SF Prep Notes
08 Physics Notes πŸ•’ Updated 2026-10-02
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Nuclei

Nuclei

1. The Nucleus

  • Discovered by Rutherford (Ξ±-scattering experiment).
  • Constituents : neutrons (n) and protons (p) – together called nucleons.
  • Neutron : neutral particle, discovered by J. Chadwick.
  • Proton : charge = +e, discovered by Goldstein.
  • Masses are nearly the same (mn slightly > mp) :
mn β‰ˆ mp β‰ˆ 1 amu (u)
1 u = 112 Γ— (mass of one 12C atom) β‰ˆ 1.66 Γ— 10βˆ’27 kg
electron nucleus proton neutron

2. Nuclide Notation

AZX
SymbolMeaning
Zatomic number = no. of protons
Amass number = no. of nucleons
N = A βˆ’ Zno. of neutrons

e.g. 126C β†’ 6 p, 6 n ;  42He β†’ 2 p, 2 n ;  23892U β†’ 92 p, 146 n

NameSameExample
IsotopesZ23592U , 23892U
IsobarsA31H , 32He
IsotonesN = A βˆ’ Z19880Hg , 19779Au
Q. Which pair are isobars? (JEE Main, 23 Jan 2026, Shift 2)
(1) 21H, 31H   (2) 23692U, 23892U   (3) 19880Hg, 19779Au   (4) 31H, 32He
(1), (2) β†’ same Z (isotopes) ; (3) β†’ N = 118 for both (isotones) ; (4) β†’ A = 3 for both.
Ans : (4)

3. Size of the Nucleus

R = R0 A1/3

R0 β‰ˆ 1.2 fm = 1.2 Γ— 10βˆ’15 m (some books use 1.25 fm) ;  1 fm = 10βˆ’15 m

Volume :

V = 43Ο€RΒ³ = 43Ο€R0Β³ A

V ∝ A

Density :

ρ = massvolume = A mp43Ο€R0Β³ A

ρ = 3 mp4Ο€ R0Β³ = constant
  • Independent of A β‡’ all nuclei have nearly the same density.
  • With R0 = 1.2 fm, mp = 1.67 Γ— 10βˆ’27 kg :
ρnucleus β‰ˆ 2.3 Γ— 1017 kg mβˆ’3

(β‰ˆ 1014 times the density of water)

4. Nuclear Force

  • Attractive ; holds nucleons together in spite of the repulsion between protons.
  • Acts between n–n, n–p and p–p (i.e. between nucleons).
  • Strongest force within nuclear dimensions (Fn β‰ˆ 100 Fe).
  • Short range – acts only inside the nucleus (range β‰ˆ 2–3 fm).
  • Charge independent – same for n–n, n–p, p–p (does not depend on the nature of nucleons).
p n p p–p also same attractive nuclear force

Key Points :

  • R ∝ A1/3 , V ∝ A , ρ independent of A.
  • Isotopes – same Z ; isobars – same A ; isotones – same N.
  • Nuclear force : strongest, short range, charge independent.
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Nuclei

5. Mass–Energy Equivalence

  • A rest mass m is equivalent to energy E (Einstein) :
E = mcΒ²

c = 3 Γ— 108 m/s (speed of light)

Energy of 1 u :

E = (1.66 Γ— 10βˆ’27)(3 Γ— 108)Β² J = 1.49 Γ— 10βˆ’10 J

   = 1.49 Γ— 10βˆ’101.6 Γ— 10βˆ’19 eV β‰ˆ 931 Γ— 106 eV

1 u β‰ˆ 931.5 MeV/cΒ²  (β‰ˆ 931 MeV/cΒ²)

i.e. (1 u) Γ— cΒ² = 931.5 MeV

6. Mass Defect (Ξ”m)

  • Rest mass of a nucleus is smaller than the sum of rest masses of its nucleons. The difference is the mass defect.
  • Ξ”m = (expected mass) βˆ’ (actual mass)

For AZX : nuclear mass mX = MX βˆ’ Z me  (MX = atomic mass)

Ξ”m = [Z mp + (A βˆ’ Z) mn] βˆ’ [MX βˆ’ Z me]

     = Z(mp + me) + (A βˆ’ Z) mn βˆ’ MX

Ξ”m = Z MH + (A βˆ’ Z) mn βˆ’ MX

MH = mp + me = mass of H-atom  (lowercase m β†’ nuclear mass, capital M β†’ atomic mass)

Ξ”m β‰ˆ Z mp + (A βˆ’ Z) mn βˆ’ MX
e.g. 126C :  Ξ”m = 6MH + 6mn βˆ’ MC β‰ˆ 6mp + 6mn βˆ’ MC

7. Binding Energy (B.E.)

  • Energy required to break the nucleus into its constituent nucleons.
  • = Energy released when the nucleus is formed from its nucleons.
B.E. = (Ξ”m) cΒ²
B.E. = [Z MH + (A βˆ’ Z) mn βˆ’ MX]in u Γ— 931.5 MeV

8. B.E. per Nucleon & Stability

Stability ∝ B.E.A  (B.E. per nucleon)
  • Higher B.E./A β‡’ more tightly bound β‡’ more stable nucleus.
B.E./A (MeV) A 8.8 4He 2H 56Fe (peak) 238U fusion β†’ ← fission 56120238 0
  • B.E./A rises sharply for light nuclei, has a maximum β‰ˆ 8.8 MeV near A = 56 (56Fe), then slowly falls (β‰ˆ 7.6 MeV for 238U).
  • For a wide range (A β‰ˆ 30–170) B.E./A β‰ˆ 8 MeV is nearly constant (nuclear force is short range – saturation).
  • 4He (β‰ˆ 7.1 MeV) lies above its neighbours (6Li β‰ˆ 5.3 MeV) β‡’ very stable.
  • Heavy nuclei β†’ move towards A β‰ˆ 56 by fission ; light nuclei β†’ by fusion. Both release energy.

9. Fission & Fusion

Fission :

  • A heavy nucleus breaks into 2 (or more) medium nuclei to become more stable.

10n + 23592U β†’ 23692U* β†’ 14456Ba + 8936Kr + 3 10n + Q

(check : A : 236 = 144 + 89 + 3 ; Z : 92 = 56 + 36) ; Q β‰ˆ 200 MeV per fission.

Fusion :

  • Light nuclei fuse to form a heavier (more stable) nucleus.

21H + 31H β†’ 42He + 10n + 17.6 MeV

(deuterium + tritium β†’ helium + neutron)

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Nuclei

10. Radioactivity

  • Spontaneous disintegration of unstable nuclei to form more (energetically) stable nuclei.
  • Types : (i) Ξ± decay (ii) Ξ²βˆ’ and Ξ²+ decay (iii) Ξ³ decay ; along with emission of neutrino / antineutrino in Ξ² decays.
ParticleChargeMassNature
Ξ±+2eβ‰ˆ 4 u42He2+ nucleus
Ξ²βˆ’βˆ’e9.1 Γ— 10βˆ’31 kgelectron 0βˆ’1e
Ξ²++e9.1 Γ— 10βˆ’31 kgpositron 0+1e
Ξ³00 (no rest mass)photon
neutrino Ξ½0β‰ˆ 0 (very tiny)emitted when a neutron is formed (p β†’ n)
antineutrino Ξ½Μ„0β‰ˆ 0 (very tiny)emitted when a neutron breaks (n β†’ p)
Penetrating power : Ξ³ > Ξ² > Ξ±
Ionising power : Ξ± > Ξ² > Ξ³

1 u = 1.66 Γ— 10βˆ’27 kg ; e = 1.6 Γ— 10βˆ’19 C

11. Neutrino & Antineutrino

  • Fundamental particles, no charge and extremely tiny mass.
  • Called "ghost particles" – they interact so weakly that trillions pass through the Earth without interacting.

12. Antiparticles

  • An antiparticle has the same mass but opposite charge (and opposite other quantum numbers) as the particle.
  • A particle and its antiparticle can annihilate – their mass converts completely into energy.
  • Electron (Ξ²βˆ’) & positron (Ξ²+) are antiparticles ; neutrino (Ξ½) & antineutrino (Ξ½Μ„) are antiparticles.
eβˆ’ + e+ β†’ Ξ³ + Ξ³  (annihilation)

13. General Nuclear Reaction

AZX β†’ A₁Z₁Y + n₁α + nβ‚‚Ξ²βˆ’ + n₃β+ + nβ‚„Ξ½ + nβ‚…Ξ½Μ„ + n₆γ + Q

  • Given X and the emitted particles, find Y using conservation of Z and A :
EmissionChange in ZChange in A
n₁ Ξ± (42He)βˆ’2nβ‚βˆ’4n₁
nβ‚‚ Ξ²βˆ’ (0βˆ’1e)+nβ‚‚0
n₃ Ξ²+ (0+1e)βˆ’n₃0
Ξ½, Ξ½Μ„, Ξ³, Q00
Z₁ = Z βˆ’ 2n₁ + nβ‚‚ βˆ’ n₃
A₁ = A βˆ’ 4n₁
e.g. 23892U emits 8 Ξ± and 6 Ξ²βˆ’ :
A₁ = 238 βˆ’ 32 = 206 ; Z₁ = 92 βˆ’ 16 + 6 = 82 β‡’ 20682Pb

14. Conserved in a Nuclear Reaction

  • Atomic number (total charge number Z)
  • Mass number A (total no. of nucleons)
  • Charge
  • Linear momentum
  • Total (mass-energy + energy) – rest mass alone is not conserved
  • Angular momentum (spin)

15. Energy Released (Q)

A + B β†’ C + D + Q

  • Q > 0 β†’ exothermic, energetically favourable (can occur spontaneously).
  • Q < 0 β†’ endothermic, energetically not favourable (needs energy input).

Key Points :

  • Ξ± : Z βˆ’ 2, A βˆ’ 4 ; Ξ²βˆ’ : Z + 1 ; Ξ²+ : Z βˆ’ 1 ; Ξ³ : no change.
  • Penetration Ξ³ > Ξ² > Ξ± ; ionisation Ξ± > Ξ² > Ξ³.
  • eβˆ’/e+ and Ξ½/Ξ½Μ„ are particle–antiparticle pairs.
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Nuclei

16. Q in terms of Mass

Q = [(mass of reactants) βˆ’ (mass of products)] cΒ²

Q = [(mA + mB) βˆ’ (mC + mD)] cΒ²

Q = [(mA + mB) βˆ’ (mC + mD)]in u Γ— 931.5 MeV

17. Q in terms of B.E.

Q = (B.E. of products) βˆ’ (B.E. of reactants)
  • Usually B.E./A is given β‡’ B.E. = B.E.A Γ— A
e.g. 21H + 31H β†’ 42He + 10n. B.E./A : 2H = 1.11 MeV, 3H = 2.83 MeV, 4He = 7.07 MeV.
B.E.(products) = 4 Γ— 7.07 = 28.28 MeV (free n has no B.E.)
B.E.(reactants) = 2 Γ— 1.11 + 3 Γ— 2.83 = 2.22 + 8.49 = 10.71 MeV
Q = 28.28 βˆ’ 10.71 β‰ˆ 17.6 MeV

18. K.E. of Products

X (at rest) β†’ Y + Z + Q

X rest Y p, KY Z p, KZ
  • Q appears as K.E. of products : KY + KZ = Q
  • Momentum conservation : 0 = pZ βˆ’ pY β‡’ pY = pZ = p
  • K = pΒ²2m β‡’ K ∝ 1m (same p)
KYKZ = mZmY
KY = mZmY + mZ Q  ,  KZ = mYmY + mZ Q

The lighter product carries most of the energy.

19. Ξ± Decay

  • Unstable nucleus emits an Ξ±-particle β‡’ mass number decreases and the nucleus moves towards stability.
  • Shown mainly by heavy nuclei (A > 210) ; B.E./A increases, so Q is positive.
AZX β†’ Aβˆ’4Zβˆ’2Y + 42He + Q

e.g. 23892U β†’ 23490Th + 42He + Q

Q = [MX βˆ’ MY βˆ’ MHe] cΒ²
B.E./AA 210 Ξ± decay β†’ smaller A, higher B.E./A

Sharing of Q (X at rest) :

Masses ∝ mass numbers : mY ∝ (A βˆ’ 4), mΞ± ∝ 4

KΞ± = A βˆ’ 4A Q  ,  KY = 4A Q

e.g. 238U : KΞ± = 234238 Q β‰ˆ 0.983 Q

  • Ξ±-particles from a given decay have a definite (discrete) energy (two-body decay).

Key Points :

  • Q = (Ξ” mass) Γ— 931.5 MeV/u = B.E.(products) βˆ’ B.E.(reactants).
  • Two-body break-up from rest : equal & opposite momenta, K ∝ 1/m.
  • KΞ± = (A βˆ’ 4)Q/A – Ξ± takes almost all of Q.
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Nuclei

20. Ξ²βˆ’ Decay

  • In Ξ² decay the N/Z ratio changes (shown by unstable nuclei).
  • In Ξ²βˆ’ decay a neutron is converted into a proton (nucleus with excess neutrons).
AZX β†’ AZ+1Y + Ξ²βˆ’ + Ξ½Μ„ + Q

Basic process : 10n β†’ 11p + 0βˆ’1e + Ξ½Μ„ + Q

e.g. 146C β†’ 147N + Ξ²βˆ’ + Ξ½Μ„ + Q

Q = [MX βˆ’ MY] cΒ²  (atomic masses)
  • Q is shared randomly between eβˆ’ and Ξ½Μ„ (three-body decay ; recoil of Y negligible) :
0 ≀ Ke ≀ Q  ,  0 ≀ EΞ½Μ„ ≀ Q

β‡’ Ξ²-particles have a continuous energy spectrum.

Stability curve (N vs Z) :

NZ N = Z stable nuclei n-rich : Ξ²βˆ’ decay p-rich : Ξ²+ / K-capture

Light stable nuclei have N β‰ˆ Z (4He, 12C) ; heavy stable nuclei have N > Z (238U : 92 p, 146 n).

21. Ξ²+ Decay

  • A proton is converted into a neutron (nucleus with excess protons).
AZX β†’ AZβˆ’1Y + Ξ²+ + Ξ½ + Q

Basic process (inside the nucleus) : 11p β†’ 10n + 0+1e + Ξ½

(a free proton cannot do this since mp < mn ; the energy comes from the nucleus)

e.g. 106C β†’ 105B + Ξ²+ + Ξ½ + Q

Q = [MX βˆ’ MY βˆ’ 2me] cΒ²

= (MX βˆ’ MY)cΒ² βˆ’ 2mecΒ² = (MX βˆ’ MY)cΒ² βˆ’ 2(0.511 MeV)

(using atomic masses ; 2mecΒ² β‰ˆ 1.02 MeV)

  • Q is shared randomly between Ξ²+ and Ξ½.

22. K-Capture (Electron Capture)

  • Nucleus captures an electron from the nearest (K) shell β‡’ a proton becomes a neutron (N/Z increases).
AZX + 0βˆ’1e β†’ AZβˆ’1Y + Ξ½ + Q

Basic : 11p + 0βˆ’1e β†’ 10n + Ξ½

e.g. 4019K + 0βˆ’1e β†’ 4018Ar + Ξ½ + Q

Q = [MX βˆ’ MY] cΒ²
Ze eβˆ’ K-shell

23. Ξ³ Decay

  • After Ξ± or Ξ² decay, the daughter nucleus is often left in an excited state ; it comes to the ground state by emitting Ξ³-photon(s).
AZY* β†’ AZY + Ξ³

e.g. AZX β†’ Aβˆ’4Zβˆ’2Y* + 42He, then Y* β†’ Y + Ξ³.  No change in A or Z.

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24. Law of Radioactive Decay

  • (Rutherford & Soddy) Rate of disintegration ∝ number of active (undecayed) nuclei present.
X→ Y+ 2Z
t = 0N000
time tNN0 βˆ’ N2(N0 βˆ’ N)

dNdt ∝ N β‡’

dNdt = βˆ’Ξ»N
  • Ξ» = decay constant – depends only on the nature of the nucleus ; not on temperature, pressure, concentration etc.

∫Nβ‚€N dNN = βˆ’βˆ«0t Ξ» dt β‡’ ln N βˆ’ ln N0 = βˆ’Ξ»t

N = N0 eβˆ’Ξ»t  (remaining)
Nβ€² = N0 βˆ’ N = N0(1 βˆ’ eβˆ’Ξ»t)  (decayed)
ln N = ln N0 βˆ’ Ξ»t  ;  t = 1Ξ» lnN0N
N vs t
N0 t
Nβ€² vs t
N0 t
ln N vs t (straight line)
ln N0 t |slope| = Ξ»

25. Activity (A or R)

  • Number of nuclei decaying per second : A = |dN/dt|
A = Ξ»N = Ξ»N0eβˆ’Ξ»t = A0 eβˆ’Ξ»t
  • Same Ξ» but more nuclei (bigger sample) β‡’ larger activity.
UnitValue
SI : becquerel (Bq)1 Bq = 1 decay/s (dps)
1 rutherford (Rd)106 dps
1 curie (Ci)3.7 Γ— 1010 dps

26. Half-life & Mean Life

  • Half-life : time in which half of the active nuclei decay.

N02 = N0eβˆ’Ξ»T β‡’ eβˆ’Ξ»T = Β½ β‡’ Ξ»T = ln 2

TΒ½ = ln 2Ξ» = 0.693Ξ»
Mean life Ο„ = 1Ξ» = TΒ½ln 2 β‰ˆ 1.44 TΒ½

After n half-lives (t = nTΒ½) :

N0 β†’ N02 β†’ N04 β†’ N08 β†’ …

N = N02n ,  A = A02n ,  Nβ€² = N0(1 βˆ’ 12n)
Q. Activity of a sample falls from 1800 sβˆ’1 to 1200 sβˆ’1 in 40 min. Find TΒ½ (ln 2 β‰ˆ 0.7, ln 3 β‰ˆ 1.1).
Ξ»t = lnA0A  and  Ξ» = ln 2TΒ½ β‡’ TΒ½ = (ln 2) tln(A0/A)
ln18001200 = ln32 = 1.1 βˆ’ 0.7 = 0.4
TΒ½ = 0.7 Γ— 400.4 = 70 min
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Nuclei

27. Writing Rate Equations

  • Rule : dN/dt of a nuclide = (rate of formation) βˆ’ (rate of decay).

Ex. 1 : Many inputs / outputs

Aλ₁ E BΞ»β‚‚ λ₃C Ξ»β‚„D
dNEdt = (λ₁NA + Ξ»β‚‚NB) βˆ’ (λ₃ + Ξ»β‚„)NE

Ex. 2 : Production at constant rate R

Factory β†’ (R per second) β†’ X β€”Ξ»β†’ Y ;  take NX = 0 at t = 0

dNXdt = R βˆ’ Ξ»NX  ,  dNYdt = Ξ»NX

Solving : NX = RΞ»(1 βˆ’ eβˆ’Ξ»t)  β†’ R/Ξ» as t β†’ ∞ (steady state)

NY = ∫0t R(1 βˆ’ eβˆ’Ξ»t) dt = R[t + 1Ξ»(eβˆ’Ξ»t βˆ’ 1)]

NY = Rt βˆ’ NX = Rt βˆ’ RΞ»(1 βˆ’ eβˆ’Ξ»t)

(total produced Rt = NX + NY)

Ex. 3 : Chain X —λ₁→ Y β€”Ξ»β‚‚β†’ Z

dNXdt = βˆ’Ξ»β‚NX ; dNYdt = λ₁NX βˆ’ Ξ»β‚‚NY ; dNZdt = Ξ»β‚‚NY

28. Parallel Decay

Q. X decays into Y (λ₁) or into Z (Ξ»β‚‚). Find the half-life of X.
dNXdt = βˆ’Ξ»β‚NX βˆ’ Ξ»β‚‚NX = βˆ’(λ₁ + Ξ»β‚‚)NX
β‡’ Ξ»eq = λ₁ + Ξ»β‚‚  ,  Teq = ln 2λ₁ + Ξ»β‚‚
Ξ»eq = λ₁ + Ξ»β‚‚  β‡’  1Teq = 1T₁ + 1Tβ‚‚

29. Successive Decay X β†’ Y β†’ Z

Q. X decays to Y (λ₁) ; Y decays to stable Z (Ξ»β‚‚). At t = 0 : NX = N0, NY = NZ = 0. Find NX, NY, NZ at time t and (NY)max.

X : dNX/dt = βˆ’Ξ»β‚NX β‡’

NX = N0eβˆ’Ξ»β‚t

Y : dNYdt + Ξ»β‚‚NY = λ₁N0eβˆ’Ξ»β‚t  (1st-order linear)

Multiply by eΞ»β‚‚t : ddt(NYeΞ»β‚‚t) = λ₁N0e(Ξ»β‚‚βˆ’Ξ»β‚)t

Integrate (0 β†’ t) : NYeΞ»β‚‚t = λ₁N0Ξ»β‚‚ βˆ’ λ₁(e(Ξ»β‚‚βˆ’Ξ»β‚)t βˆ’ 1)

NY = λ₁N0Ξ»β‚‚ βˆ’ λ₁(eβˆ’Ξ»β‚t βˆ’ eβˆ’Ξ»β‚‚t)
NZ = N0 βˆ’ NX βˆ’ NY
  • NX continuously decreases ; NZ continuously increases ; NY first increases, then decreases.
N0t tm NX NY NZ

Maximum NY : dNY/dt = 0

λ₁NX = Ξ»β‚‚NY β‡’ λ₁N0eβˆ’Ξ»β‚t = λ₂λ₁N0Ξ»β‚‚ βˆ’ λ₁(eβˆ’Ξ»β‚t βˆ’ eβˆ’Ξ»β‚‚t)

β‡’ Ξ»β‚‚eβˆ’Ξ»β‚‚t = λ₁eβˆ’Ξ»β‚t β‡’ e(Ξ»β‚‚βˆ’Ξ»β‚)t = λ₂λ₁

tm = 1Ξ»β‚‚ βˆ’ λ₁ lnλ₂λ₁

Then NY = λ₁λ₂NX = λ₁λ₂N0eβˆ’Ξ»β‚tm , with eβˆ’Ξ»β‚tm = (λ₁/Ξ»β‚‚)λ₁/(Ξ»β‚‚βˆ’Ξ»β‚)

(NY)max = N0 (λ₁/Ξ»β‚‚)Ξ»β‚‚/(Ξ»β‚‚ βˆ’ λ₁)

Check : λ₁ = Ξ», Ξ»β‚‚ = 2Ξ» β‡’ tm = ln2/Ξ», (NY)max = N0(Β½)Β² = N0/4 βœ“

Key Points :

  • N = N0eβˆ’Ξ»t ; TΒ½ = 0.693/Ξ» ; Ο„ = 1/Ξ» ; A = Ξ»N.
  • Parallel decay : Ξ»'s add ; series chain : write formation βˆ’ decay.
  • NY is maximum when λ₁NX = Ξ»β‚‚NY.