Nuclei - Class 12 Physics

Class 12 Physics | Unit VIII

Chapter 13: Nuclei

Nuclear Properties • Mass Defect • Binding Energy • Radioactivity • Fission & Fusion

1. Composition & Size of Nucleus

Nucleus: Contains protons (Z) and neutrons (N). Together called nucleons (A = Z + N). A = mass number, Z = atomic number.

Nuclear radius: R = R₀A^1/3 where R₀ = 1.2 fm = 1.2×10⁻¹⁵ m
Nuclear volume: V = (4/3)πR³ ∝ A
Nuclear density: ρ = 3m_p/(4πR₀³) ≈ 2.3×10¹⁷ kg/m³ (constant for all nuclei!)

Nuclear density is same for all nuclei → nucleons are uniformly distributed and tightly packed.
1 fm = 10⁻¹⁵ m (fermi). Nucleus radius: ~1–10 fm depending on A.

1.1 Important Terminology

Term	Definition	Example
Isotopes	Same Z, different A (same element)	¹H, ²H (D), ³H (T)
Isobars	Same A, different Z (diff elements)	³₁H and ³₂He
Isotones	Same N, different Z	¹₄C and ¹₅N (both N=8)
Isomers	Same Z and A, different energy states	Nuclear excited states

2. Mass Defect & Binding Energy

Mass defect (Δm): The mass of a nucleus is always LESS than the sum of masses of its individual nucleons. This “missing mass” is converted to binding energy via E=mc².

Mass defect:
Δm = [Zm_p + (A−Z)m_n] − M_nucleus

Binding Energy:
BE = Δm × c² = Δm × 931.5 MeV/u
1 u (atomic mass unit) = 1.66×10⁻²⁷ kg = 931.5 MeV/c²

Binding Energy per nucleon:
BE/A = measure of nuclear stability. Higher BE/A = more stable nucleus.

2.1 BE/A Curve — Key Features

Maximum BE/A ≈ 8.75 MeV/nucleon at A ≈ 56 (⁵⁶Fe — most stable nucleus).
Light nuclei (A<20): BE/A increases rapidly. Notable peaks at ⁴He, ¹²C, ¹⁶O (magic numbers).
Medium nuclei (A = 30–120): BE/A ≈ 8.0–8.75 (relatively flat — most stable region).
Heavy nuclei (A>120): BE/A decreases slowly to ~7.6 MeV for uranium.
Implication: Light nuclei can gain stability by fusion (moving right → higher BE/A). Heavy nuclei gain stability by fission (moving left → higher BE/A). Both release energy!

⚠️ NEET: Δm = [Zm_p+(A−Z)m_n]−M. BE = Δm×931.5 MeV. 1u = 931.5 MeV/c². Fe-56 = most stable. Higher BE/A = more stable. Density of ALL nuclei = 2.3×10¹⁷ kg/m³.

3. Nuclear Force

Nuclear force: The strong attractive force that holds nucleons together inside the nucleus. It is the strongest force in nature.

Short range: Effective only up to ~2–3 fm. Beyond this, it drops to zero rapidly.
Charge independent: Same between p-p, n-n, and p-n pairs.
Strongest force: ~100× stronger than electromagnetic force at nuclear distances.
Attractive but becomes repulsive at very short distances (<0.7 fm) → prevents nucleons from collapsing.
Saturated: Each nucleon interacts only with its nearest neighbours (not all other nucleons). This explains constant nuclear density and near-constant BE/A.
NOT a central force (unlike gravity and Coulomb). Depends on spin and other quantum properties.

4. Radioactivity

Radioactivity: Spontaneous disintegration of unstable nuclei with emission of radiation (α, β, γ). Discovered by Henri Becquerel (1896). Further studied by Marie and Pierre Curie.

4.1 Types of Radioactive Decay

Property	α-decay	β⁻-decay	γ-decay
Particle	⁴₂He (helium nucleus)	e⁻ + antineutrino	Photon (γ-ray)
Charge	+2e	−e	0
Mass	4u	~0	0
Z change	Z → Z−2	Z → Z+1	No change
A change	A → A−4	No change	No change
Penetration	Low (paper stops)	Medium (Al sheet stops)	High (lead/concrete)
Ionisation	Highest	Medium	Lowest
Speed	~c/10	~0.9c	c

4.2 Nuclear Reactions

α-decay: ^A_ZX → ^A−4_Z−2Y + ⁴₂He
β⁻-decay: ^A_ZX → ^A_Z+1Y + e⁻ + ν̄ (neutron → proton + electron + antineutrino)
β⁺-decay: ^A_ZX → ^A_Z−1Y + e⁺ + ν (proton → neutron + positron + neutrino)
γ-decay: ^A_ZX* → ^A_ZX + γ (excited nucleus → ground state + photon)

5. Radioactive Decay Law

N(t) = N₀ e^−λt (exponential decay)

Activity: A = λN = A₀e^−λt = dN/dt
Units: 1 Becquerel (Bq) = 1 decay/s | 1 Curie (Ci) = 3.7×10¹⁰ decays/s

Half-life: T_1/2 = ln2/λ = 0.693/λ
After n half-lives: N = N₀/2ⁿ = N₀(1/2)ⁿ

Mean life: τ = 1/λ = T_1/2/0.693 = 1.44×T_1/2

λ = decay constant (probability of decay per unit time)

After 1 half-life: 50% remains. After 2: 25%. After 3: 12.5%. After n: (1/2)ⁿ.
Radioactive decay is a statistical (probabilistic) process. We cannot predict when a particular nucleus will decay.
Half-life is independent of initial amount and external conditions (temperature, pressure, etc.).

⚠️ NEET (Very frequently asked!): N=N₀e^−λt. T_1/2=0.693/λ. After n half-lives: N₀/2ⁿ. Mean life = 1.44×T_1/2. Activity A=λN. 1 Ci=3.7×10¹⁰ Bq.

6. Nuclear Fission

Nuclear fission: A heavy nucleus splits into two lighter nuclei (of comparable mass) with release of energy and 2–3 neutrons. Discovered by Hahn and Strassmann (1938).

Example: ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n + ~200 MeV

Energy released per fission of U-235 ≈ 200 MeV
The released neutrons can cause further fission → chain reaction.

Critical mass: Minimum mass of fissile material needed for a sustained chain reaction.
Controlled fission: Nuclear reactor (moderator: heavy water/graphite slows neutrons; control rods: cadmium/boron absorb excess neutrons).
Uncontrolled fission: Atomic bomb.
Fissile materials: U-235, Pu-239, U-233.

7. Nuclear Fusion

Nuclear fusion: Two light nuclei combine to form a heavier nucleus with release of energy. Requires extremely high temperature (~10⁷ K) → called thermonuclear reaction.

Proton-proton cycle (Sun):
4 ¹H → ⁴He + 2e⁺ + 2ν + 26.7 MeV

Energy per fusion ≈ 26.7 MeV (for 4 protons → 1 He)

Energy per nucleon from fusion (~6.7 MeV/nucleon) > fission (~0.85 MeV/nucleon). Fusion releases MORE energy per unit mass!
Fusion is energy source of all stars including our Sun.
Hydrogen bomb uses uncontrolled fusion (triggered by fission bomb).
Controlled fusion not yet achieved on Earth (plasma confinement is the challenge).

7.1 Fission vs Fusion

Feature	Fission	Fusion
Nuclei	Heavy → lighter	Light → heavier
Energy/reaction	~200 MeV	~26.7 MeV
Energy/nucleon	~0.85 MeV	~6.7 MeV
Condition	Neutron bombardment	~10⁷ K temperature
Waste	Radioactive (dangerous)	Helium (safe)
Control	Achieved (reactors)	Not yet achieved

🎓 NEET Previous Year Questions

Q1. [NEET 2022] After 3 half-lives, what fraction of radioactive substance remains?

Answer N/N₀ = (1/2)³ = 1/8. So 7/8 has decayed.

Q2. [NEET 2021] The binding energy per nucleon is maximum for:

Answer Fe-56 (iron). BE/A ≈ 8.75 MeV/nucleon. Most stable nucleus.

Q3. [NEET 2020] In β⁻ decay, what happens inside the nucleus?

Answer A neutron converts to a proton: n → p + e⁻ + ν̄. Z increases by 1, A unchanged.

Q4. [NEET 2019] The radius of a nucleus with mass number A is proportional to:

Answer R = R₀A^1/3. So R ∝ A^1/3.

Q5. [NEET 2017] Half-life of a substance is 30 min. How long for 75% to decay?

Answer 75% decayed → 25% left = 1/4 = (1/2)². So n=2 half-lives = 60 minutes.

💡 Rapid Revision

R=R₀A^1/3 | ρ=2.3×10¹⁷ kg/m³ (constant!) | 1u=931.5 MeV/c²
Δm=[Zm_p+(A−Z)m_n]−M | BE=Δm×931.5 MeV | Fe-56: max BE/A
α: Z−2, A−4 | β⁻: Z+1, A same | γ: no change in Z or A
N=N₀e^−λt | T_1/2=0.693/λ | τ=1.44T_1/2 | After n T_1/2: N₀/2ⁿ
Fission: heavy→light (~200 MeV) | Fusion: light→heavy (~26.7 MeV)
Nuclear force: strongest, short range (~2–3 fm), charge independent

Numericals - Nuclei - Class 12

CLASS 12 PHYSICS | NCERT SOLUTIONS

Chapter 13 — Nuclei

22 NCERT Exercise & Exemplar Questions — Step-by-Step Solutions

📝 NCERT Exercise Questions (Q1 – Q12)

3 MarksQ1 (Ex 13.1). Find the radius of ¹⁹⁷Au and ¹⁰⁷Ag nuclei. (R₀=1.2 fm)

✓ Solution
R = R₀A^1/3
Au (A=197): R = 1.2×(197)^1/3 = 1.2×5.818 = 6.98 fm

Ag (A=107): R = 1.2×(107)^1/3 = 1.2×4.745 = 5.69 fm

3 MarksQ2 (Ex 13.2). Find the binding energy per nucleon of ⁵⁶Fe. Given: m(Fe) = 55.934939 u, m_p=1.007825 u, m_n=1.008665 u.

✓ Solution
Z=26, A=56, N=30
Δm = 26×1.007825 + 30×1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939 = 56.4634 − 55.934939 = 0.528461 u
BE = 0.528461×931.5 = 492.26 MeV

BE/A = 492.26/56 = 8.79 MeV/nucleon (highest — most stable!)

3 MarksQ3 (Ex 13.3). A radioactive isotope has half-life of 5 years. How long for its activity to reduce to 3.125% of original?

✓ Solution
3.125% = 3.125/100 = 1/32 = (1/2)⁵
So n = 5 half-lives

t = 5×T_1/2 = 5×5 = 25 years

3 MarksQ4 (Ex 13.4). α-particle has KE = 5.5 MeV. Find distance of closest approach to gold nucleus (Z=79).

✓ Solution
At closest approach: KE = k×(2e)(Ze)/d ⇒ d = 2kZe²/KE
d = 2×9×10⁹×79×(1.6×10⁻¹⁹)²/(5.5×10⁶×1.6×10⁻¹⁹)
= 2×9×10⁹×79×2.56×10⁻³⁸/(8.8×10⁻¹³)

d = 4.13×10⁻¹⁴ m ≈ 41.3 fm

3 MarksQ5 (Ex 13.5). The half-life of ⁹⁰Sr is 28 years. Find disintegration rate of 15 mg sample.

✓ Solution
N = (15×10⁻³/90)×6.023×10²³ = 1.004×10²⁰ atoms
λ = 0.693/T_1/2 = 0.693/(28×365.25×24×3600) = 0.693/8.83×10⁸ = 7.847×10⁻¹⁰ s⁻¹

A = λN = 7.847×10⁻¹⁰×1.004×10²⁰ = 7.878×10¹⁰ decays/s ≈ 7.88×10¹⁰ Bq

= 7.88×10¹⁰/3.7×10¹⁰ = 2.13 Ci

3 MarksQ6 (Ex 13.6). Find energy released in α-decay: ²²⁶Ra → ²²²Rn + ⁴He. Masses: Ra=226.02540 u, Rn=222.01750 u, He=4.00260 u.

✓ Solution
Δm = m(Ra) − [m(Rn) + m(He)] = 226.02540 − (222.01750 + 4.00260)
= 226.02540 − 226.02010 = 0.0053 u

Q = 0.0053×931.5 = 4.94 MeV

3 MarksQ7 (Ex 13.7). How many α and β particles are emitted when ²³²Th decays to ²⁰⁸Pb?

✓ Solution
Change in A: 232−208 = 24. Each α reduces A by 4 → number of α = 24/4 = 6
Change in Z: Th(90)→Pb(82). 6α would reduce Z by 12 → Z = 90−12 = 78.
But final Z=82 → need 82−78 = 4 increases ⇒ 4 β⁻ particles

6 α-particles and 4 β⁻-particles

3 MarksQ8 (Ex 13.8). Calculate the energy released in the fusion reaction: 4 ¹H → ⁴He + 2e⁺ + 2ν. Masses: H=1.007825 u, He=4.002603 u.

✓ Solution
Δm = 4×1.007825 − 4.002603 = 4.0313 − 4.002603 = 0.028697 u
(Positron mass included in atomic masses)

Q = 0.028697×931.5 = 26.72 MeV

3 MarksQ9 (Ex 13.9). Calculate the energy released per fission of ²³⁵U if products are ¹⁴¹Ba + ⁹²Kr + 3n. Masses: U=235.04393, Ba=140.91440, Kr=91.92630, n=1.00867 u.

✓ Solution
Δm = [m(U)+m(n)] − [m(Ba)+m(Kr)+3m(n)]
= (235.04393+1.00867) − (140.91440+91.92630+3×1.00867)
= 236.0526 − (140.91440+91.92630+3.02601) = 236.0526 − 235.86671 = 0.18589 u

Q = 0.18589×931.5 = 173.2 MeV (≈200 MeV including γ-rays and KE)

3 MarksQ10 (Ex 13.10). A radioactive sample has activity of 1000 disintegrations/min. After 5 min, activity is 500 disintegrations/min. Find (a) decay constant, (b) half-life.

✓ Solution
A = A₀e^−λt ⇒ 500 = 1000×e^−5λ
e^−5λ = 0.5 ⇒ −5λ = ln(0.5) = −0.693
(a) λ = 0.693/5 = 0.1386 per min

(b) T_1/2 = 0.693/λ = 0.693/0.1386 = 5 min

(Activity halved in 5 min → T_1/2=5 min, as expected.)

3 MarksQ11 (Ex 13.11). Nuclear density is constant for all nuclei. Calculate its value.

✓ Solution
ρ = mass/volume = Am_p/[(4/3)πR³] = Am_p/[(4/3)πR₀³A]
= m_p/[(4/3)πR₀³] (A cancels!)
= 1.67×10⁻²⁷/[(4/3)π(1.2×10⁻¹⁵)³]
= 1.67×10⁻²⁷/7.238×10⁻⁴⁵

ρ = 2.3×10¹⁷ kg/m³ (independent of A!)

3 MarksQ12 (Ex 13.12). The nucleus ²³Na has mass 22.9898 u. Find mass defect and binding energy per nucleon. (m_p=1.00783 u, m_n=1.00867 u)

✓ Solution
Z=11, N=12, A=23
Δm = 11×1.00783 + 12×1.00867 − 22.9898
= 11.08613 + 12.10404 − 22.9898 = 23.19017 − 22.9898 = 0.20037 u
BE = 0.20037×931.5 = 186.64 MeV

BE/A = 186.64/23 = 8.11 MeV/nucleon

🌟 Additional / Exemplar Questions (Q13 – Q22)

3 MarksQ13. A sample initially has 2.0×10¹⁶ radioactive atoms. After 24 hours, 0.125×10¹⁶ remain. Find half-life.

✓ Solution
N/N₀ = 0.125/2.0 = 1/16 = (1/2)⁴ ⇒ n=4 half-lives

T_1/2 = 24/4 = 6 hours

3 MarksQ14. Compare the radii of ²⁷Al and ¹²⁵Te nuclei.

✓ Solution
R ∝ A^1/3
R_Al/R_Te = (27/125)^1/3 = (27/125)^1/3

= 3/5 = 0.6

Te nucleus has radius 5/3 times that of Al.

3 MarksQ15. The decay constant of a radioactive sample is 0.0693 per day. Find half-life, mean life, and activity if N = 10¹⁰.

✓ Solution
T_1/2 = 0.693/λ = 0.693/0.0693 = 10 days
τ = 1/λ = 1/0.0693 = 14.43 days (= 1.44×T_1/2 ✓)

A = λN = 0.0693/(86400)×10¹⁰ = 8.02×10⁻⁷×10¹⁰ = 8020 decays/s ≈ 8.02 kBq

3 MarksQ16. How many α and β particles are emitted in: ²³⁸U → ²⁰⁶Pb?

✓ Solution
ΔA = 238−206 = 32. Number of α = 32/4 = 8
Z changes: U(92)→Pb(82). 8α reduce Z by 16 → Z=92−16=76. Need 82−76=6 β⁻

8 α-particles and 6 β⁻-particles

3 MarksQ17. Calculate the energy equivalent of 1 atomic mass unit in MeV.

✓ Solution
1 u = 1.66054×10⁻²⁷ kg
E = mc² = 1.66054×10⁻²⁷×(3×10⁸)²
= 1.66054×10⁻²⁷×9×10¹⁶ = 1.4945×10⁻¹⁰ J

= 1.4945×10⁻¹⁰/1.6×10⁻¹⁹ MeV = 931.5 MeV

3 MarksQ18. Activity of a sample decreases from 1200 Bq to 150 Bq in 9 hours. Find half-life.

✓ Solution
A/A₀ = 150/1200 = 1/8 = (1/2)³ ⇒ n=3

T_1/2 = 9/3 = 3 hours

3 MarksQ19. Find the Q-value of the reaction: ²H + ³H → ⁴He + n. Masses: ²H=2.014102 u, ³H=3.016049 u, ⁴He=4.002603 u, n=1.008665 u.

✓ Solution
Δm = (2.014102+3.016049) − (4.002603+1.008665)
= 5.030151 − 5.011268 = 0.018883 u

Q = 0.018883×931.5 = 17.59 MeV

Positive Q → energy is released (exothermic fusion reaction).

3 MarksQ20. Mean life of a radioactive sample is 100 s. Find probability that a nucleus will decay in 200 s.

✓ Solution
λ = 1/τ = 1/100 = 0.01 s⁻¹
Probability of survival in time t: P(survive) = e^−λt = e^−0.01×200 = e⁻² = 0.1353

Probability of decay = 1 − e⁻² = 1 − 0.1353 = 0.8647 ≈ 86.5%

3 MarksQ21. If the mass of a proton is 1.0073 u and neutron is 1.0087 u, find the BE/A of ⁴He (mass = 4.0015 u).

✓ Solution
Z=2, N=2
Δm = 2×1.0073 + 2×1.0087 − 4.0015 = 2.0146 + 2.0174 − 4.0015 = 0.0305 u
BE = 0.0305×931.5 = 28.41 MeV

BE/A = 28.41/4 = 7.1 MeV/nucleon

3 MarksQ22. How much energy is released if all atoms in 1 kg of U-235 undergo fission? (200 MeV per fission)

✓ Solution
Number of atoms = (1000/235)×6.023×10²³ = 2.563×10²⁴
Total energy = 2.563×10²⁴×200 MeV = 5.126×10²⁶ MeV
= 5.126×10²⁶×1.6×10⁻¹³ J = 8.2×10¹³ J

E = 8.2×10¹³ J ≈ 82 TJ (≈ same as 20,000 tonnes of TNT!)

✍ Score Guide — 22 Questions
NCERT Ex 13.1–13.12: 3 marks × 12 = 36 | Additional Q13–Q22: 3 marks × 10 = 30 | Total: 66 marks

Formula Capsule - Nuclei - Class 12

CLASS 12 PHYSICS | FORMULA CAPSULE

Nuclei

Chapter 13 — Complete Formula Sheet for NEET/JEE

📅 Section A — Nuclear Properties

Quantity	Formula	Key Note
Radius	R = R₀A^1/3	R₀=1.2 fm; R ∝ A^1/3
Volume	V = (4/3)πR³ ∝ A	V ∝ A (linear)
Density	ρ ≈ 2.3×10¹⁷ kg/m³	Constant for ALL nuclei!
Isotopes	Same Z, different A	Same element, diff mass
Isobars	Same A, different Z	Different elements
Isotones	Same N, different Z	Same neutron number

📅 Section B — Mass Defect & Binding Energy

Quantity	Formula	Key Note
Mass defect	Δm = Zm_p+(A−Z)m_n−M	Always positive
Binding energy	BE = Δm×931.5 MeV	1u = 931.5 MeV/c²
BE per nucleon	BE/A	Higher = more stable
Most stable	Fe-56: BE/A ≈ 8.75 MeV	Peak of BE/A curve
Q-value	Q = (Δm_reactants−Δm_products)c²	+Q = exothermic

📅 Section C — Radioactive Decay

Quantity	Formula	Key Note
Decay law	N = N₀e^−λt	Exponential decay
Activity	A = λN = A₀e^−λt	Bq or Ci
Half-life	T_1/2 = 0.693/λ	50% remains after T_1/2
After n half-lives	N = N₀/2ⁿ	n = t/T_1/2
Mean life	τ = 1/λ = 1.44 T_1/2	τ > T_1/2 always

📅 Section D — Decay Rules & α/β Count

Decay	ΔZ	ΔA	Particle
α-decay	−2	−4	⁴He
β⁻-decay	+1	0	e⁻ + ν̄
β⁺-decay	−1	0	e⁺ + ν
γ-decay	0	0	γ-photon

To find α & β count	Formula
Number of α	n_α = ΔA/4
Number of β⁻	n_β = Z_final − (Z_initial − 2n_α)

📅 Section E — Fission & Fusion

Property	Fission	Fusion
Process	Heavy → lighter	Light → heavier
Energy/reaction	~200 MeV (U-235)	~26.7 MeV (4H→He)
Energy/nucleon	~0.85 MeV	~6.7 MeV (higher!)
Condition	Neutron bombardment	~10⁷ K
Example	Nuclear reactor, A-bomb	Sun, H-bomb

🧠 Memory Tricks

“α = −2, −4 | β = +1, 0” Alpha: Z drops by 2, A drops by 4. Beta⁻: Z rises by 1, A unchanged. Gamma: nothing changes (just energy loss). For α/β counting: α from ΔA, then β adjusts Z.

“Half = Halve: 1/2, 1/4, 1/8, 1/16...” After 1 half-life: 1/2. After 2: 1/4. After 3: 1/8. After n: 1/2ⁿ. Quick trick: 3.125% = 1/32 = (1/2)⁵ = 5 half-lives. 6.25% = (1/2)⁴ = 4.

“τ = 1.44 × T_1/2” Mean life is always 1.44 times the half-life. Or: T_1/2 = 0.693τ. λ = 0.693/T_1/2 = 1/τ. These three are interconnected by ln2 = 0.693.

“1u = 931.5 MeV” Golden conversion! Mass in u × 931.5 = energy in MeV. m_p ≈ m_n ≈ 1u ≈ 931.5 MeV. BE = Δm × 931.5 MeV. Always use this for quick calculations.

“Fe-56 = King of Stability” Iron-56 has highest BE/A (~8.75 MeV). Left of Fe: fusion releases energy. Right of Fe: fission releases energy. Both move toward Fe on the BE/A curve!

“Nuclear density = constant!” ρ = 2.3×10¹⁷ kg/m³ for ALL nuclei (A cancels in the formula). A teaspoon of nuclear matter would weigh ~10¹² kg! Nuclear force saturates.

🔢 Critical Values

1u = 931.5 MeV/c² 1u = 1.66×10⁻²⁷ kg R₀ = 1.2 fm ρ = 2.3×10¹⁷ kg/m³ m_p = 1.00783 u m_n = 1.00867 u 1 Ci = 3.7×10¹⁰ Bq U-235 fission ≈ 200 MeV 4H→He fusion ≈ 26.7 MeV

❌ Common Mistakes to Avoid

β⁻ is NOT an orbital electron: In β⁻ decay, the electron comes from INSIDE the nucleus (n→p+e⁻+ν̄), not from the electron cloud.
γ changes neither Z nor A: γ-decay is just energy release from excited nucleus. No transmutation occurs.
Half-life ≠ mean life: τ = 1.44×T_1/2. Mean life is LONGER than half-life. Don't use them interchangeably.
Fission energy per reaction > fusion: But fusion energy per nucleon is HIGHER. Per kg of fuel, fusion gives much more energy.
Nuclear density is constant: Don't say “heavier nuclei are denser”. ALL nuclei have the same density (~2.3×10¹⁷ kg/m³).
BE/A ≠ BE: Fe-56 has highest BE/A but NOT highest total BE. Heavy nuclei have higher total BE but lower BE/A.

📱 Practice MCQs for this topic inside our App