JEE Main chapters · Class 12 · Physics

Atoms and Nuclei: JEE Main Physics formulas and notes

Atoms and Nuclei is a high-scoring and quick chapter in JEE Mains: 2 or 3 questions almost every year, mostly short numerical questions. The favourites are the energy levels of hydrogen and hydrogen-like ions, spectral series and the number of lines, binding energy and the Q value, radioactive decay with half-life and activity, and the number of α and β particles. Learn the formulas of Bohr and of decay by heart and practise the ratio questions; about 7 hours spread over 5 days is enough.

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Rutherford: nucleus 10⁻¹⁵ m, atom 10⁻¹⁰ m; r₀ = 2Ze²/(4πε₀K) = 2Z × 1.44/K(MeV) fm; b = (r₀/2) cot(θ/2); N ∝ 1/sin⁴(θ/2)
Bohr: mvr = nh/2π; r = 0.529 n²/Z Å; v = 2.19 × 10⁶ Z/n m/s; E = −13.6 Z²/n² eV; K : U : E = 1 : −2 : −1
T ∝ n³/Z², I ∝ Z²/n³, μ ∝ n, 2πr = nλ (λ = 3.32 n Å in hydrogen)
1/λ = RZ²(1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹, 1/R = 91.2 nm, λ(nm) = 1240/E(eV)
Series: Lyman n₁ = 1 (UV), Balmer 2 (visible), Paschen 3, Brackett 4, Pfund 5 (all IR); lines from level n: n(n − 1)/2
Ionisation 13.6 Z² eV; first excitation 10.2 Z² eV; a photon needs an exact match, an electron can give part; hν > 13.6 leaves K = hν − 13.6
A = Z + N; R = 1.2 A^1/3 fm; density 2.3 × 10¹⁷ kg/m³ independent of A; 1 u = 931.5 MeV; nuclear force strong, short range, charge independent
Δm = Zmₚ + (A − Z)mₙ − M; BE = Δm × 931.5 MeV; curve peaks near A = 56 (8.8 MeV); Q = ΣBE(products) − ΣBE(reactants)
Fission ≈ 200 MeV (k < 1, = 1, > 1; moderator, control rods); fusion needs 10⁷ K, d + t gives 17.6 MeV
N = N₀e^−λt = N₀/2^t/T; T = 0.693/λ; τ = 1.44 T; A = λN; 1 Ci = 3.7 × 10¹⁰ Bq; nα = ΔA/4; Kα = Q(A − 4)/A; parallel λ = λ₁ + λ₂

Formulas with solved examples

Rutherford model and alpha scattering

Thomson and Rutherford models

Thomson: a positive sphere of radius about 10⁻¹⁰ m with electrons embedded in it
Rutherford (Geiger and Marsden, α rays on gold foil): most α particles go straight through, a few are scattered through large angles, about 1 in 8000 turns back
so the positive charge and nearly all the mass sit in a tiny nucleus of size about 10⁻¹⁵ m
atom size about 10⁻¹⁰ m
the electrons move around the nucleus
failures: an electron going round a nucleus should radiate and spiral in, and the model cannot explain the line spectrum

How to apply. Link every observation with a conclusion: straight through means mostly empty space; large angle means a small, heavy, positive centre; back-scattering means a single strong repulsion.

Example. If the atom were 100 m wide the nucleus would be 100 × 10⁻⁵ m = 1 mm across.

Distance of closest approach

head-on, the α particle stops where all its kinetic energy has become electric potential energy: K = (1)/(4πε₀)((2e)(Ze))/(r₀)
r₀ = (2Ze²)/(4πε₀K) = (2Z × 1.44 MeV fm)/(K in MeV)
r₀ ∝ (1)/(K) ∝ (1)/(v²)
for a proton of the same K the distance is half, since its charge is e instead of 2e

How to apply. Write K = kq₁q₂/r₀ and solve. If the speed is doubled, K becomes four times and r₀ becomes one fourth. A heavier projectile with the same K does not change r₀; only its charge does.

Example. A 5 MeV α particle on gold (Z = 79): r₀ = 2 × 79 × 1.44 ÷ 5 = 45.5 fm, much larger than the nuclear radius of gold (about 7 fm).

Impact parameter and angle of scattering

b = (r₀)/(2) cot(θ)/(2) = (kq₁q₂)/(2K) cot(θ)/(2)
b = 0 gives θ = 180° (the α particle returns along its path)
θ = 90° gives b = (r₀)/(2)
a large b gives a small θ
b is the perpendicular distance from the nucleus to the original line of motion

How to apply. Find r₀ first and then use b = (r₀/2) cot(θ/2). The smaller the impact parameter, the larger the deflection.

Example. Same 5 MeV α on gold, θ = 90°: b = 45.5 ÷ 2 = 22.8 fm.

Number of particles scattered

N(θ) ∝ (1)/(sin⁴(θ/2))
N ∝ Z²
N ∝ foil thickness
N ∝ (1)/(K²)
very few particles are scattered at large angles, most at small angles
α particles of 5 MeV to 8 MeV are used, so that they reach the nucleus

How to apply. Take the ratio of the two values of sin⁴(θ/2). Do not add the other factors if they are the same for both angles.

Example. N₆₀ ÷ N₁₂₀ = sin⁴60° ÷ sin⁴30° = 0.5625 ÷ 0.0625 = 9, so nine times as many α particles are counted at 60° as at 120°.

Bohr model of hydrogen

Postulates of Bohr

an electron moves in certain stationary circular orbits without radiating
the Coulomb force gives the centripetal force: (mv²)/(r) = (kZe²)/(r²)
the angular momentum is quantised: mvr = (nh)/(2π) (n = 1, 2, 3 …)
radiation is emitted or absorbed only in a jump between two orbits: hf = Ei − Ef
valid only for one-electron species: H, He⁺, Li²⁺
it fails for atoms with more electrons and does not explain the fine structure or the intensity of lines

How to apply. Combine the two equations (force and angular momentum) to get r, v and E. Do not use the postulates for helium atom or lithium atom.

Example. The angular momentum in n = 3 is 3h ÷ 2π = 3 × 6.63 × 10⁻³⁴ ÷ 6.28 = 3.17 × 10⁻³⁴ J s.

Radius of the nth orbit

rₙ = a₀(n²)/(Z)
a₀ = (ε₀h²)/(πme²) = 0.529 Å
rₙ ∝ (n²)/(Z)
radii of hydrogen in the ratio 1 : 4 : 9 : 16
r₁ of hydrogen = 0.529 Å, of He⁺ = 0.265 Å, of Li²⁺ = 0.176 Å

How to apply. Put in the numbers a₀ = 0.529 Å, n and Z. For ratios write rₙ ∝ n²/Z only.

Example. Li²⁺ in n = 2: r = 0.529 × 4 ÷ 3 = 0.705 Å. Hydrogen in n = 3: r = 0.529 × 9 = 4.76 Å.

Speed in the nth orbit

vₙ = (Ze²)/(2ε₀hn) = 2.19 × 10⁶ (Z)/(n) m/s = (Z)/(n)αc with α = (1)/(137)
vₙ ∝ (Z)/(n)
in hydrogen the speed in n = 1 is c ÷ 137
the speed is much lower than c, so no relativity is needed

How to apply. The speed is the quickest quantity to get: 2.19 × 10⁶ times Z over n.

Example. He⁺ (Z = 2) in n = 2: v = 2.19 × 10⁶ × 2 ÷ 2 = 2.19 × 10⁶ m/s.

Energy of the nth level

Eₙ = −13.6 (Z²)/(n²) eV
kinetic energy K = +13.6 (Z²)/(n²) eV
potential energy U = −27.2 (Z²)/(n²) eV
total E = −K = (U)/(2)
E is negative, so the electron is bound; E = 0 at n = ∞ (free electron at rest)
Eₙ in joule: −2.18 × 10⁻¹⁸ Z²/n²

How to apply. Learn the three ratios K : U : E = +1 : −2 : −1. An electron’s energy rises (becomes less negative) as n increases.

Example. Li²⁺ in the ground state: E = −13.6 × 9 = −122.4 eV, K = 122.4 eV and U = −244.8 eV.

Period, current and magnetic moment

T = (2πr)/(v) ∝ (n³)/(Z²)
f ∝ (Z²)/(n³)
I = ef ∝ (Z²)/(n³)
orbital magnetic moment μ = IA = (e)/(2m)L = n(eh)/(4πm) ∝ n (Bohr magneton 9.27 × 10⁻²⁴ A m²)
field at the nucleus B = (μ₀I)/(2r) ∝ (Z³)/(n⁵)

How to apply. Write each quantity in terms of r and v first, then substitute r ∝ n²/Z and v ∝ Z/n. Take the ratio of two cases.

Example. The time period in n = 2 against n = 1 in hydrogen is 2³ : 1 = 8 : 1. In the ground state of hydrogen T = 2π × 0.529 × 10⁻¹⁰ ÷ 2.19 × 10⁶ = 1.5 × 10⁻¹⁶ s.

De Broglie wave in a Bohr orbit

2πrₙ = nλ
λ = (h)/(mv)
this gives mvr = (nh)/(2π)
λₙ = (2πrₙ)/(n) = 3.32 n Å for hydrogen
the number of waves around the orbit is n

How to apply. The idea of de Broglie explains why only certain orbits are allowed: a whole number of wavelengths must fit.

Example. Hydrogen in n = 3: λ = 3.32 × 3 = 9.96 Å, and three wavelengths fit around the orbit of circumference 2π × 4.76 = 29.9 Å.

Hydrogen spectrum

Rydberg formula

(1)/(λ) = RZ²((1)/(n₁²) − (1)/(n₂²)) with n₂ > n₁
R = 1.097 × 10⁷ m⁻¹
(1)/(R) = 91.2 nm
Rhc = 13.6 eV
for hydrogen Z = 1
energy of the photon E = 13.6 Z²((1)/(n₁²) − (1)/(n₂²)) eV, and λ(nm) = (1240)/(E in eV)

How to apply. Put n₁ for the lower level (the series) and n₂ for the upper level. Do the bracket as a fraction before inverting.

Example. Hα line (n₂ = 3 to n₁ = 2): 1/λ = 1.097 × 10⁷ × 5/36, so λ = 656 nm (red).

Series of hydrogen

Lyman: n₁ = 1, ultraviolet
Balmer: n₁ = 2, visible (and near ultraviolet)
Paschen: n₁ = 3, infrared
Brackett: n₁ = 4, infrared
Pfund: n₁ = 5, infrared
longest wavelength: n₂ = n₁ + 1
shortest wavelength (series limit): n₂ = ∞, λ = (n₁²)/(R)
Lyman 121.6 nm to 91.2 nm
Balmer 656.3 nm to 364.6 nm
Paschen 1875 nm to 820.6 nm

How to apply. Learn which n₁ gives which series and the position in the spectrum. Only the Balmer series has visible lines in hydrogen.

Example. Balmer: longest 656.3 nm, shortest 364.6 nm, ratio 9 : 5. Lyman longest ÷ Balmer longest = (4/3) ÷ (36/5) = 5 : 27.

Number of spectral lines

electrons in many atoms jumping down from the level n to all lower levels: number of lines = (n(n − 1))/(2)
for one atom the largest number of photons in one cascade from the level n is n − 1
from n = 2: 1 line, n = 3: 3, n = 4: 6, n = 5: 10, n = 6: 15
lines in a particular series ending on n₁ from the level n: n − n₁

How to apply. Count the pairs of levels. If the question asks for lines in the visible region or in one series, count only the jumps to the required lower level.

Example. Gas excited to n = 5 gives 5 × 4 ÷ 2 = 10 lines; of them 4 belong to the Lyman series.

Hydrogen-like ions and equal lines

for Z: (1)/(λ) = Z²R(…)
Eₙ ∝ Z² so every wavelength of the ion is 1/Z² times that of hydrogen for the same n₁ and n₂
He⁺ lines with n₂ and n₁ both even (4 → 2, 6 → 4 …) have the same wavelength as hydrogen lines
ratio of the wavelengths of the same transition in H and He⁺ is 4 : 1

How to apply. Match Z²(1/n₁² − 1/n₂²) of the ion with 1/n₁′² − 1/n₂′² of hydrogen. Put n₁ = 2n₁′, n₂ = 2n₂′ for He⁺.

Example. He⁺ from n = 2 to n = 1: E = 13.6 × 4 × 3/4 = 40.8 eV, λ = 1240 ÷ 40.8 = 30.4 nm, a quarter of 121.6 nm. He⁺ 4 → 2 gives 10.2 eV, the same as hydrogen 2 → 1.

Energy levels and transitions

Excitation and ionisation energy

ionisation energy of the ground state: 13.6 Z² eV (hydrogen 13.6 eV, He⁺ 54.4 eV, Li²⁺ 122.4 eV)
ionisation energy from the level n: (13.6Z²)/(n²) eV
first excitation energy (n = 1 to 2): 10.2 Z² eV
excitation potential 10.2 V, ionisation potential 13.6 V for hydrogen
binding energy of the electron in a level = −E

How to apply. Energy to reach a level = difference of the two energies. Ionisation is the transition to n = ∞ (E = 0).

Example. Hydrogen in n = 2 needs 3.4 eV to ionise. He⁺ in n = 3 needs 54.4 ÷ 9 = 6.04 eV.

Absorption of a photon and collision with an electron

a photon is absorbed only if hf = Ef − Ei exactly (all of its energy is taken)
a photon with hf > 13.6 eV ionises the atom and the free electron has K = hf − 13.6 eV
an electron of energy K can excite the atom if K ≥ the gap and keeps the rest (K − gap)
gap n = 1 → 2: 10.2 eV, → 3: 12.09 eV, → 4: 12.75 eV, → ∞: 13.6 eV

How to apply. A photon needs an exact match for a jump between levels, but any energy above 13.6 eV for ionisation. An electron can give any part of its energy that makes a jump allowed.

Example. A 15 eV photon on ground-state hydrogen: ionisation, K = 15 − 13.6 = 1.4 eV. An electron of 11 eV can raise the atom to n = 2 and leaves 0.8 eV, but cannot reach n = 3.

Emission after excitation

after excitation to the level n the atom may emit every line between the levels up to n
the shortest wavelength is from n to 1 and the longest is from n to n − 1
the highest energy photon: 13.6 Z²(1 − 1/n²) eV
the photon wavelength λ = (1240)/(E) nm

How to apply. Find the highest level that can be reached with the given energy, then count the lines and compute the wavelengths as asked.

Example. Hydrogen bombarded with 12.5 eV electrons: 12.09 ≤ 12.5 < 12.75, so n = 3 is reached. Three lines come out: 3 → 1 (102.6 nm), 2 → 1 (121.6 nm), 3 → 2 (656 nm).

Recoil of the atom

momentum is conserved: the photon momentum p = (E)/(c) is taken by the atom in the opposite direction
recoil speed v = (E)/(mc)
recoil energy (p²)/(2m) = (E²)/(2mc²), which is far smaller than E (about 10⁻⁸ of E)
the photon energy is therefore slightly less than the energy difference of the levels

How to apply. Calculate the momentum of the photon and divide it by the mass of the atom. The recoil is negligible in questions on energy.

Example. Hydrogen emitting a 10.2 eV photon: p = 10.2 × 1.6 × 10⁻¹⁹ ÷ 3 × 10⁸ = 5.4 × 10⁻²⁷ kg m/s, v = 5.4 × 10⁻²⁷ ÷ 1.67 × 10⁻²⁷ = 3.3 m/s.

Nucleus: composition and size

Composition and kinds of nuclei

ᴬ_ZX: Z = atomic number (protons), A = mass number (nucleons), N = A − Z neutrons
isotopes: same Z, different A (¹H, ²H, ³H)
isobars: same A, different Z (³H and ³He)
isotones: same N
isomers: same A and Z in different energy states
the isotopes have the same chemical properties and different mass

How to apply. Read the symbol in this way: top left is A, bottom left is Z. Neutrons = A − Z.

Example. ²³⁸₉₂U has 92 protons and 238 − 92 = 146 neutrons. ²³⁵U is an isotope of it; ¹⁴C and ¹⁴N are isobars.

Nuclear radius

R = R₀A^1/3 with R₀ = 1.2 fm (1 fm = 10⁻¹⁵ m)
R ∝ A^1/3
volume ∝ A
(R₁)/(R₂) = ((A₁)/(A₂))^1/3
a nucleus of A = 8 has twice the radius of A = 1

How to apply. Take the cube root of A. For ratios write only R ∝ A^1/3; doubling the radius means eight times the nucleons.

Example. A = 64: R = 1.2 × 4 = 4.8 fm. A = 27 and A = 64 have radii in the ratio 3 : 4.

Nuclear density

ρ = (mass)/(volume) = (A × 1.66 × 10⁻²⁷)/((4/3)π(R₀A^1/3)³) = 2.3 × 10¹⁷ kg/m³
independent of A (same for all nuclei)
about 10¹⁴ times the density of water
the density does not depend on the size, so the nucleons are packed in the same way in every nucleus
number density of nucleons about 1.4 × 10⁴⁴ per m³

How to apply. Cancel A: the ratio of the densities of two nuclei is 1. The charge density, however, depends on Z/A.

Example. One cubic centimetre of nuclear matter: 2.3 × 10¹⁷ × 10⁻⁶ = 2.3 × 10¹¹ kg, 230 million tonnes.

Atomic mass unit and mass-energy

1 u = (1)/(12) of the mass of ¹²C atom = 1.66 × 10⁻²⁷ kg = 931.5 MeV/c²
E = mc²
mass of proton 1.00728 u, neutron 1.00866 u, electron 0.00055 u (0.511 MeV/c²)
1 eV = 1.6 × 10⁻¹⁹ J

How to apply. The energy equivalent of a mass in u is mass × 931.5 MeV. For a mass in kg use c² = 9 × 10¹⁶ m²/s².

Example. 1 mg converted completely gives 10⁻⁶ × 9 × 10¹⁶ = 9 × 10¹⁰ J.

Nuclear force

strong, attractive and acts over a short range of a few femtometre
about the same between p–p, n–n and p–n (charge independent)
saturated: a nucleon feels only its neighbours
not a central force and depends on spin
much stronger than the Coulomb force at 1 fm, but decays quickly beyond a few fm
in heavy nuclei the Coulomb repulsion of the protons grows as Z², so N/Z rises from 1 (light) to about 1.5 (heavy)

How to apply. To compare forces choose the distance: gravitational ≪ electromagnetic ≪ nuclear at 1 fm. Use short range and saturation to explain why BE per nucleon becomes constant.

Example. ²⁰⁸Pb has N/Z = 126 ÷ 82 = 1.54, while ¹²C has N/Z = 1.

Mass defect and binding energy

Δm = Zmₚ + (A − Z)mₙ − M(nucleus)
using atomic masses: Δm = Z m(¹H) + (A − Z)mₙ − M(atom), because the electron masses cancel
BE = Δm × 931.5 MeV (Δm in u)
BE is the energy needed to separate the nucleus completely into its free nucleons
the mass of a nucleus is less than the sum of its nucleons

How to apply. Use the mass of the hydrogen atom for the protons if the mass of the atom is given. Do not use the electron mass again.

Example. ⁴He: Δm = 2 × 1.007825 + 2 × 1.008665 − 4.002603 = 0.030377 u, so BE = 0.030377 × 931.5 = 28.3 MeV.

Binding energy per nucleon

BE/A = (BE)/(A)
higher BE/A means a more stable nucleus
⁴He 7.07 MeV, ¹²C 7.68 MeV, ⁵⁶Fe 8.8 MeV (near the greatest), ²³⁸U 7.6 MeV
most nuclei in the middle (A = 30 to 170) have BE/A about 8 MeV
²H has only 1.11 MeV

How to apply. Compare nuclei with BE/A, not with the total BE. A heavier nucleus has the larger BE and may be less stable.

Example. ²H: BE = 2.22 MeV, so BE/A = 1.11 MeV. ⁴He: 28.3 ÷ 4 = 7.07 MeV.

Binding energy curve

BE/A rises steeply for A < 20 (peaks at ⁴He, ¹²C, ¹⁶O), has a broad maximum of about 8.8 MeV near A = 56 and decreases slowly to 7.6 MeV at A = 238
fusion of light nuclei (A < 56) and fission of heavy nuclei (A > 56) both move towards the peak and release energy
nuclei near A = 56 are the most stable
the curve is nearly flat from A = 30 to 170, which shows that the nuclear force is saturated

How to apply. Energy is released whenever a process raises the BE per nucleon. State the direction: A below 56, fusion; A above 56, fission.

Example. Fission of A = 240 (BE/A = 7.6 MeV) into two nuclei of A = 120 (8.5 MeV): the gain is 240 × 0.9 = 216 MeV.

Q value of a reaction

Q = (Σ mass of reactants − Σ mass of products)c²
Q = Σ(BE of products) − Σ(BE of reactants)
Q > 0: energy is released (exothermic)
Q < 0: energy must be supplied (endothermic)
Q is shared as kinetic energy of the products

How to apply. With the BE use the total BE of each side (BE per nucleon times A). With masses subtract the sum of the products from the reactants.

Example. ²H + ³H → ⁴He + n: BE of the products = 28.3 MeV, of the reactants 2.22 + 8.48 = 10.70 MeV, so Q = 17.6 MeV.

Separation energy of a nucleon

energy to remove one neutron: Sₙ = BE(A, Z) − BE(A − 1, Z)
it is larger for a nucleus with a closed shell and an even number of nucleons
for ⁴He the removal of one neutron costs far more than the average of 7.07 MeV

How to apply. Take the difference between the binding energy of the nucleus and that of the nucleus left behind.

Example. BE of ³He is 7.72 MeV and of ⁴He 28.30 MeV: removing a neutron from ⁴He needs 28.30 − 7.72 = 20.6 MeV.

Fission and fusion

Nuclear fission

a heavy nucleus splits into two medium fragments with two or three neutrons after it absorbs a slow neutron
²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n + about 200 MeV (A: 236 → 141 + 92 + 3; Z: 92 → 56 + 36)
energy per fission about 200 MeV, about 0.85 MeV per nucleon
most of it appears as kinetic energy of the fragments
fission fragments are neutron-rich and radioactive

How to apply. Check A and Z on both sides. For the energy from a mass of fuel, find the number of nuclei (m ÷ M × 6.02 × 10²³) and multiply by 200 MeV.

Example. 1 kg of ²³⁵U fissioned completely: N = 1000 ÷ 235 × 6.02 × 10²³ = 2.56 × 10²⁴, E = 2.56 × 10²⁴ × 200 × 1.6 × 10⁻¹³ = 8.2 × 10¹³ J, about 20 000 tonnes of TNT.

Chain reaction and the reactor

each fission frees 2 or 3 neutrons, which cause more fission
multiplication factor k = (neutrons in a generation)/(neutrons in the previous generation)
k < 1: the reaction dies out; k = 1: steady (a reactor); k > 1: it grows (a bomb)
moderator (heavy water, graphite) slows neutrons to thermal energies
control rods (cadmium, boron) absorb neutrons
coolant carries the heat away
a critical mass is needed

How to apply. Match the parts with their function. For the growth after n generations use kⁿ.

Example. With k = 1.05 the number of neutrons after 100 generations is 1.05¹⁰⁰ = 131 times larger.

Nuclear fusion

light nuclei join to form a heavier one with larger BE/A
²H + ²H → ³He + n + 3.27 MeV
²H + ³H → ⁴He + n + 17.6 MeV
in the Sun four ¹H become one ⁴He with the release of about 26.7 MeV
it needs a temperature of 10⁷ K to 10⁸ K so that the nuclei overcome the Coulomb repulsion (thermonuclear reaction)
energy per nucleon is about four times that of fission

How to apply. Remember the energy of one reaction, then divide by the number of nucleons to compare with fission.

Example. ²H + ³H: 17.6 ÷ 5 = 3.5 MeV per nucleon, against 0.85 MeV per nucleon in the fission of ²³⁵U.

Power from fission and the Sun

power P = (number of fissions per second) × 200 MeV
fissions per second = (P)/(200 MeV in joule)
the Sun radiates 3.8 × 10²⁶ W, so the mass it loses is (P)/(c²) = 4 × 10⁹ kg every second
a reactor that is 30 % efficient needs three times as many fissions for the same electric power

How to apply. Convert MeV into joule (× 1.6 × 10⁻¹³) and divide the power by this energy.

Example. A reactor of thermal power 1000 MW needs 10⁹ ÷ (200 × 1.6 × 10⁻¹³) = 3.1 × 10¹⁹ fissions every second.

Radioactivity and decay law

Law of radioactive decay

the number of nuclei decaying per second is proportional to the number present: (dN)/(dt) = −λN
N = N₀e^−λt
λ = decay constant (probability of decay per unit time, s⁻¹)
decay is random and spontaneous, independent of temperature, pressure and chemical state
the mass also decays as m = m₀e^−λt

How to apply. Use the exponential form when t is not a whole number of half-lives. In a graph of ln N against t, the slope is −λ.

Example. λ = 0.01 s⁻¹ and t = 100 s: N ÷ N₀ = e⁻¹ = 0.37.

Half-life and mean life

T = (0.693)/(λ) = (ln 2)/(λ)
τ = (1)/(λ) = 1.44 T
N = N₀(1/2)^t/T
after n half-lives N = N₀/2ⁿ
at t = τ, N = N₀/e = 0.37 N₀
the sum of the lives of all nuclei divided by N₀ gives τ
63 % has decayed at t = τ and 50 % at t = T

How to apply. Use the powers of 1/2 when the time is a multiple of T. For other times write t/T as a decimal.

Example. T = 10 min, t = 25 min: N ÷ N₀ = (1/2)^2.5 = 0.177, 17.7 % is left.

Activity

A = −(dN)/(dt) = λN = A₀e^−λt
A = A₀/2^t/T
unit becquerel (Bq) = 1 decay per second
1 curie (Ci) = 3.7 × 10¹⁰ Bq
1 rutherford = 10⁶ Bq
activity is proportional to the number of nuclei, so it falls with the same half-life
the activity of a given amount is larger for a smaller T

How to apply. Find λ in s⁻¹ from T, then A = λN. A sample with equal N and smaller T has more activity.

Example. N = 10²⁰ nuclei, T = 1 hour: λ = 0.693 ÷ 3600 = 1.93 × 10⁻⁴ s⁻¹, A = 1.93 × 10¹⁶ Bq = 5.2 × 10⁵ Ci.

Fraction decayed and mass left

fraction decayed = 1 − e^−λt = 1 − 2^−t/T
after 1, 2, 3, 4 half-lives: ½, ¾, ⅞, 15/16 decayed
time to fall to a fraction f: t = T log₂(1/f) = (1)/(λ) ln(N₀/N)
mass left m = m₀/2^t/T
the number decayed in a time is N(t₁) − N(t₂)

How to apply. Find the number of half-lives n = t/T first. For the decayed amount subtract the remaining amount from the starting amount.

Example. T = 5 days, 20 g, 15 days: n = 3, m = 20 ÷ 8 = 2.5 g left and 17.5 g decayed. 90 % decayed needs t = T log₂10 = 3.32 T.

Alpha, beta and gamma

α: helium nucleus ⁴₂He, A falls by 4, Z falls by 2; discrete energies of 4 MeV to 9 MeV; most ionising, least penetrating (stopped by paper), speed about 10⁷ m/s
β⁻: n → p + e⁻ + ν̄, Z rises by 1, A same; continuous energy spectrum (the neutrino shares the energy); penetrates a few mm of aluminium
β⁺: p → n + e⁺ + ν, Z falls by 1
γ: photon from an excited nucleus, A and Z same; no charge; most penetrating (cm of lead); speed c
in an electric or magnetic field α and β bend in opposite ways, γ does not bend

How to apply. A decay equation must balance A and Z. The electron in β⁻ decay is created in the nucleus; it was not in it before.

Example. ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He. ¹⁴₆C → ¹⁴₇N + e⁻ + ν̄.

Number of α and β particles in a series

nα = (A − A′)/(4)
nβ (β⁻) = Z′ − Z + 2nα
each α lowers A by 4 and Z by 2; each β⁻ raises Z by 1
the daughter of an α decay lies two places to the left in the periodic table

How to apply. First find nα from the fall in A, then find nβ from the net change of Z.

Example. ²³⁸U → ²⁰⁶Pb: nα = (238 − 206) ÷ 4 = 8; Z: 92 − 16 + nβ = 82, so nβ = 6.

Energy shared in α decay

Q = (M parent − M daughter − M α)c²
momentum conservation: the daughter and the α particle have equal and opposite momentum
Kα = Q(A − 4)/(A)
Kd = Q(4)/(A)
the α particle carries almost all of Q

How to apply. Use p² = 2mK for both and the ratio of masses 4 : (A − 4).

Example. ²²⁶Ra → ²²²Rn + α with Q = 4.87 MeV: Kα = 4.87 × 222 ÷ 226 = 4.78 MeV.

Successive decay

parent → daughter → …: (dN₂)/(dt) = λ₁N₁ − λ₂N₂
secular equilibrium (T₁ ≫ T₂): λ₁N₁ = λ₂N₂, that is the activities are equal and (N₁)/(N₂) = (λ₂)/(λ₁) = (T₁)/(T₂)
the daughter activity first rises and then follows the parent

How to apply. At equilibrium equate the two activities and use N ∝ T.

Example. ²²⁶Ra (T = 1600 y) and ²²²Rn (T = 3.8 d) in equilibrium: N(Ra) ÷ N(Rn) = 1600 × 365 ÷ 3.8 = 1.5 × 10⁵.

Parallel decay

a nucleus can decay by two routes: λ = λ₁ + λ₂
(1)/(T) = (1)/(T₁) + (1)/(T₂), so T = (T₁T₂)/(T₁ + T₂)
the fraction by the route 1 is (λ₁)/(λ) = (T₂)/(T₁ + T₂)
the half-life of the sample is shorter than either T₁ or T₂

How to apply. Add the decay constants, not the half-lives.

Example. T₁ = 20 min and T₂ = 30 min: T = 20 × 30 ÷ 50 = 12 min and 30 ÷ 50 = 60 % decay by route 1.

Radiocarbon dating

living matter has a constant ratio of ¹⁴C to ¹²C because of the exchange with the air; after death the ¹⁴C decays with T = 5730 years
age t = (T)/(0.693) ln(A₀)/(A) = (T)/(0.693) ln(N₀)/(N)
the method is useful for ages of a few thousand to about 50 000 years

How to apply. Compare the activity of the sample with that of a living sample of the same mass.

Example. If the activity of a bone is one fourth of that of living matter, two half-lives have passed: 2 × 5730 = 11 460 years.

Most asked question types

How do I solve “Rutherford’s experiment and its conclusions” questions in JEE Main?

Statements about the α scattering experiment are given. Choose the correct inference about the nucleus and the atom. 1. Most α pass straight: the atom is mostly empty. 2. Few large deflections: a small, heavy, positive nucleus. 3. Size: nucleus 10⁻¹⁵ m, atom 10⁻¹⁰ m. Example: About 1 in 8000 returns: a nucleus of charge Ze and of size 10⁻¹⁵ m.

How do I solve “Distance of closest approach” questions in JEE Main?

The kinetic energy of an α particle (or a proton) and the atomic number of the target are given. Find the distance of closest approach, or the energy for a given distance. 1. K = k q₁q₂/r₀. 2. r₀ = 2Z × 1.44 ÷ K(MeV) fm for an α particle. 3. Doubling v makes r₀ one fourth. Example: 5 MeV α on gold: 45.5 fm.

How do I solve “Impact parameter and angle of scattering” questions in JEE Main?

The impact parameter and the energy are given. Find the angle, or find b for 90° or 180°. 1. Find r₀ first. 2. b = (r₀/2) cot(θ/2). 3. b = 0 gives 180°. Example: 5 MeV α on gold, θ = 90°: b = 22.8 fm.

How do I solve “Radius, speed and energy of an electron in Bohr’s orbit” questions in JEE Main?

The level n and the atom (H, He⁺, Li²⁺) are given. Find the radius, the speed, the kinetic or total energy, or compare two cases. 1. r = 0.529 n²/Z Å. 2. v = 2.19 × 10⁶ Z/n m/s. 3. E = −13.6 Z²/n² eV; K = −E. Example: Li²⁺, n = 2: r = 0.705 Å. He⁺, n = 2: v = 2.19 × 10⁶ m/s.

How do I solve “Period, current and magnetic moment in the nth orbit” questions in JEE Main?

The time period, the equivalent current, the magnetic moment or the angular momentum is asked for two orbits or two ions as a ratio. 1. T ∝ n³/Z², I ∝ Z²/n³. 2. μ = n eh/4πm, L = nh/2π. 3. Put rₙ ∝ n²/Z and vₙ ∝ Z/n. Example: Hydrogen n = 2 against n = 1: T is 8 times.

How do I solve “Wavelength of a line from the Rydberg formula” questions in JEE Main?

Two levels of hydrogen (or of an ion) are given. Find the wavelength or the frequency of the emitted line. 1. 1/λ = RZ²(1/n₁² − 1/n₂²). 2. R = 1.097 × 10⁷ m⁻¹. 3. Or E = 13.6 Z²(…) and λ = 1240/E nm. Example: 3 → 2 in hydrogen: 656 nm.

How do I solve “Longest and shortest wavelength of a series” questions in JEE Main?

A series (Lyman, Balmer, Paschen) is named. Find the longest wavelength, the shortest one, or the ratio. 1. Longest: n₂ = n₁ + 1. 2. Shortest: n₂ = ∞, λ = n₁²/R. 3. Balmer 656.3 nm to 364.6 nm. Example: Balmer: longest ÷ shortest = 9 : 5.

How do I solve “Number of spectral lines” questions in JEE Main?

The atoms are excited to the level n. Find the number of different lines, or the lines in the visible or in one series. 1. n(n − 1)/2 for all jumps. 2. n − n₁ for the lines ending on n₁. 3. Visible lines are those that end on n = 2. Example: n = 5: 10 lines in all.

How do I solve “Ionisation and excitation energy” questions in JEE Main?

The atom or ion and its state are given. Find the energy to excite it to a given level or to remove the electron, or the potential in volt. 1. Ionisation from n: 13.6 Z²/n² eV. 2. Excitation = Ef − Ei. 3. Potential in volt = energy in eV. Example: He⁺ ground state: 54.4 eV; hydrogen n = 2: 3.4 eV.

How do I solve “Photon absorbed or emitted in a transition” questions in JEE Main?

A photon is absorbed or emitted in a jump. Find its energy, its wavelength, or the final level. 1. hf = |Ef − Ei|. 2. λ(nm) = 1240/E(eV). 3. For absorption, the photon must have exactly the gap energy. Example: 4 → 2 in hydrogen: 2.55 eV, 486 nm.

How do I solve “Excitation by collision with an electron or a photon” questions in JEE Main?

A beam of electrons or photons of given energy hits hydrogen. Find the levels reached, the lines seen, or the energy left. 1. List the gaps: 10.2, 12.09, 12.75, 13.6 eV. 2. The highest gap not above the energy is the level reached. 3. Ionisation by a photon leaves K = hf − 13.6. Example: 12.5 eV electrons: n = 3, three lines.

How do I solve “Hydrogen-like ion with a line equal to that of hydrogen” questions in JEE Main?

A transition in He⁺ or Li²⁺ is to be matched with a transition in hydrogen, or the wavelength in the ion is asked. 1. Z²(1/n₁² − 1/n₂²) for the ion. 2. Equate it with 1/n₁′² − 1/n₂′² for hydrogen. 3. He⁺: even levels double those of hydrogen. Example: He⁺ 4 → 2 has the same wavelength (121.6 nm) as H 2 → 1.

How do I solve “Isotopes, isobars and composition of the nucleus” questions in JEE Main?

A symbol or a decay is given. Find the number of protons, neutrons and nucleons, or name isotopes, isobars and isotones. 1. Z = protons, N = A − Z. 2. Same Z: isotopes. Same A: isobars. 3. Same N: isotones. Example: ²³⁸U: 92 protons, 146 neutrons.

How do I solve “Nuclear radius, volume and density” questions in JEE Main?

The mass number is given. Find the radius, or the ratio of radii, volumes and densities of two nuclei. 1. R = 1.2 A^1/3 fm. 2. V ∝ A, so ρ is the same. 3. ρ = 2.3 × 10¹⁷ kg/m³. Example: A = 64: 4.8 fm. A = 27 and 64: radii 3 : 4.

How do I solve “Binding energy from masses” questions in JEE Main?

The masses of a nucleus (or atom) and of the nucleons are given. Find the mass defect, the binding energy and the binding energy per nucleon. 1. Δm = Zmₚ + (A − Z)mₙ − M. 2. BE = 931.5 Δm MeV. 3. Divide by A for BE per nucleon. Example: ⁴He: Δm = 0.030377 u, BE = 28.3 MeV, 7.07 MeV per nucleon.

How do I solve “Binding energy curve” questions in JEE Main?

Statements about the curve of BE per nucleon against A are given. Say which nuclei are stable, where fission or fusion releases energy, and read the energy released. 1. The peak is at A = 56 (8.8 MeV). 2. Moving towards the peak releases energy. 3. Energy = A × (change of BE/A). Example: A = 240 (7.6) into two of A = 120 (8.5): 216 MeV.

How do I solve “Q value and energy released in a nuclear reaction” questions in JEE Main?

The masses or the binding energies of the reactants and the products are given. Find Q or the energy released, and the kinetic energy shared. 1. Q = (Σm reactants − Σm products)c². 2. Or Σ BE of products − Σ BE of reactants. 3. Q > 0 means energy is released. Example: ²H + ³H → ⁴He + n: 17.6 MeV.

How do I solve “Fission, chain reaction and power of a reactor” questions in JEE Main?

The energy per fission and the fuel or the power is given. Find the energy from a mass, the rate of fission, or the multiplication factor. 1. N = (m/M) × 6.02 × 10²³. 2. Energy = N × 200 MeV. 3. k < 1, = 1, > 1. Example: 1 kg of ²³⁵U: 8.2 × 10¹³ J.

How do I solve “Fraction left, half-life and mean life” questions in JEE Main?

The half-life and a time are given. Find the fraction left, the time for a given fraction, or the mean life. 1. N = N₀/2^t/T. 2. T = 0.693/λ, τ = 1.44 T. 3. The time for a fraction f: t = T log₂(1/f). Example: T = 10 min, t = 25 min: 17.7 % is left.

How do I solve “Activity and the number of nuclei” questions in JEE Main?

The activity, the number of nuclei or the mass is given. Find the activity, the decay constant or the activity after some time. 1. A = λN. 2. A = A₀/2^t/T. 3. 1 Ci = 3.7 × 10¹⁰ Bq. Example: 10²⁰ nuclei, T = 1 h: 1.9 × 10¹⁶ Bq.

How do I solve “α and β particles and the change of A and Z” questions in JEE Main?

A parent and a final nucleus are given. Find the number of α and β particles emitted, or the daughter nucleus after a decay. 1. nα = ΔA/4. 2. Z′ = Z − 2nα + nβ. 3. Gamma emission changes nothing. Example: ²³⁸U → ²⁰⁶Pb: 8 α and 6 β.

How do I solve “Parallel and successive decay” questions in JEE Main?

A nucleus decays by two routes, or a daughter is formed from a parent. Find the effective half-life, the fraction by one route, or the ratio of the numbers at equilibrium. 1. Parallel: λ = λ₁ + λ₂. 2. Fraction by route 1: λ₁/λ. 3. Equilibrium: λ₁N₁ = λ₂N₂. Example: T₁ = 20 min, T₂ = 30 min: T = 12 min.

Common myths and the truth

Myth: In Rutherford’s experiment all α particles are deflected a lot. Truth: Most go straight through; only about 1 in 8000 is deflected through a large angle.
Myth: The electron in a Bohr orbit radiates because it accelerates. Truth: Bohr assumed no radiation in a stationary orbit; light is emitted only in a jump between orbits.
Myth: The total energy of the electron is positive because the kinetic energy is positive. Truth: E = −K is negative, since the potential energy is −2K. A negative E means the electron is bound.
Myth: The speed of the electron increases with n. Truth: v ∝ Z/n, so it decreases as the orbit becomes bigger.
Myth: The ionisation energy of hydrogen is the energy of the ground state, −13.6 eV. Truth: It is the energy needed to remove the electron, +13.6 eV.
Myth: Any photon of more than 10.2 eV can excite the atom from n = 1 to n = 2. Truth: A photon is absorbed only at the exact gap of 10.2 eV (or at more than 13.6 eV to ionise).
Myth: A heavier nucleus is always more stable because its binding energy is bigger. Truth: Stability depends on BE per nucleon, which is highest near iron.
Myth: The density of a large nucleus is greater than that of a small one. Truth: R³ ∝ A, so the density is constant for all nuclei.
Myth: After two half-lives all the nuclei have decayed. Truth: One fourth is still left after two half-lives; a half-life is the time for half of the remainder to decay.
Myth: Heating the sample or compressing it changes the rate of radioactive decay. Truth: λ does not depend on temperature, pressure or chemical form.

Mistakes to avoid

The energy of a level is negative; the ionisation energy is positive and equals −E.
Excitation energy is a difference of two levels, not the energy of the upper level.
A photon is absorbed only for an exact gap, but an electron can pass on a part of its energy.
Number of lines from the level n is n(n − 1)/2; one atom gives at most n − 1 photons.
For He⁺ and Li²⁺ the radius has Z in the denominator and the energy has Z² in the numerator.
The nuclear density is the same for all nuclei; do not divide by A again.
Use the mass of the hydrogen atom (not the proton) with atomic masses in the mass defect, so that the electrons cancel.
The half-life of a mixture in parallel decay is less than both; add the λ, not the T.

Simulations for Atoms and Nuclei

Bohr’s atomic model — Choose an orbit and calculate the photon released in a transition to the ground state.
Bohr orbits and electron waves — Choose the orbit n and nuclear charge Z. Exactly n de Broglie wavelengths fit around each allowed orbit.
Hydrogen energy-level ladder — Pick a jump and calculate the photon it emits.
Spectral lines of hydrogen-like ions — Choose a one-electron ion and a transition. Higher nuclear charge shifts every line towards shorter…
Hydrogen emission spectrum — Select a transition and calculate the emitted spectral line.
Rutherford alpha scattering — Send an alpha particle past a nucleus and vary the impact parameter.
Activity of a radioactive sample — Pick an isotope and a sample mass. Count the atoms and calculate how many decay each second.
Alpha decay and nuclear recoil — Start with a stationary nucleus and see how the alpha particle and daughter share momentum and energy.
Binding energy per nucleon — Slide across the mass number and read where nuclei are most stable.
Neutron multiplication in fission — Track the average number of neutrons through successive generations.
Energy from nuclear fusion — Fuse light nuclei and calculate the energy released from the mass defect.
Mass defect and binding energy — Convert a nuclear mass defect into binding energy.
Radioactive decay — See individual nuclei disappear while the population follows a half-life curve.
Size and density of nuclei — Build nuclei of increasing mass number. The radius grows as A^(1/3), so the density stays the same.
Radiometric dating — Measure the remaining fraction of a radioactive parent and estimate the sample age.

More: Class 12 Physics simulations · all JEE Main chapters.