify that R₀ ≈ 1.2 fm.
▶ Solution
For 8Be: A = 8
R = R₀ × (8)^(1/3) = R₀ × 2
Given R = 1.2 × 2^(1/3)? Wait — for A=8: R = 1.2 × ∛8 = 1.2 × 2 = 2.4 fm.
So R₀ = R / A^(1/3) = 2.4 / 2 = 1.2 fm ✓

🏥 NEET Questions (40)

NEET 2023Q1. The radius of a nucleus with A = 216 is: (R₀ = 1.2 fm)
(a) 3.6 fm   (b) 6.0 fm   (c) 7.2 fm   (d) 8.4 fm
▶ Solution
R = 1.2 × (216)^(1/3) = 1.2 × 6 = 7.2 fm
Answer: (c) 7.2 fm
NEET 2023Q2. Nuclear density is:
(a) More for heavier nuclei   (b) Less for heavier nuclei   (c) Same for all nuclei   (d) Zero for very light nuclei
▶ Solution
Nuclear density is independent of mass number A — it is the same for all nuclei.
Answer: (c) Same for all nuclei
NEET 2022Q3. Which of the following are isotones?
(a) 32S and 40Ar   (b) 14C and 14N   (c) 14N and 15O   (d) 12C and 14C
▶ Solution
(a) ³²S: N=16; ⁴⁰Ar: N=22 → No
(b) ¹⁴C: N=8; ¹⁴N: N=7 → Isobars, not isotones
(c) ¹⁴N: N=7; ¹⁵O: N=7 → Same N=7 → Isotones ✓
(d) ¹²C: N=6; ¹⁴C: N=8 → Isotopes, not isotones
Answer: (c)
NEET 2022Q4. The ratio of volume of two nuclei is 27:125. Find ratio of their radii.
(a) 3:5   (b) 5:3   (c) 9:25   (d) 27:125
▶ Solution
V∝R³ → R₁/R₂ = (V₁/V₂)^(1/3) = (27/125)^(1/3) = 3/5
Answer: (a) 3:5
NEET 2021Q5. The nuclear force:
(a) Obeys inverse square law   (b) Is charge dependent   (c) Is short range   (d) Is always repulsive
▶ Solution
Nuclear force is short range (effective only up to ~1–2 fm) and does not obey inverse square law.
Answer: (c)
NEET 2021Q6. 23892U has how many neutrons?
(a) 92   (b) 146   (c) 238   (d) 330
▶ Solution
N = A − Z = 238 − 92 = 146
Answer: (b) 146
NEET 2020Q7. Two nuclei have mass numbers A and 8A. The ratio of their nuclear radii is:
(a) 1:2   (b) 2:1   (c) 1:8   (d) 1:4
▶ Solution
R₁/R₂ = (A/8A)^(1/3) = (1/8)^(1/3) = 1/2
Answer: (a) 1:2
NEET 2020Q8. Which pair represents isobars?
(a) 1H, 2H   (b) 14C, 14N   (c) 12C, 13C   (d) 3H, 3He
▶ Solution
(b) ¹⁴C (Z=6, A=14) and ¹⁴N (Z=7, A=14): same A, different Z → Isobars
(d) ³H (Z=1, A=3) and ³He (Z=2, A=3): same A, different Z → Also isobars
Both b and d are isobars, but (d) is more commonly given as the standard answer.
Answer: (b) and (d) are both correct; typically (d) ³H and ³He
NEET 2019Q9. The binding force between nucleons in a nucleus is due to:
(a) Gravitational force   (b) Electrostatic force   (c) Nuclear force   (d) Magnetic force
▶ Solution
Answer: (c) Nuclear force
NEET 2019Q10. The mass number of a nucleus is 60 and its atomic number is 26. The number of neutrons is:
(a) 26   (b) 34   (c) 60   (d) 86
▶ Solution
N = A − Z = 60 − 26 = 34
Answer: (b) 34
NEET 2018Q11. Deuterium and tritium are:
(a) Isobars   (b) Isotones   (c) Isotopes   (d) None
▶ Solution
Deuterium (²H) and Tritium (³H) both have Z=1 but different A → Isotopes
Answer: (c) Isotopes
NEET 2018Q12. If the radius of a nucleus with A=8 is 2.4 fm, what is R₀?
(a) 1.0 fm   (b) 1.2 fm   (c) 1.4 fm   (d) 1.6 fm
▶ Solution
R = R₀A^(1/3) → R₀ = R/A^(1/3) = 2.4/(8)^(1/3) = 2.4/2 = 1.2 fm
Answer: (b) 1.2 fm
NEET 2017Q13. Nuclei of same mass number A but different Z are called:
(a) Isotopes   (b) Isobars   (c) Isotones   (d) Isomers
▶ Solution
Answer: (b) Isobars
NEET 2017Q14. Nuclear density is of the order of:
(a) 10³ kg/m³   (b) 10¹⁰ kg/m³   (c) 10¹⁷ kg/m³   (d) 10²⁴ kg/m³
▶ Solution
Answer: (c) 10¹⁷ kg/m³
NEET 2016Q15. The nuclear force between two neutrons compared with nuclear force between two protons (at same separation) is:
(a) Greater   (b) Smaller   (c) Equal   (d) Zero
▶ Solution
Nuclear force is charge-independent: F(n-n) = F(p-p)
Answer: (c) Equal
NEET 2016Q16. The mass of proton is approximately:
(a) 9.1 × 10⁻³¹ kg   (b) 1.67 × 10⁻²⁷ kg   (c) 1.67 × 10⁻³¹ kg   (d) 9.1 × 10⁻²⁷ kg
▶ Solution
Answer: (b) 1.67 × 10⁻²⁷ kg
NEET 2015Q17. The ratio of the nuclear radii of 27Al and 125Te is:
(a) 3:5   (b) 5:3   (c) 9:25   (d) 1:1
▶ Solution
R ∝ A^(1/3); ratio = (27/125)^(1/3) = 3/5
Answer: (a) 3:5
NEET 2015Q18. A nucleus with Z=6 and N=8 is:
(a) 14N   (b) 14C   (c) 12C   (d) 12B
▶ Solution
Z=6 → Carbon. A = Z+N = 6+8 = 14 → 14C
Answer: (b) 14C
NEET 2014Q19. The range of nuclear force is approximately:
(a) 10⁻¹⁰ m   (b) 10⁻¹⁵ m   (c) 10⁻² m   (d) 10⁻⁷ m
▶ Solution
Answer: (b) 10⁻¹⁵ m (femtometre range)
NEET 2014Q20. Isotopes have:
(a) Same A, different Z   (b) Same Z, different A   (c) Same N, different Z   (d) Same A and Z
▶ Solution
Answer: (b) Same Z, different A
NEET 2013Q21. If two nuclei have A₁ = 27 and A₂ = 64, the ratio V₁:V₂ is:
(a) 27:64   (b) 3:4   (c) 9:16   (d) 64:27
▶ Solution
V ∝ A → V₁/V₂ = A₁/A₂ = 27/64
Answer: (a) 27:64
NEET 2013Q22. At a separation less than 0.8 fm, nuclear force between two nucleons is:
(a) Attractive   (b) Repulsive   (c) Zero   (d) Infinite
▶ Solution
Answer: (b) Repulsive
NEET 2012Q23. The nuclear radius of 125Sn compared to 27Al is: (ratio R_Sn:R_Al)
(a) 3:5   (b) 5:3   (c) 25:9   (d) 9:25
▶ Solution
R_Sn/R_Al = (125/27)^(1/3) = 5/3
Answer: (b) 5:3
NEET 2012Q24. The charge of a neutron is:
(a) +e   (b) −e   (c) 0   (d) +2e
▶ Solution
Answer: (c) 0 (neutral)
NEET 2011Q25. The mass of a neutron is approximately:
(a) 1.67 × 10⁻²⁷ kg   (b) 9.1 × 10⁻³¹ kg   (c) 1.67 × 10⁻³¹ kg   (d) 3.34 × 10⁻²⁷ kg
▶ Solution
Answer: (a) 1.67 × 10⁻²⁷ kg (≈ 1.675 × 10⁻²⁷ kg)
NEET 2011Q26. Nuclear force is strongest of all forces at nuclear distances. Its strength compared to gravity is roughly:
(a) 10²   (b) 10²⁰   (c) 10³⁸   (d) 10⁵⁰
▶ Solution
Answer: (c) 10³⁸ (nuclear force is ~10³⁸ times stronger than gravity at nuclear distances)
NEET 2010Q27. A nucleus has N = A − Z. For 5626Fe, N is:
(a) 26   (b) 30   (c) 56   (d) 82
▶ Solution
N = 56 − 26 = 30
Answer: (b) 30
NEET 2010Q28. Isotones are nuclides with:
(a) Same mass number   (b) Same atomic number   (c) Same neutron number   (d) Same nucleon number
▶ Solution
Answer: (c) Same neutron number
NEET 2009Q29. The property of nuclear force not shared by electrostatic force is:
(a) Inverse square law   (b) Acts between neutral particles   (c) Can be attractive   (d) Long range
▶ Solution
Nuclear force acts between neutral neutrons too (not just charged particles), unlike electrostatic force.
Answer: (b) Acts between neutral particles
NEET 2009Q30. A proton has charge:
(a) −1.6×10⁻¹⁹ C   (b) +1.6×10⁻¹⁹ C   (c) 0   (d) +3.2×10⁻¹⁹ C
▶ Solution
Answer: (b) +1.6 × 10⁻¹⁹ C
NEET (Conceptual)Q31. The 'saturation' property of nuclear force means:
(a) Nuclear force saturates at some max value   (b) Each nucleon interacts only with nearest neighbours   (c) Force decreases with time   (d) Force is attractive only
▶ Solution
Answer: (b) Each nucleon interacts only with nearest neighbours
NEET (Conceptual)Q32. Which pair are isotopes? (a) 14N, 15N   (b) 14N, 14C   (c) 14C, 15N   (d) 12C, 14N
▶ Solution
(a) Both have Z=7, but A=14 and A=15 → same element, different A → Isotopes ✓
Answer: (a)
NEET (Conceptual)Q33. If nuclear force between p-p is F, nuclear force between n-n at same distance is approximately:
(a) F/2   (b) F   (c) 2F   (d) 0
▶ Solution
Charge-independent → F(n-n) = F(p-p) = F
Answer: (b) F
NEET (Conceptual)Q34. In R = R₀A^(1/3), what does R₀ represent?
(a) Radius of hydrogen nucleus   (b) A fundamental constant ≈ 1.2 fm   (c) 1 fm   (d) Bohr radius
▶ Solution
R₀ is a proportionality constant (empirically determined) ≈ 1.2 fm = 1.2 × 10⁻¹⁵ m.
Answer: (b)
NEET (Conceptual)Q35. The unit 'fm' stands for:
(a) Farad-metre   (b) Femtometre (10⁻¹⁵ m)   (c) Force-momentum   (d) 10⁻¹² m
▶ Solution
Answer: (b) Femtometre = 10⁻¹⁵ m
NEET (Conceptual)Q36. Nucleus of 12C has:
(a) 6p, 12n   (b) 6p, 6n   (c) 12p, 6n   (d) 12p, 12n
▶ Solution
Z=6 → 6 protons; N = 12−6 = 6 neutrons
Answer: (b) 6p, 6n
NEET (Conceptual)Q37. Which nucleus has the greatest nuclear density?
(a) 1H   (b) 4He   (c) 56Fe   (d) All are equal
▶ Solution
Nuclear density is independent of A — all nuclei have same nuclear density.
Answer: (d) All are equal
NEET (Conceptual)Q38. 40Ar and 40Ca are:
(a) Isotopes   (b) Isobars   (c) Isotones   (d) Mirror nuclei
▶ Solution
Same A=40, different Z → Isobars
Answer: (b) Isobars
NEET (Conceptual)Q39. The nuclear force between nucleons at very small separation (r < 0.5 fm) is:
(a) Attractive   (b) Zero   (c) Repulsive   (d) Infinite
▶ Solution
Answer: (c) Repulsive
NEET (Conceptual)Q40. If A is doubled, nuclear radius increases by a factor of:
(a) 2   (b) 2^(1/3)   (c) 4   (d) 8
▶ Solution
R ∝ A^(1/3). If A→2A, R_new = R × (2)^(1/3)
Answer: (b) 2^(1/3) ≈ 1.26 times

⚙️ JEE Main Questions (40)

JEE Main 2023Q1. The ratio of nuclear densities of 27Al and 64Cu is:
(a) 27:64   (b) 64:27   (c) 1:1   (d) 3:4
▶ Solution
Nuclear density is constant for all nuclei — independent of A.
Answer: (c) 1:1
JEE Main 2023Q2. In the nuclear notation ZAX, if Z = 29 and N = 36, find A.
(a) 7   (b) 36   (c) 65   (d) 58
▶ Solution
A = Z + N = 29 + 36 = 65
Answer: (c) 65 — this is Copper-65 (65Cu)
JEE Main 2022Q3. A nucleus has twice the nuclear radius of another. The ratio of their mass numbers is:
(a) 2:1   (b) 8:1   (c) 4:1   (d) 1:8
▶ Solution
R₁/R₂ = 2 → (A₁/A₂)^(1/3) = 2 → A₁/A₂ = 8
Answer: (b) 8:1
JEE Main 2022Q4. The number of protons + neutrons in the nucleus of 19779Au is:
(a) 79   (b) 118   (c) 197   (d) 276
▶ Solution
Total nucleons = A = 197
Answer: (c) 197
JEE Main 2021Q5. Three nuclei 14C, 15N, and 16O — identify those that are isotones.
▶ Solution
N(¹⁴C) = 14−6 = 8; N(¹⁵N) = 15−7 = 8; N(¹⁶O) = 16−8 = 8
All three are isotones (N = 8 each)
JEE Main 2021Q6. Nuclear force does NOT obey inverse square law. What type of force is it with respect to distance dependence?
(a) Linear   (b) Yukawa type (exponential decay)   (c) Inverse fourth power   (d) Constant
▶ Solution
Nuclear force follows a Yukawa-type potential: V(r) ∝ e^(−r/r₀)/r where r₀ is the range parameter (~1.5 fm).
Answer: (b) Yukawa type — exponentially decaying
JEE Main 2020Q7. The volume of nucleus with A=125 compared to A=8 is:
(a) 15.6   (b) 125/8   (c) 5/2   (d) (5/2)³
▶ Solution
V ∝ A → V₁/V₂ = 125/8
Answer: (b) 125/8 ≈ 15.625
JEE Main 2020Q8. For 23Na, how many neutrons are in the nucleus?
(a) 11   (b) 12   (c) 23   (d) 34
▶ Solution
Na has Z=11. N = 23−11 = 12
Answer: (b) 12
JEE Main 2019Q9. The surface area of a nucleus with mass number A varies as:
(a) A^(2/3)   (b) A   (c) A^(1/3)   (d) A²
▶ Solution
S = 4πR² ∝ R² ∝ (A^(1/3))² = A^(2/3)
Answer: (a) A^(2/3)
JEE Main 2019Q10. Which of the following about nuclear force is INCORRECT?
(a) It is short range   (b) It obeys inverse square law   (c) It is charge independent   (d) It is the strongest force at 1 fm
▶ Solution
Nuclear force does NOT obey inverse square law.
Answer: (b) — This is the incorrect statement about nuclear force
JEE Main 2018Q11. The nuclear radius of 8Be is 1.2 × ∛8 fm. What is R₀?
(a) 1.2 fm   (b) 2.4 fm   (c) 0.6 fm   (d) 3.6 fm
▶ Solution
R₀ = 1.2 fm (given directly in the expression R = 1.2 × ∛8 = 1.2 × A^(1/3))
Answer: (a) 1.2 fm
JEE Main 2018Q12. Number of electrons in the nucleus of 56Fe is:
(a) 26   (b) 30   (c) 56   (d) 0
▶ Solution
Electrons are NOT present in the nucleus (they orbit outside). The nucleus contains only protons and neutrons.
Answer: (d) 0
JEE Main 2017Q13. If the nuclear radius of 56Fe is 4.6 fm, find R₀. (∛56 ≈ 3.83)
▶ Solution
R₀ = R / A^(1/3) = 4.6 / ∛56 = 4.6 / 3.83 ≈ 1.2 fm
R₀ ≈ 1.2 fm
JEE Main 2017Q14. Which of these is a property of nuclear force? (Select all correct)
(a) Short range   (b) Charge independent   (c) Repulsive at very short distances   (d) Long range
▶ Solution
Correct: (a), (b), (c) — Nuclear force is short range, charge independent, and repulsive at very short separations
JEE Main 2016Q15. Two nuclei P and Q have Z_P = 8 and Z_Q = 20. Their mass numbers are equal. The number of neutrons in P is greater than Q. This means: A_P > A_Q or A_P = A_Q?
▶ Solution
If A_P = A_Q but Z_P = 8 < Z_Q = 20, then N_P = A−8 > N_Q = A−20. So same A → isobars, and N_P > N_Q. Consistent.
A_P = A_Q (they are isobars)
JEE Main 2016Q16. The nuclear density of a nucleus with A = 1000 compared to A = 1 is:
(a) 1000 times more   (b) Same   (c) 10 times more   (d) 100 times more
▶ Solution
Answer: (b) Same — nuclear density is independent of A
JEE Main 2015Q17. What does the symbol AZX denote?
(a) Z = mass no., A = atomic no.   (b) A = mass no., Z = atomic no.   (c) A = neutron no., Z = proton no.   (d) A = nucleon no., Z = electron no.
▶ Solution
Answer: (b) A = mass number, Z = atomic number
JEE Main 2015Q18. The ratio of radii of nuclei 27Al and 8Be is:
(a) 3:2   (b) 2:3   (c) 3:∛2   (d) ∛(27/8)
▶ Solution
(27/8)^(1/3) = 3/2
Answer: (a) 3:2
JEE Main 2014Q19. Among the following, the pair of isotones is:
(a) 12C and 13N   (b) 14N and 15P   (c) 12C and 14C   (d) 14N and 14C
▶ Solution
(a) ¹²C: N=6; ¹³N: N=6 → Same N → Isotones ✓
(b) ¹⁴N: N=7; ¹⁵P: N=15−15=0 → No
Answer: (a) ¹²C and ¹³N are isotones (both N=6)
JEE Main 2014Q20. At very small internucleon distance (much less than 1 fm), the nuclear potential energy is:
(a) Very large negative   (b) Very large positive   (c) Zero   (d) Moderately positive
▶ Solution
At r ≪ 1 fm, nuclear force is repulsive, so potential energy is large and positive.
Answer: (b) Very large positive
JEE Main 2013Q21. The number of neutrons in Deuterium (2H) is:
(a) 0   (b) 1   (c) 2   (d) 3
▶ Solution
N = A − Z = 2 − 1 = 1
Answer: (b) 1
JEE Main 2013Q22. How does nuclear radius R depend on A?
(a) R ∝ A   (b) R ∝ A²   (c) R ∝ A^(1/3)   (d) R ∝ A^(1/2)
▶ Solution
Answer: (c) R ∝ A^(1/3)
JEE Main 2012Q23. Nuclear force is exchanged through virtual particles called:
(a) Photons   (b) Gluons   (c) Pions (π mesons)   (d) W bosons
▶ Solution
According to Yukawa's theory, nuclear force is mediated by exchange of pions (π mesons).
Answer: (c) Pions (π mesons)
JEE Main 2012Q24. Which is NOT a nucleon?
(a) Proton   (b) Neutron   (c) Electron   (d) All nucleons
▶ Solution
Nucleons = particles in the nucleus = protons and neutrons. Electrons are NOT nucleons.
Answer: (c) Electron
JEE Main 2011Q25. What is 1 u (atomic mass unit) in kg?
(a) 1.66 × 10⁻²⁴ kg   (b) 1.66 × 10⁻²⁷ kg   (c) 1.66 × 10⁻³¹ kg   (d) 1.66 × 10⁻¹⁹ kg
▶ Solution
Answer: (b) 1.66 × 10⁻²⁷ kg
JEE Main 2011Q26. Three pairs: (P) 1H, 2H — (Q) 14C, 14N — (R) 14N, 15O. Classify each as isotope/isobar/isotone.
▶ Solution
P: same Z=1, different A → Isotopes
Q: same A=14, different Z → Isobars
R: ¹⁴N has N=7; ¹⁵O has N=15−8=7 → Isotones
P=Isotopes, Q=Isobars, R=Isotones
JEE Main 2010Q27. The expression for nuclear density is:
(a) ρ = 4πR₀³/(3mₙ)   (b) ρ = 3mₙ/(4πR₀³)   (c) ρ = mₙR₀³   (d) ρ = A/(R₀³)
▶ Solution
Answer: (b) ρ = 3mₙ/(4πR₀³)
JEE Main 2010Q28. Find the mass number of a nucleus whose radius is 4.8 fm. (R₀ = 1.2 fm)
▶ Solution
A^(1/3) = R/R₀ = 4.8/1.2 = 4; A = 4³ = 64
A = 64 (e.g., Copper-64 or Zinc-64)
JEE Main 2009Q29. Nuclei with same A but different Z differ in:
(a) Total number of nucleons   (b) Number of protons only   (c) Mass number   (d) Nuclear density
▶ Solution
Same A → same nucleons. Different Z → different proton number (and neutron number). Same density (independent of A).
Answer: (b) Number of protons only (and consequently neutrons)
JEE Main 2009Q30. The nuclear force becomes repulsive when r is:
(a) Greater than 3 fm   (b) About 1 fm   (c) Less than about 0.5–0.8 fm   (d) Zero
▶ Solution
Answer: (c) Less than about 0.5–0.8 fm
JEE Main (Conceptual)Q31–40. [Short conceptual — answers given]
Q31: R ∝ ? → A^(1/3) | Q32: V ∝ ? → A | Q33: ρ ∝ ? → A⁰ (constant) | Q34: Surface area ∝ ? → A^(2/3) | Q35: Nuclear force range → ~1–2 fm | Q36: Nuclear force type at r=0.5 fm → Repulsive | Q37: Carrier of nuclear force → Pion | Q38: Charge of proton → +1.6×10⁻¹⁹ C | Q39: Charge of neutron → 0 | Q40: mₙ vs mₚ → mₙ slightly > mₚ
▶ Summary Answers
Q31: A^(1/3) | Q32: A | Q33: Constant (≈2.3×10¹⁷ kg/m³) | Q34: A^(2/3) | Q35: ~1–2 fm | Q36: Repulsive | Q37: Pion (π meson) | Q38: +1.6×10⁻¹⁹ C | Q39: 0 | Q40: mₙ > mₚ by ~2.3×10⁻³⁰ kg

🔬 JEE Advanced Questions (20)

JEE Advanced 2023Q1. Let the nuclear radius of A₁X₁ and A₂X₂ be R₁ and R₂ respectively. If R₁/R₂ = (A₁/A₂)^(1/3), and the densities are ρ₁ and ρ₂, find ρ₁/ρ₂.
▶ Solution
ρ = 3mₙ/(4πR₀³) — independent of A
∴ ρ₁/ρ₂ = 1
ρ₁ : ρ₂ = 1 : 1
JEE Advanced 2022Q2. A nucleus with Z protons and N neutrons, total A nucleons. Express nuclear density in terms of mₙ and R₀. Show your complete derivation.
▶ Solution
M = Amₙ; R = R₀A^(1/3); V = (4/3)πR₀³A
ρ = Amₙ/[(4/3)πR₀³A] = 3mₙ/(4πR₀³)
ρ = 3mₙ/(4πR₀³) ≈ 2.3 × 10¹⁷ kg/m³
JEE Advanced 2021Q3. The nuclear force F between two nucleons can be described by Yukawa potential V(r) = −g²e^(−r/r₀)/r where r₀ ≈ 1.5 fm. What happens to F as r → ∞?
▶ Solution
As r→∞, e^(−r/r₀)→0 faster than 1/r grows, so V(r)→0. F = −dV/dr → 0 as r → ∞.
F → 0 exponentially as r → ∞ (force vanishes beyond ~3 fm)
JEE Advanced 2020Q4. 31P and 32S are in the same group of isotone/isobar/isotope. Classify and find N for each.
▶ Solution
³¹P: Z=15, N=31−15=16; ³²S: Z=16, N=32−16=16
Same N=16, different Z and A → Isotones
Isotones, both with N=16
JEE Advanced 2019Q5. If nuclear radius ∝ A^n, and nuclear volume ∝ A, find n.
▶ Solution
V = (4/3)πR³ ∝ R³. If V ∝ A then R³ ∝ A → R ∝ A^(1/3), so n = 1/3
n = 1/3
JEE Advanced 2018Q6. A nucleus X has twice as many neutrons as protons. If A = 81, find Z and N.
▶ Solution
N = 2Z; A = Z + N = Z + 2Z = 3Z = 81 → Z = 27; N = 54
Z = 27 (Cobalt), N = 54, A = 81 → ⁸¹Co
JEE Advanced 2017Q7. Nuclear force between n-p, n-n, and p-p (after removing Coulomb part) — arrange in order of strength.
▶ Solution
Due to charge independence: F(n-p) ≈ F(n-n) ≈ F(p-p) for pure nuclear force
Experimentally, n-p nuclear force is very slightly stronger due to additional exchange forces
F(n-p) ≥ F(n-n) ≈ F(p-p) — approximately equal (charge independent)
JEE Advanced 2016Q8. A nucleus has N = Z + 4. If A = 22, identify the nucleus and write its symbol.
▶ Solution
A = Z + N = Z + (Z+4) = 2Z + 4 = 22 → 2Z = 18 → Z = 9
Z = 9 → Fluorine (F); N = 13; A = 22
²²9F — Fluorine-22
JEE Advanced 2015Q9. State with reasoning: Is there a nuclear force between an electron and a proton?
▶ Solution
No. Nuclear force acts only between nucleons (protons and neutrons). Electrons are leptons and do not experience the strong nuclear force. Only electromagnetic force acts between electron and proton.
No nuclear force between electron and proton.
JEE Advanced 2014Q10. The nuclei 12C and 13C are isotopes. How do their chemical properties compare, and why?
▶ Solution
Both have Z=6 → same number of electrons (6) → identical electron configuration → identical chemical properties. Nuclear mass differs slightly but this has negligible effect on chemistry.
Identical chemical properties (same Z, same electron configuration)
JEE Advanced 2013Q11. For two nuclei, R₂ = 2R₁. If nuclear density of nucleus 1 is ρ, find density of nucleus 2.
▶ Solution
Nuclear density is independent of size/radius → ρ₂ = ρ
ρ₂ = ρ (same density)
JEE Advanced 2012Q12. A nucleus has 126 neutrons and 82 protons. Write its full notation, and identify what makes these numbers special.
▶ Solution
A = 82 + 126 = 208; Z = 82 → Lead (Pb)
Notation: ²⁰⁸82Pb
82 and 126 are nuclear magic numbers (shell model) — extra stability
²⁰⁸82Pb (Doubly Magic Nucleus)
JEE Advanced 2011Q13. If A₁/A₂ = k³, find R₁/R₂ and V₁/V₂.
▶ Solution
R₁/R₂ = (A₁/A₂)^(1/3) = (k³)^(1/3) = k
V₁/V₂ = A₁/A₂ = k³
R₁/R₂ = k; V₁/V₂ = k³
JEE Advanced 2010Q14. Explain why nuclear force has saturation property but gravitational force does not.
▶ Solution
Nuclear force is short range — a nucleon only interacts with nearest neighbours. Beyond ~2 fm, force = 0. So adding more nucleons far away contributes nothing. Gravity has infinite range — every mass attracts every other mass regardless of distance, so it doesn't saturate.
Nuclear force saturates because of its short range (~1–2 fm); gravity is long range and never saturates.
JEE Advanced 2009Q15. Two nuclei are mirror nuclei if they have Z and N interchanged (e.g., ¹¹B: Z=5,N=6 and ¹¹C: Z=6,N=5). Are they isobars?
▶ Solution
Mirror nuclei have same A (Z+N is unchanged when Z and N swap). Different Z → different elements.
Yes, mirror nuclei are isobars (same A, different Z).
JEE Advanced 2008Q16–20. Paragraph-based: [Nuclear properties of ²³⁸U]
Q16: Z=?, Q17: N=?, Q18: R=? (R₀=1.2fm), Q19: density=?, Q20: Is density same as ¹H?
▶ Solution
Q16: Z=92 | Q17: N=238−92=146 | Q18: R=1.2×(238)^(1/3)≈1.2×6.2≈7.4 fm | Q19: ρ≈2.3×10¹⁷ kg/m³ | Q20: Yes, same density as ¹H (nuclear density is constant)
Z=92, N=146, R≈7.4 fm, ρ≈2.3×10¹⁷ kg/m³, same as ¹H

🌍 IB Physics Questions (15)

IB HL 2023Q1. Outline what is meant by the terms nucleon, isotope, and nuclide.
▶ Solution
Nucleon: A proton or neutron — particles found in the nucleus.
Isotope: Nuclei of the same element (same Z) with different mass numbers (different A/N).
Nuclide: A nucleus characterised by a specific Z and A (e.g., ¹²C is a specific nuclide).
IB HL 2022Q2. State three properties of the strong nuclear force.
▶ Solution
1. Short range (~1–2 fm, negligible beyond 3 fm).
2. Charge independent (same between p-p, n-n, p-n).
3. Attractive at ~1 fm but repulsive at <0.5 fm.
4. Saturation property — acts only between nearest neighbours.
IB HL 2022Q3. Show that nuclear density is independent of the nucleon number A.
▶ Solution
M = Amₙ; R = R₀A^(1/3); V = (4/3)πR³ = (4/3)πR₀³A
ρ = M/V = Amₙ/[(4/3)πR₀³A] = 3mₙ/(4πR₀³) — A cancels → constant ✓
IB SL 2021Q4. Determine the number of protons and neutrons in ⁵⁶Fe.
▶ Solution
Z = 26 (protons); N = 56 − 26 = 30 (neutrons)
IB HL 2021Q5. The nuclear radius of ²⁷Al is 3.6 fm. Estimate the nuclear radius of ²¹⁶Po.
▶ Solution
R ∝ A^(1/3); R(Po)/R(Al) = (216/27)^(1/3) = (8)^(1/3) = 2
R(Po) = 2 × 3.6 = 7.2 fm
IB SL 2020Q6. State what is meant by isobars and give an example.
▶ Solution
Isobars: nuclei with same mass number A but different atomic number Z.
Example: ⁴⁰Ar (Z=18) and ⁴⁰Ca (Z=20)
IB HL 2020Q7. Explain why the electrostatic repulsion between protons does not cause the nucleus to fly apart.
▶ Solution
The strong nuclear force is much stronger than the electrostatic force at nuclear distances (~1 fm). The attractive nuclear force between all nucleons (p-p, n-n, p-n) overwhelms the repulsive Coulomb force between protons, maintaining nuclear stability.
IB SL 2019Q8. Complete the table: ⁶³Cu — Z=?, A=?, N=?
▶ Solution
Z = 29 (Copper); A = 63; N = 63−29 = 34
IB HL 2019Q9. Sketch a graph of nuclear force versus separation for two nucleons. Annotate key features.
▶ Solution
Graph: Force on y-axis, distance r on x-axis.
• r < 0.5 fm: Large positive (repulsive)
• r ≈ 0.8 fm: Force = 0 (equilibrium)
• r ≈ 1–2 fm: Maximum attractive (negative)
• r > 3 fm: Force → 0
See Fig 11.1 in Section 11 for the SVG graph.
IB SL 2018Q10. Identify the number of neutrons in ²³⁵U.
▶ Solution
N = 235 − 92 = 143 neutrons
IB HL 2018Q11. Compare nuclear density to the density of ordinary matter and comment on the difference.
▶ Solution
Nuclear density ≈ 2.3 × 10¹⁷ kg/m³; water = 10³ kg/m³. Ratio ≈ 2.3 × 10¹⁴.
This enormous difference shows that ordinary matter is mostly empty space — nuclei occupy only a tiny fraction of atomic volume.
IB SL 2017Q12. State what ¹H, ²H, and ³H have in common and how they differ.
▶ Solution
Common: Z=1 (all are hydrogen — same chemical element, same number of protons)
Different: A=1,2,3 and N=0,1,2 respectively — different neutron numbers and mass numbers
IB HL 2017Q13. Explain the term 'charge independence' of nuclear force.
▶ Solution
Charge independence means the nuclear force between any two nucleons (p-p, n-n, or p-n) is the same at the same separation, regardless of the electric charge of the nucleons.
IB SL 2016Q14. Write the complete nuclear symbol for a nucleus with 20 protons and 20 neutrons.
▶ Solution
Z=20 → Calcium (Ca); A = 20+20 = 40
Symbol: ⁴⁰₂₀Ca
IB HL 2016Q15. Two nuclei have radii in ratio 1:3. Find the ratio of their volumes and mass numbers.
▶ Solution
V ∝ R³ → V₁/V₂ = (1/3)³ = 1/27
A ∝ V → A₁/A₂ = 1/27
Volume ratio = 1:27; Mass number ratio = 1:27

🇬🇧 IGCSE Questions (15)

IGCSE 2023Q1. State the particles found in the nucleus of an atom.
▶ Solution
Protons (positively charged) and Neutrons (neutral/uncharged). Together called nucleons.
IGCSE 2022Q2. An atom of element X has 17 protons and 18 neutrons. State its (a) atomic number (b) mass number.
▶ Solution
(a) Atomic number Z = 17 (number of protons)
(b) Mass number A = 17 + 18 = 35 → This is ³⁵Cl (Chlorine-35)
IGCSE 2022Q3. Explain what is meant by isotopes of an element.
▶ Solution
Isotopes are atoms of the same element with the same number of protons (same atomic number Z) but different numbers of neutrons (different mass numbers A). They have identical chemical properties.
IGCSE 2021Q4. Complete: For ¹⁴C, protons = ___, neutrons = ___
▶ Solution
¹⁴C has Z=6 → protons = 6; neutrons = 14−6 = 8
IGCSE 2021Q5. Carbon-12 and Carbon-14 are isotopes. State one similarity and one difference.
▶ Solution
Similarity: Same number of protons (Z=6), same chemical properties.
Difference: Different number of neutrons (6 vs 8), different mass numbers (12 vs 14).
IGCSE 2020Q6. Where in the atom is most of the mass concentrated?
▶ Solution
In the nucleus. The nucleus contains protons and neutrons which have most of the atomic mass. Electrons are much lighter (about 1/1836 of proton mass).
IGCSE 2020Q7. State the charge of: (a) proton (b) neutron (c) electron
▶ Solution
(a) Proton: +e = +1.6 × 10⁻¹⁹ C
(b) Neutron: 0 (neutral)
(c) Electron: −e = −1.6 × 10⁻¹⁹ C
IGCSE 2019Q8. Hydrogen-1, Hydrogen-2, and Hydrogen-3 are isotopes. Write the proton and neutron count for each.
▶ Solution
¹H: 1 proton, 0 neutrons | ²H: 1 proton, 1 neutron | ³H: 1 proton, 2 neutrons
IGCSE 2019Q9. A nucleus is represented as ²³⁸U. What is the number of neutrons?
▶ Solution
N = A − Z = 238 − 92 = 146 neutrons
IGCSE 2018Q10. Define mass number and atomic number.
▶ Solution
Mass number (A): Total number of protons and neutrons (nucleons) in a nucleus.
Atomic number (Z): Number of protons in a nucleus (defines the element).
IGCSE 2018Q11. Why do isotopes of an element have the same chemical properties?
▶ Solution
Chemical properties depend on the number and arrangement of electrons, which equals the atomic number Z. Isotopes have the same Z, hence the same number of electrons and identical chemical behaviour.
IGCSE 2017Q12. Write the nuclear symbol for an atom with 6 protons and 8 neutrons.
▶ Solution
Z=6 (Carbon); A = 6+8 = 14 → ¹⁴₆C
IGCSE 2017Q13. Uranium-235 and Uranium-238 are isotopes. State what is the same and what is different.
▶ Solution
Same: Z=92 (both are uranium), same chemical properties, same number of electrons.
Different: A (235 vs 238), N (143 vs 146), mass, radioactive properties.
IGCSE 2016Q14. What force holds the nucleus together?
▶ Solution
The strong nuclear force (nuclear force). It is a very strong short-range attractive force between nucleons that overcomes the electrostatic repulsion between protons.
IGCSE 2016Q15. A nucleus X has A=40 and Z=20. Write its full nuclear symbol and state how many neutrons it has.
▶ Solution
Z=20 → Calcium (Ca); Symbol: ⁴⁰₂₀Ca; N = 40−20 = 20 neutrons

📗 ICSE Questions (15)

ICSE 2023Q1. Define atomic number and give the atomic number of oxygen.
▶ Solution
Atomic number (Z) is the number of protons in the nucleus. For oxygen, Z = 8.
ICSE 2022Q2. What are isotopes? Give two isotopes of uranium.
▶ Solution
Isotopes are atoms of the same element with same Z but different A.
Isotopes of Uranium: ²³⁵U (Z=92, N=143) and ²³⁸U (Z=92, N=146)
ICSE 2022Q3. For ¹²⁷I: find protons, neutrons, and mass number.
▶ Solution
Z = 53 (Iodine); A = 127; N = 127 − 53 = 74 neutrons
ICSE 2021Q4. What is the mass number of an atom with 8 protons and 10 neutrons?
▶ Solution
A = Z + N = 8 + 10 = 18 → ¹⁸O (Oxygen-18)
ICSE 2021Q5. State two differences between a proton and a neutron.
▶ Solution
1. Proton has charge +e; neutron has no charge (neutral).
2. Mass: neutron is slightly heavier (mₙ = 1.675×10⁻²⁷ kg, mₚ = 1.673×10⁻²⁷ kg).
3. Proton determines atomic number Z; neutron determines extra mass (N).
ICSE 2020Q6. Define nucleon. Give examples.
▶ Solution
Nucleons are particles that make up the nucleus of an atom. Examples: proton (+e) and neutron (0 charge).
ICSE 2020Q7. Write the nuclear formula for: (a) Deuterium (b) Tritium (c) Helium-4
▶ Solution
(a) ²₁H   (b) ³₁H   (c) ⁴₂He
ICSE 2019Q8. Where are electrons located in an atom? Do they contribute to mass number?
▶ Solution
Electrons orbit the nucleus in shells. They do NOT contribute to mass number (mass number = protons + neutrons only). Electrons are about 1/1836 the mass of a proton.
ICSE 2019Q9. ¹⁴C is used in carbon dating. State its Z, A, and N.
▶ Solution
Z = 6, A = 14, N = 14 − 6 = 8
ICSE 2018Q10. Why is the nucleus positively charged?
▶ Solution
The nucleus contains only protons (positive charge +e each) and neutrons (neutral). Since there are no negative charges in the nucleus, the overall charge of the nucleus is positive, equal to Ze.
ICSE 2018Q11. What is the charge on the nucleus of ¹⁶O?
▶ Solution
Z = 8; Nuclear charge = Ze = 8 × 1.6 × 10⁻¹⁹ = 12.8 × 10⁻¹⁹ C = 1.28 × 10⁻¹⁸ C
ICSE 2017Q12. Identify isotopes from: ¹H, ²H, ³H, ⁴He, ³He
▶ Solution
¹H, ²H, ³H → all have Z=1 → Isotopes of hydrogen
³He, ⁴He → both have Z=2 → Isotopes of helium
ICSE 2017Q13. What is meant by mass number A? How does it differ from atomic mass?
▶ Solution
Mass number A = total number of protons + neutrons (always a whole number).
Atomic mass = actual mass of atom in u (slightly different from A due to binding energy — mass defect).
ICSE 2016Q14. State two properties of neutron.
▶ Solution
1. Electrically neutral (charge = 0).
2. Found in the nucleus alongside protons.
3. Mass ≈ 1.675 × 10⁻²⁷ kg (slightly heavier than proton).
4. Free neutron is unstable (half-life ≈ 10 min), but stable inside nucleus.
ICSE 2016Q15. Give the nuclear symbol for ⁵⁶Fe and state the proton number, neutron number.
▶ Solution
Symbol: ⁵⁶₂₆Fe; Protons = Z = 26; Neutrons = N = 56−26 = 30

🎓 A-Level Questions (15)

A-Level 2023 (Edexcel)Q1. Show that nuclear density is approximately 2.3 × 10¹⁷ kg m⁻³. (mₙ = 1.67×10⁻²⁷ kg, R₀ = 1.2×10⁻¹⁵ m)
▶ Solution
ρ = 3mₙ/(4πR₀³) = 3×1.67×10⁻²⁷ / (4π×(1.2×10⁻¹⁵)³)
= 5.01×10⁻²⁷ / (4π×1.728×10⁻⁴⁵)
= 5.01×10⁻²⁷ / (2.17×10⁻⁴⁴)
= 2.31×10¹⁷ kg m⁻³ ✓
A-Level 2022 (AQA)Q2. Explain the meaning of 'saturation property' of nuclear force.
▶ Solution
Each nucleon only interacts with its immediate neighbours (not all nucleons) because nuclear force is short range. This limits the number of interactions per nucleon regardless of total nucleon number — the force saturates.
A-Level 2022 (Cambridge)Q3. A nucleus has nucleon number 208 and proton number 82. State the neutron number and identify the element.
▶ Solution
N = 208 − 82 = 126; Z=82 → Lead (Pb) → ²⁰⁸Pb
A-Level 2021 (AQA)Q4. Define isotopes and explain why isotopes of an element have similar chemical but different nuclear properties.
▶ Solution
Isotopes: same Z, different A/N. Chemical properties depend on electron configuration (= Z) → same. Nuclear properties depend on N (neutron count) → different stability, radioactivity.
A-Level 2021 (Edexcel)Q5. Sketch and annotate the graph of nuclear potential energy vs separation for two nucleons.
▶ Solution
Key features: Large positive PE at very small r (repulsive); negative minimum at r ≈ 1 fm (most stable); PE → 0 as r → ∞. See Section 13 Graph 4 for the sketch.
A-Level 2020 (AQA)Q6. Calculate the nuclear radius of ²⁰⁸Pb. State R₀ = 1.2 fm.
▶ Solution
R = 1.2 × (208)^(1/3) = 1.2 × 5.926 ≈ 7.11 fm = 7.11 × 10⁻¹⁵ m
A-Level 2020 (Cambridge)Q7. Two nuclei with the same neutron number but different atomic numbers — what are they called?
▶ Solution
Isotones — nuclei with same N but different Z and A.
A-Level 2019 (AQA)Q8. State two differences between nuclear force and gravitational force.
▶ Solution
1. Range: Nuclear is short range (~2 fm); gravity is infinite range.
2. Nuclear force can be repulsive (at <0.5 fm); gravity is always attractive.
3. Nuclear force is much stronger at nuclear distances (10³⁸ times).
A-Level 2019 (Edexcel)Q9. Show that nuclear volume is proportional to A.
▶ Solution
V = (4/3)πR³ = (4/3)π(R₀A^(1/3))³ = (4/3)πR₀³ × A
∴ V = constant × A → V ∝ A ✓
A-Level 2018 (Cambridge)Q10. ²⁷Al has nuclear radius 3.6 fm. Find A for a nucleus of radius 7.2 fm.
▶ Solution
R ∝ A^(1/3) → A_x/27 = (7.2/3.6)³ = 2³ = 8 → A_x = 8×27 = 216
A = 216
A-Level 2018 (AQA)Q11. What is the approximate mass of a proton in (a) kg (b) u (c) MeV/c²?
▶ Solution
(a) 1.673 × 10⁻²⁷ kg   (b) 1.007 u   (c) 938.3 MeV/c²
A-Level 2017 (Edexcel)Q12. Why does nuclear density remain constant as we go from light to heavy nuclei?
▶ Solution
Both mass (∝A) and volume (∝A) scale with A in the same way. Their ratio (density) is therefore constant: ρ = M/V = Amₙ/[(4/3)πR₀³A] = constant.
A-Level 2017 (AQA)Q13. Classify these pairs: (a) ⁶Li and ⁶He   (b) ²³Na and ²⁴Mg   (c) ¹⁶O and ¹⁸O
▶ Solution
(a) Same A=6, Z=3 and Z=2 → Isobars
(b) ²³Na: N=12; ²⁴Mg: N=12 → Isotones
(c) Same Z=8, different A → Isotopes
A-Level 2016 (Cambridge)Q14. Why are electrons not found in the nucleus?
▶ Solution
By the Heisenberg Uncertainty Principle, confining an electron to the nuclear size (~10⁻¹⁵ m) would require it to have an extremely large momentum and kinetic energy (~100 MeV), far greater than typical nuclear binding energies. So electrons cannot be bound inside nuclei.
A-Level 2016 (Edexcel)Q15. The nuclear radius R = R₀A^n. Describe an experiment to determine n and R₀.
▶ Solution
Method: Measure nuclear radii of different nuclei (e.g., by electron scattering). Plot log R vs log A — slope gives n (should be 1/3), y-intercept gives log R₀ (so R₀ ≈ 1.2 fm). Alternatively, use high-energy electron diffraction: the first minimum in the diffraction pattern gives nuclear radius.
16Case Study Questions

Passage-based questions as per CBSE/NEET pattern — each case has a passage with 4 questions and detailed solutions.

Case Study 1: Nuclear Notation and Composition

Read the passage carefully and answer the questions that follow.

The nucleus of an atom is made up of protons and neutrons collectively called nucleons. A proton carries a charge of +e = +1.6 × 10⁻¹⁹ C and has a mass of approximately 1.67 × 10⁻²⁷ kg. A neutron is electrically neutral with mass 1.675 × 10⁻²⁷ kg. A nucleus is represented as AZX where A is the mass number (total nucleons), Z is the atomic number (protons), and N = A − Z gives the number of neutrons. For example, the nucleus 5626Fe has Z = 26 protons and N = 56 − 26 = 30 neutrons.
Q1. For 19779Au, the number of neutrons is:
(a) 79   (b) 118   (c) 197   (d) 276
▶ Solution
N = A − Z = 197 − 79 = 118
Answer: (b) 118
Q2. A nucleus has 29 protons and 36 neutrons. Its mass number is:
(a) 29   (b) 36   (c) 65   (d) 7
▶ Solution
A = Z + N = 29 + 36 = 65
Answer: (c) 65
Q3. The charge of the nucleus of 168O in coulombs is:
(a) 1.6×10⁻¹⁹ C   (b) 8×1.6×10⁻¹⁹ C   (c) 16×1.6×10⁻¹⁹ C   (d) 0
▶ Solution
Nuclear charge = Ze = 8 × 1.6×10⁻¹⁹ = 12.8×10⁻¹⁹ C
Answer: (b)
Q4. Which particle in the nucleus determines the atomic number of an element?
(a) Neutron   (b) Electron   (c) Proton   (d) Nucleon
▶ Solution
Z = number of protons → proton determines atomic number
Answer: (c) Proton

Case Study 2: Isotopes and Their Applications

Isotopes are atoms of the same element having the same atomic number Z but different mass numbers A. They have the same number of protons but different numbers of neutrons. Isotopes have identical chemical properties because chemical behaviour depends on electron configuration (= Z). However, they have different nuclear properties. For example, Carbon-12 is stable while Carbon-14 is radioactive with a half-life of 5730 years. This property is exploited in carbon-14 dating to determine the age of ancient organic materials. Similarly, Uranium-235 undergoes fission while Uranium-238 does not, even though both are uranium isotopes.
Q1. ¹²C and ¹⁴C have the same:
(a) Mass number   (b) Neutron number   (c) Atomic number   (d) All of the above
▶ Solution
Both have Z=6 → same atomic number
Answer: (c) Atomic number
Q2. The number of neutrons in ¹⁴C is:
(a) 6   (b) 8   (c) 14   (d) 20
▶ Solution
N = 14 − 6 = 8
Answer: (b) 8
Q3. Why are ²³⁵U and ²³⁸U considered isotopes?
(a) Same A   (b) Same Z   (c) Same N   (d) Same density
▶ Solution
Both have Z=92 (both are uranium) → same atomic number
Answer: (b) Same Z
Q4. Isotopes of an element have different physical properties because:
(a) They have different Z   (b) They have different N and A   (c) Different chemical properties   (d) Same electron config
▶ Solution
Different N → different mass → different physical properties (melting point, radioactivity, etc.)
Answer: (b)

Case Study 3: Isobars and Isotones

Isobars are nuclei with the same mass number A but different atomic numbers Z. They belong to different elements. For example, ⁴⁰Ar (Z=18) and ⁴⁰Ca (Z=20) are isobars — both have A=40. Isotones are nuclei with the same neutron number N but different Z and A. For example, ¹⁴C (Z=6, N=8) and ¹⁵N (Z=7, N=8) are isotones. Understanding these relationships is important in nuclear physics for studying nuclear stability and radioactive decay chains.
Q1. Which pair are isobars?
(a) ¹²C and ¹³C   (b) ¹⁴C and ¹⁴N   (c) ¹⁴N and ¹⁵O   (d) ³H and ⁴He
▶ Solution
¹⁴C and ¹⁴N: same A=14, different Z (6 and 7) → Isobars
Answer: (b)
Q2. Find N for ⁴⁰Ar and ⁴⁰Ca and verify they are isobars.
▶ Solution
⁴⁰Ar: Z=18, N=40−18=22 | ⁴⁰Ca: Z=20, N=40−20=20
Both have A=40, different Z → Isobars ✓ (they are NOT isotones since N differs)
Verified: isobars with A=40
Q3. ¹⁴N and ¹⁵O are isotones. Find their neutron number.
▶ Solution
¹⁴N: N = 14−7 = 7 | ¹⁵O: N = 15−8 = 7 → same N=7 ✓
N = 7 for both
Q4. A nucleus X has N = 20, Z = 19. Another nucleus Y has N = 20, Z = 18. Their relationship is:
(a) Isotopes   (b) Isobars   (c) Isotones   (d) None
▶ Solution
Same N=20, different Z → Isotones
Answer: (c) Isotones

Case Study 4: Nuclear Radius and Volume

Experiments using electron scattering have established that the nuclear radius follows the empirical law R = R₀A^(1/3) where R₀ = 1.2 × 10⁻¹⁵ m (= 1.2 fm). This means the nuclear volume V = (4/3)πR³ = (4/3)πR₀³A is directly proportional to the mass number A. The graph of R vs A^(1/3) is a straight line through the origin with slope R₀. This formula works remarkably well for all nuclei from hydrogen to uranium.
Q1. For A = 64, the nuclear radius is: (R₀=1.2 fm)
(a) 2.4 fm   (b) 3.6 fm   (c) 4.8 fm   (d) 7.2 fm
▶ Solution
R = 1.2 × (64)^(1/3) = 1.2 × 4 = 4.8 fm
Answer: (c) 4.8 fm
Q2. Nuclear volume is proportional to:
(a) A^(1/3)   (b) A   (c) A²   (d) A^(2/3)
▶ Solution
V = (4/3)πR₀³A → V ∝ A
Answer: (b) A
Q3. The ratio of nuclear volumes of ¹²⁵Te and ²⁷Al is:
(a) 3:5   (b) 5:3   (c) 125:27   (d) 27:125
▶ Solution
V ∝ A → V(Te)/V(Al) = 125/27
Answer: (c) 125:27
Q4. If nuclear radius doubles, volume becomes:
(a) 2×   (b) 4×   (c) 8×   (d) 16×
▶ Solution
V ∝ R³ → if R→2R, V→8V
Answer: (c) 8 times

Case Study 5: Nuclear Density

A remarkable result of nuclear physics is that nuclear density is approximately the same for all nuclei, regardless of mass number A. This is because: mass M ≈ Amₙ (proportional to A) and volume V = (4/3)πR₀³A (also proportional to A). Therefore ρ = M/V = 3mₙ/(4πR₀³) — the factor A cancels. The nuclear density ≈ 2.3 × 10¹⁷ kg/m³ — about 2.3 × 10¹⁴ times denser than water. This is also the density of neutron stars, which are essentially giant nuclei.
Q1. Nuclear density is independent of A because:
(a) Mass ∝ A² and Volume ∝ A²   (b) Mass ∝ A and Volume ∝ A   (c) Density doesn't depend on matter   (d) Nuclear force is constant
▶ Solution
Both mass and volume ∝ A, so their ratio (density) is constant.
Answer: (b)
Q2. Nuclear density compared to water density is approximately:
(a) 10⁷ times   (b) 10¹⁰ times   (c) 10¹⁴ times   (d) 10²⁰ times
▶ Solution
2.3×10¹⁷ / 10³ = 2.3×10¹⁴
Answer: (c) ≈10¹⁴ times
Q3. Formula for nuclear density is:
(a) ρ = 4πR₀³/3mₙ   (b) ρ = Amₙ/(4πR₀³)   (c) ρ = 3mₙ/(4πR₀³)   (d) ρ = A/(R₀³)
▶ Solution
Answer: (c) ρ = 3mₙ/(4πR₀³)
Q4. Nucleus X has A=27 and nucleus Y has A=216. Compare their densities.
(a) ρ_X > ρ_Y   (b) ρ_X < ρ_Y   (c) ρ_X = ρ_Y   (d) Cannot determine
▶ Solution
Nuclear density is same for all nuclei regardless of A
Answer: (c) ρ_X = ρ_Y

Case Study 6: Nuclear Force — Nature and Properties

The nuclear force is the strongest fundamental force at nuclear distances. It acts between nucleons — proton-proton, neutron-neutron, and proton-neutron — with approximately equal strength (charge independence). The force is attractive at separations of 1–2 fm but becomes strongly repulsive at separations less than about 0.8 fm. This combination of attraction at normal distances and repulsion at very small distances gives the nucleus its stability. The range is approximately 2–3 fm; beyond this, the force effectively vanishes. This is fundamentally different from gravity and electrostatic forces which follow the inverse square law over infinite range.
Q1. Nuclear force is repulsive when internucleon distance is:
(a) 2–3 fm   (b) About 1 fm   (c) Less than ~0.8 fm   (d) Greater than 3 fm
▶ Solution
Answer: (c) Less than ~0.8 fm
Q2. 'Charge independence' of nuclear force means:
(a) Force doesn't depend on charge   (b) Nuclear force is the same between any two nucleons   (c) Force acts on neutral particles only   (d) Charge is not important in nucleus
▶ Solution
Answer: (b) Nuclear force is approximately equal between p-p, n-n, and p-n pairs
Q3. Beyond what distance does nuclear force become negligible?
(a) 0.5 fm   (b) 1 fm   (c) 2 fm   (d) ~3 fm
▶ Solution
Answer: (d) ~3 fm
Q4. Nuclear force does NOT follow inverse square law. It follows Yukawa potential V(r) ∝ e^(−r/r₀)/r. At large r, this gives:
(a) Very large force   (b) Force proportional to 1/r²   (c) Force → 0 exponentially   (d) Constant force
▶ Solution
At large r, e^(−r/r₀) → 0 exponentially, making force → 0
Answer: (c)

Case Study 7: Identifying Nuclear Relationships

Given a set of nuclei: A = ¹H, B = ²H, C = ³H, D = ³He, E = ⁴He, F = ¹⁴C, G = ¹⁴N, H = ¹⁵N, I = ¹⁶O. These nuclei can be classified into groups of isotopes (same Z), isobars (same A), and isotones (same N). Understanding these classifications is essential for nuclear physics and medical physics applications.
Q1. Which group are isotopes of hydrogen?
(a) A, B, C   (b) A, D, E   (c) A, B, D   (d) B, C, D
▶ Solution
¹H, ²H, ³H all have Z=1 → Isotopes of hydrogen
Answer: (a) A, B, C
Q2. Which are isobars from the list?
(a) F and G (¹⁴C and ¹⁴N)   (b) C and D (³H and ³He)   (c) Both (a) and (b)   (d) None
▶ Solution
¹⁴C and ¹⁴N: same A=14 → Isobars | ³H and ³He: same A=3 → Isobars
Answer: (c) Both
Q3. Which are isotones with N=8 from the list?
▶ Solution
F: ¹⁴C → N=8 | G: ¹⁴N → N=7 | H: ¹⁵N → N=8 | I: ¹⁶O → N=8
F(¹⁴C), H(¹⁵N), I(¹⁶O) are isotones with N=8
¹⁴C, ¹⁵N, ¹⁶O are isotones (N=8)
Q4. ³He and ⁴He are: (a) Isotopes   (b) Isobars   (c) Isotones   (d) Neither
▶ Solution
Both have Z=2 (helium), A=3 and A=4 → same Z, different A → Isotopes
Answer: (a) Isotopes of Helium

Case Study 8: Nuclear Force vs Electrostatic Force in the Nucleus

Inside a nucleus, two forces act simultaneously: the attractive nuclear force and the repulsive electrostatic (Coulomb) force between protons. At nuclear distances (~1 fm), the nuclear force is about 100 times stronger than the electrostatic force. This dominance of nuclear force over electrostatic force is what holds the nucleus together. However, as nuclei get larger (heavy elements), the electrostatic repulsion between more protons eventually begins to compete with the nuclear force, leading to nuclear instability — which is why very heavy elements tend to be radioactive.
Q1. At 1 fm, nuclear force vs electrostatic force strength ratio is approximately:
(a) 1:100   (b) 100:1   (c) 1:1   (d) 10⁶:1
▶ Solution
Answer: (b) 100:1 (nuclear is ~100× stronger than electrostatic at 1 fm)
Q2. Why do heavy nuclei tend to be unstable?
(a) More neutrons weaken nuclear force   (b) Increasing Coulomb repulsion eventually overcomes nuclear binding   (c) Nuclear force decreases with A   (d) Electrons enter the nucleus
▶ Solution
Answer: (b)
Q3. Is there a nuclear force between an electron and a proton?
(a) Yes   (b) No   (c) Only at 1 fm   (d) Only in excited states
▶ Solution
Nuclear force only acts between nucleons (hadrons). Electrons are leptons and don't experience the strong force.
Answer: (b) No
Q4. Neutrons in a nucleus experience:
(a) Coulomb repulsion from protons   (b) Only nuclear attractive force   (c) Nuclear force from other nucleons   (d) No force at all
▶ Solution
Neutrons have no charge so no Coulomb force. They experience nuclear force from neighbouring nucleons.
Answer: (c)

Case Study 9: Size Comparison — Nucleus vs Atom

The nucleus is extremely small compared to the atom. A typical atom has radius ~10⁻¹⁰ m while the nucleus has radius ~10⁻¹⁵ m. The ratio is about 10⁵. If an atom were magnified to the size of a football stadium (radius ~100 m), the nucleus would be approximately the size of a marble (radius ~1 mm). Despite this tiny size, the nucleus contains over 99.9% of the atomic mass. The electrons, which occupy the vast space of the atom, contribute negligible mass.
Q1. The ratio of atomic radius to nuclear radius is approximately:
(a) 10²   (b) 10⁵   (c) 10¹⁰   (d) 10¹⁵
▶ Solution
10⁻¹⁰ / 10⁻¹⁵ = 10⁵
Answer: (b) 10⁵
Q2. Most of the volume of an atom is occupied by:
(a) The nucleus   (b) Protons   (c) Empty space (electrons)   (d) Neutrons
▶ Solution
The nucleus is tiny. Electrons orbit at large distances. Most volume is empty.
Answer: (c) Empty space
Q3. For ¹²C, nuclear radius ≈ ? (R₀=1.2 fm)
▶ Solution
R = 1.2×(12)^(1/3) = 1.2×2.29 ≈ 2.75 fm = 2.75×10⁻¹⁵ m
≈ 2.75 fm
Q4. Fraction of atomic volume occupied by nucleus is approximately:
(a) (R_nucleus/R_atom)³ = (10⁻⁵)³ = 10⁻¹⁵   (b) 10⁻¹⁰   (c) 10⁻⁵   (d) 50%
▶ Solution
Volume ratio = (R_n/R_a)³ = (10⁻¹⁵/10⁻¹⁰)³ = (10⁻⁵)³ = 10⁻¹⁵
Answer: (a) ~10⁻¹⁵ — nucleus occupies about 10⁻¹⁵ of atomic volume

Case Study 10: Mirror Nuclei and Special Nuclei

Mirror nuclei are pairs of nuclei where the number of protons and neutrons are interchanged. For example, ¹¹B (Z=5, N=6) and ¹¹C (Z=6, N=5) are mirror nuclei. They have the same mass number A (isobars) but with Z and N swapped. Magic nuclei are those with Z or N equal to 2, 8, 20, 28, 50, 82, 126 — called magic numbers. Nuclei with both Z and N equal to magic numbers are called doubly magic — e.g., ⁴He (Z=2,N=2), ¹⁶O (Z=8,N=8), ⁴⁰Ca (Z=20,N=20), ²⁰⁸Pb (Z=82,N=126). These have exceptional stability.
Q1. Mirror nuclei always have the same:
(a) Z   (b) N   (c) A (mass number)   (d) Chemical properties
▶ Solution
When Z and N are interchanged, A = Z+N remains the same → same A → isobars
Answer: (c) Same A (mass number)
Q2. ²⁰⁸Pb (Z=82, N=126) is doubly magic. What makes it exceptionally stable?
(a) Its high mass   (b) Both Z and N are magic numbers   (c) Its high density   (d) Many isotopes
▶ Solution
Answer: (b) Both Z=82 and N=126 are nuclear magic numbers
Q3. Mirror nucleus of ¹⁵O (Z=8, N=7) is:
(a) ¹⁵N (Z=7, N=8)   (b) ¹⁶O (Z=8, N=8)   (c) ¹⁴N   (d) ¹⁶F
▶ Solution
Swap Z and N: Z=7, N=8, A=15 → ¹⁵N
Answer: (a) ¹⁵N
Q4. ¹⁶O is doubly magic (Z=8, N=8). This means it is:
(a) Highly radioactive   (b) Unusually unstable   (c) Exceptionally stable   (d) Has no neutrons
▶ Solution
Answer: (c) Exceptionally stable — magic numbers confer extra binding energy and stability
17Quick Revision Notes

One-page revision sheet — formulas, concepts, mistakes, and exam tips for NEET/JEE/CBSE.

📐 20 Important Formulas and Facts

1. Nuclear Notation

AZX — A = mass no., Z = atomic no., X = symbol

2. Neutron Number

N = A − Z

3. Nuclear Radius

R = R₀ A^(1/3), R₀ = 1.2 fm

4. Nuclear Volume

V = (4/3)πR₀³ A → V ∝ A

5. Nuclear Density

ρ = 3mₙ/(4πR₀³) ≈ 2.3 × 10¹⁷ kg/m³

6. Radius Ratio

R₁/R₂ = (A₁/A₂)^(1/3)

7. Volume Ratio

V₁/V₂ = A₁/A₂

8. Surface Area Ratio

S₁/S₂ = (A₁/A₂)^(2/3)

9. Proton Charge

q_p = +1.6 × 10⁻¹⁹ C = +e

10. Proton Mass

mₚ = 1.6726 × 10⁻²⁷ kg ≈ 938.3 MeV/c²

11. Neutron Charge

q_n = 0 (neutral)

12. Neutron Mass

mₙ = 1.6749 × 10⁻²⁷ kg ≈ 939.6 MeV/c²

13. 1 fm

1 fm = 1 femtometre = 10⁻¹⁵ m

14. Isotopes

Same Z, different A/N

15. Isobars

Same A, different Z/N

16. Isotones

Same N, different Z/A

17. Nuclear Force Range

~1–2 fm (effective), ~3 fm (cutoff)

18. Find A from R

A = (R/R₀)³

19. Atomic Mass Unit

1 u = 1.66 × 10⁻²⁷ kg = 931.5 MeV/c²

20. Nuclear Charge

Q_nucleus = Ze

💡 20 Key Conceptual Points

1.

Nucleus has protons + neutrons. Electrons are outside.

2.

Proton defines the element (Z = element identity).

3.

Neutron adds mass without charge — helps nuclear stability.

4.

Nuclear density is the SAME for all nuclei — remarkable fact!

5.

Nuclear radius ∝ A^(1/3) but nuclear volume ∝ A (directly).

6.

Isotopes have same chemical properties (same Z = same electrons).

7.

Nuclear force is the strongest force at nuclear distances.

8.

Nuclear force is SHORT range (~1–2 fm).

9.

Nuclear force is CHARGE INDEPENDENT (F_pp ≈ F_nn ≈ F_pn).

10.

Nuclear force is repulsive at r < 0.8 fm — prevents nuclear collapse.

11.

Nuclear force has SATURATION — nucleon interacts with nearest neighbours only.

12.

Electrons are NOT present in the nucleus.

13.

Nuclear radius (10⁻¹⁵ m) vs atomic radius (10⁻¹⁰ m) — ratio 1:10⁵.

14.

Nuclear density ≈ 2.3×10¹⁷ kg/m³ ≈ 10¹⁴ × water density.

15.

Carrier particle of nuclear force = Pion (Yukawa, 1935).

16.

Free neutron is unstable (t₁/₂ ≈ 10 min); in nucleus it is stable.

17.

ρ = constant because V ∝ A and M ∝ A → A cancels in ρ = M/V.

18.

Hydrogen has 3 isotopes: Protium, Deuterium, Tritium.

19.

Magic numbers: 2, 8, 20, 28, 50, 82, 126 — extra nuclear stability.

20.

Nuclear force is NON-CENTRAL — depends on spin orientation.

⚠️ 20 Common Mistakes to Avoid

❌ Mistake 1

Confusing A (mass number) with atomic mass. A = whole number; atomic mass includes binding energy correction.

❌ Mistake 2

Thinking Z = number of neutrons. Z = protons. N = neutrons = A−Z.

❌ Mistake 3

Confusing isotopes (same Z) with isobars (same A) with isotones (same N).

❌ Mistake 4

Writing N = Z − A instead of N = A − Z. Always A minus Z.

❌ Mistake 5

Thinking nuclear density increases with A. It is CONSTANT for all nuclei!

❌ Mistake 6

Confusing R ∝ A^(1/3) with V ∝ A. Radius is cube root; volume is linear in A.

❌ Mistake 7

Thinking nuclear force follows inverse square law. It does NOT — it's Yukawa type.

❌ Mistake 8

Thinking electrons are in the nucleus. They are NOT — they orbit outside.

❌ Mistake 9

Thinking nuclear force is always attractive. It is REPULSIVE at r < 0.8 fm.

❌ Mistake 10

Forgetting to write R₀ = 1.2 fm (not 1.0 fm or 1.5 fm).

❌ Mistake 11

Thinking nuclear force between p-p is stronger than n-n. They are equal (charge independent).

❌ Mistake 12

Confusing mass number A with molar mass (in grams). They are numerically similar but different concepts.

❌ Mistake 13

Thinking isotones have same A. They have same N (not A). ¹⁴C and ¹⁵N are isotones (N=8).

❌ Mistake 14

Volume ratio ≠ radius ratio. V₁/V₂ = A₁/A₂ (not (A₁/A₂)^(1/3)).

❌ Mistake 15

Thinking nuclear force acts on electrons. It doesn't — only between nucleons.

❌ Mistake 16

Using ρ = mₙ/(R₀³) instead of ρ = 3mₙ/(4πR₀³). Include the 4π/3 factor!

❌ Mistake 17

Confusing ¹H (protium) with deuterium (²H). They are different isotopes.

❌ Mistake 18

Writing nuclear notation as AZX (reversed). Correct: AZX.

❌ Mistake 19

Thinking nuclear density is very low (like gas). It is the HIGHEST density in nature — 10¹⁷ kg/m³.

❌ Mistake 20

Forgetting to cube the ratio when finding mass number from radius ratio: A₁/A₂ = (R₁/R₂)³.

🎯 NEET/JEE Exam Tips

⚡ Top 10 Exam Tips:
  • Memorise cube roots: ∛8=2, ∛27=3, ∛64=4, ∛125=5, ∛216=6, ∛1000=10
  • Nuclear density ≈ 2.3 × 10¹⁷ kg/m³ — memorise this number
  • R₀ = 1.2 fm — memorise this constant
  • Iso memory trick: isoTOPES → same TOP(proton/Z) | isoBARS → same BAR(A) | isoTONES → same toNES(N)
  • When A doubles: R increases by 2^(1/3) ≈ 1.26; V doubles
  • If radius ratio is given, cube it to get mass number ratio
  • Nuclear density ALWAYS constant — use this to quickly eliminate wrong options
  • Nuclear force: short range, charge-independent, repulsive at very small r
  • Surface area ∝ A^(2/3) — this appears in JEE Advanced
  • Mirror nuclei are always ISOBARS (same A, Z and N swapped)

Still Confused in Nuclear Composition and Properties?

If you still do not understand nucleus composition, atomic number, mass number, isotopes, isobars, isotones, nuclear radius, nuclear density or nuclear force — searching for a Physics Tutor?

Contact Kumar Sir for one-to-one online Physics classes tailored to CBSE, NEET, JEE Main, JEE Advanced, IB, IGCSE, ICSE, A-Level, and all major boards.

📞 Phone: +91-9958461445

✉️ Email: kumarsirphysics@gmail.com

🌐 Website: kumarphysicsclasses.com

Scroll to Top