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State Board · Class 12 · Physics

Lesson 15: Structure of Atoms and Nuclei

7 concepts · 56 practice questions with answers · Free to read

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Thomson's model, α-particle scattering and Rutherford's nuclear model

From daily life

A ball bouncing back from a tiny hidden obstacle — α particles bounce off the nucleus.

Simple definition

The nucleus is the small, massive, positive core of the atom.

Rutherford's scattering experimentαmost go straightfew deflect1 in 8000 backb = kZe² ÷ E · cot(θ/2)

In the figure : α particles scatter; the nucleus is tiny.

Rule 1: Thomson's model — The atom is a positive sphere with embedded electrons (plum pudding). Net charge is zero.

  • Easy : Atomic radius ≈ 10⁻¹⁰ m.
  • Medium : In this model α particles should deflect only slightly.
  • Exam (CET) : In Thomson's model the positive charge is? (1) spread throughout (2) at the centre (3) in orbits (4) absent → answer 1)

Rule 2: Alpha scattering — α particles on a thin gold foil : most go straight, a few deflect widely, about 1 in 8000 bounce back ⇒ a small, massive, positive nucleus.

  • Easy : α of 5.3 MeV on gold Z = 79 : closest approach r₀ = k·2Z·e² ÷ E = 4.29 × 10⁻¹⁴ m.
  • Medium : r₀ ≈ 4.3 × 10⁻¹⁴ m ≪ atomic radius.
  • Exam (CET) : Back-scattered α's suggest? (1) a small dense nucleus (2) electrons (3) photons (4) neutrons → answer 1)

Rule 3: Rutherford's model and its flaws — Electrons orbit the nucleus. Flaws : (1) an accelerating electron should radiate and spiral in, (2) it cannot explain line spectra.

  • Easy : Predicted lifetime ≈ 10⁻¹¹ s.
  • Medium : Hence Bohr's model.
  • Exam (CET) : Rutherford's model fails to explain? (1) stability and spectra (2) nucleus (3) Z (4) density → answer 1)

Derivation : Distance of closest approach

  • Step 1 : Head-on α : initial KE = E.
  • Step 2 : At the turning point all KE becomes potential energy.
  • Step 3 : E = k (2e)(Ze) ÷ r₀.
  • Step 4 : r₀ = 2 k Z e² ÷ E.
Remember

r₀ = 2kZe²/E; nucleus ≈ 10⁻¹⁴ m; atom ≈ 10⁻¹⁰ m.

Common mistake / CET trap

Most α's go straight — the atom is mostly empty.

Shortcut

Memory clue : "most straight, few back".

Spot the difference

ModelNucleusFlaw
Thomsonnoscattering
Rutherfordyesstability, spectra
Bohryesmulti-electron atoms

Practice questions

Easy
  1. 1Who did the α-scattering experiment?
  2. 2Thomson's model?
  3. 3Rutherford's conclusions?
Medium
  1. 4E = 4 MeV, Z = 29 : r₀?
  2. 5α's KE doubled : r₀?
  3. 6Z doubled : r₀?
Challenging
  1. 71 in 8000 returns : meaning?
  2. 8Limit of Rutherford's model?
Show answers (try them yourself first)
  1. 1Rutherford (Geiger–Marsden)
  2. 2Electrons in a positive sphere
  3. 3A small, massive, positive nucleus
  4. 4r₀ = 2.09 × 10⁻¹⁴ m
  5. 5Halves
  6. 6Doubles
  7. 7The nucleus is very small
  8. 8Stability and spectra unexplained

Atomic spectra and the Rydberg formula

From daily life

The red glow of a neon sign comes from neon's own spectral lines.

Simple definition

A line spectrum has lines at definite wavelengths.

Hydrogen spectral seriesLymanBalmerPaschenBrackettPfund1/λ = R (1/n₁² − 1/n₂²)R = 1.097 × 10⁷ m⁻¹

In the figure : Five series; Balmer is visible.

Rule 1: Rydberg formula — 1/λ = R (1/n₁² − 1/n₂²), n₂ > n₁, R = 1.097 × 10⁷ m⁻¹.

  • Easy : Balmer first line (3 → 2) : λ = 656 nm (red).
  • Medium : Lyman first line (2 → 1) : λ = 121.5 nm (UV).
  • Exam (CET) : 1/λ ∝ ? (1) (1/n₁² − 1/n₂²) (2) n₁n₂ (3) n₂ − n₁ (4) n₁² → answer 1)

Rule 2: Spectral series — Lyman n₁ = 1 (UV), Balmer 2 (visible), Paschen 3, Brackett 4, Pfund 5 (IR).

  • Easy : Balmer shortest (∞ → 2) : λ = 4/R = 365 nm.
  • Medium : Paschen first : λ = 1875 nm.
  • Exam (CET) : The visible series? (1) Balmer (2) Lyman (3) Paschen (4) Pfund → answer 1)

Rule 3: Emission and absorption spectra — Emission : bright lines from a hot gas; absorption : dark lines in a continuous spectrum at the same λ. The lines are an element's fingerprint.

  • Easy : Fraunhofer lines in sunlight (absorption).
  • Medium : Sodium's yellow line at 589 nm.
  • Exam (CET) : Absorption lines are? (1) dark (2) bright (3) coloured (4) absent → answer 1)

Derivation : Series limit

  • Step 1 : As n₂ → ∞ : 1/n₂² → 0.
  • Step 2 : 1/λ = R/n₁².
  • Step 3 : Balmer (n₁ = 2) : λ_min = 4/R.
  • Step 4 : = 4 ÷ 1.097 × 10⁷ = 365 nm.
Remember

1/λ = R(1/n₁² − 1/n₂²); Balmer n₁ = 2.

Common mistake / CET trap

Always n₂ > n₁; R is in m⁻¹ so λ comes in metres — convert to nm.

Shortcut

Memory clue : "Lyman UV, Balmer visible, Paschen IR".

Spot the difference

Seriesn₁Region
Lyman1UV
Balmer2visible
Paschen3IR
Brackett4IR
Pfund5far IR

Practice questions

Easy
  1. 1Rydberg formula?
  2. 2Which series is visible?
  3. 3Value of R?
Medium
  1. 44 → 2 : λ?
  2. 53 → 1 : λ?
  3. 6Balmer series limit?
Challenging
  1. 7Shortest λ of Paschen?
  2. 84 → 3 : which series?
Show answers (try them yourself first)
  1. 11/λ = R(1/n₁² − 1/n₂²)
  2. 2Balmer
  3. 31.097 × 10⁷ m⁻¹
  4. 4λ = 486 nm
  5. 5λ = 102.6 nm
  6. 6365 nm
  7. 7λ = 820 nm
  8. 8Paschen (n₁ = 3)

Bohr's model : postulates, orbit radius and speed, de Broglie's explanation

From daily life

You can stand only on the steps of a staircase — an electron only in definite orbits.

Simple definition

The Bohr model gives fixed orbits with quantised angular momentum.

Bohr model : orbits and transitionn=1n=2n=3hf = E₂ − E₁m v r = n h ÷ 2πr_n = 0.529 n² ÷ Z Å

In the figure : Orbits n = 1, 2, 3; a photon on a transition.

Rule 1: Bohr's postulates — (1) Electrons move in fixed orbits without radiating, (2) angular momentum m v r = n h ÷ 2π, (3) in a transition h f = E_i − E_f.

  • Easy : n = 1 : L = h/2π = 1.05 × 10⁻³⁴ J s.
  • Medium : n = 3 : L = 3h/2π.
  • Exam (CET) : Angular momentum? (1) nh/2π (2) nh (3) h/n (4) n²h → answer 1)

Rule 2: Orbit radius and speed — r_n = 0.529 n² ÷ Z Å; v_n = 2.19 × 10⁶ Z ÷ n m/s.

  • Easy : H, n = 3 : r = 4.76 Å; v = 7.30 × 10⁵ m/s.
  • Medium : He⁺ (Z = 2), n = 2 : r = 1.058 Å.
  • Exam (CET) : r_n ∝ ? (1) n² (2) n (3) 1/n (4) n³ → answer 1)

Rule 3: de Broglie's explanation — A standing wave on the orbit : 2π r = n λ ⇒ m v r = n h/2π — the angular-momentum rule follows from waves.

  • Easy : n = 2 : two full wavelengths fit the orbit.
  • Medium : Use λ = h/mv.
  • Exam (CET) : Number of waves on the orbit? (1) n (2) n² (3) 1/n (4) 2n → answer 1)

Derivation : r_n = 0.529 n²/Z Å

  • Step 1 : Centripetal : m v² ÷ r = k Z e² ÷ r².
  • Step 2 : m v r = n h/2π ⇒ v = n h ÷ 2π m r.
  • Step 3 : Solve for r : r = n² h² ÷ 4π² m k Z e².
  • Step 4 : Numerically r_n = 0.529 n² ÷ Z Å.
Remember

mvr = nh/2π; r_n = 0.529n²/Z Å; v_n = 2.19×10⁶Z/n.

Common mistake / CET trap

r_n ∝ n² but v_n ∝ 1/n — do not mix them.

Shortcut

Memory clue : "radius squares, speed inverts".

Spot the difference

QuantityFormulaWith n
r0.529 n²/Z Å∝ n²
v2.19×10⁶ Z/n∝ 1/n
Lnh/2π∝ n

Practice questions

Easy
  1. 1Bohr's postulates?
  2. 2r_n ∝ ?
  3. 3Unit of angular momentum?
Medium
  1. 4H, n = 2 : r?
  2. 5H, n = 2 : v?
  3. 6Li²⁺ (Z = 3), n = 1 : r?
Challenging
  1. 7Ratio of radii n = 3 to n = 1?
  2. 8n = 4 : waves on the orbit?
Show answers (try them yourself first)
  1. 1Fixed orbits, mvr = nh/2π, hf = ΔE
  2. 2n²
  3. 3J s
  4. 4r = 2.116 Å
  5. 5v = 1.095 × 10⁶ m/s
  6. 6r = 0.176 Å
  7. 79 : 1
  8. 84

Energy levels and transitions of hydrogen-like atoms

From daily life

Stepping down a ladder releases energy; an electron stepping down emits a photon.

Simple definition

An energy level is a definite electron energy; a transition is a jump between levels.

Hydrogen energy levelsn=∞0 eVn=4−0.85 eVn=3−1.51 eVn=2−3.4 eVn=1−13.6 eVLymanBalmerE_n = −13.6 Z² ÷ n² eV

In the figure : Levels E_n; arrows show transitions.

Live figure

Energy levels and transitions of hydrogen-like atoms

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Rule 1: Energy levels — E_n = −13.6 Z² ÷ n² eV; negative ⇒ bound. n = 1 ground state, n > 1 excited; ionisation energy 13.6 eV (H).

  • Easy : H : E₁ = −13.6, E₂ = -3.4, E₃ = -1.51 eV.
  • Medium : He⁺ ground state : −13.6 × 4 = -54.4 eV.
  • Exam (CET) : E_n ∝ ? (1) −1/n² (2) n² (3) 1/n (4) n → answer 1)

Rule 2: Transitions — h f = E_i − E_f; λ(nm) = 1240 ÷ ΔE(eV). Higher → lower : emission; lower → higher : absorption.

  • Easy : 3 → 2 : ΔE = |1.89| eV ⇒ λ = 656 nm.
  • Medium : 2 → 1 : ΔE = 10.2 eV ⇒ λ = 121.6 nm.
  • Exam (CET) : Smallest transition energy? (1) 3 → 2 (2) 2 → 1 (3) 4 → 1 (4) ∞ → 1 → answer 1)

Rule 3: Excitation and ionisation — First excitation energy of H : 13.6 − 3.4 = 10.2 eV; ionisation 13.6 eV. From level n up to n(n − 1)/2 lines are possible.

  • Easy : From n = 4 : 6 lines.
  • Medium : From n = 5 : 10 lines.
  • Exam (CET) : Lines from n = 3? (1) 3 (2) 2 (3) 6 (4) 1 → answer 1)

Derivation : E_n = −13.6 Z²/n² eV

  • Step 1 : E = KE + PE = ½ k Z e²/r − k Z e²/r = −k Z e² ÷ 2r.
  • Step 2 : With r = 0.529 n²/Z Å ⇒ E = −k Z² e² ÷ (2 × 0.529 n² Å).
  • Step 3 : k e² ÷ 0.529 Å = 27.2 eV.
  • Step 4 : E_n = −13.6 Z² ÷ n² eV.
Remember

E_n = −13.6Z²/n² eV; λ(nm) = 1240/ΔE; lines = n(n−1)/2.

Common mistake / CET trap

Take ΔE positive (larger − smaller); watch signs because E is negative.

Shortcut

Memory clue : "down emits, up absorbs".

Spot the difference

TransitionΔE (eV)λ (nm)
2 → 110.2121.5
3 → 21.89656
4 → 22.55486

Practice questions

Easy
  1. 1Formula for E_n?
  2. 2Ionisation energy of H?
  3. 3When is a photon emitted?
Medium
  1. 4n = 4 → 2 : λ?
  2. 5He⁺ n = 2 : E?
  3. 6Total lines from n = 4?
Challenging
  1. 7Absorption n = 1 → 3 : photon energy?
  2. 8Ionisation energy of Li²⁺?
Show answers (try them yourself first)
  1. 1−13.6Z²/n² eV
  2. 213.6 eV
  3. 3Higher → lower
  4. 4λ = 486 nm
  5. 5E = -13.6 eV
  6. 66
  7. 712.1 eV
  8. 8122.4 eV

The nucleus : constituents, size, density and nuclear force

From daily life

A thimbleful of nuclear matter would weigh millions of tonnes.

Simple definition

A nucleus is made of protons and neutrons, collectively nucleons.

Nuclear size and nuclear forceAR = R₀ A^(1/3)U(r) : attractive ~ 1 fmr

In the figure : R = R₀A^(1/3); a short-range force.

Rule 1: Constituents and notation — A nucleus has Z protons and N neutrons; A = Z + N. Isotopes : same Z; isobars : same A; isotones : same N.

  • Easy : ¹⁴₆C : Z = 6, N = 8.
  • Medium : ²³⁸₉₂U : N = 146.
  • Exam (CET) : Isotopes have the same? (1) Z (2) A (3) N (4) mass → answer 1)

Rule 2: Size and density — R = R₀ A^{1/3}, R₀ = 1.2 fm. Density ≈ 2.3 × 10¹⁷ kg/m³, independent of A.

  • Easy : A = 64 ⇒ R = 1.2 × 4 = 4.8 fm.
  • Medium : Density = 2.29 × 10¹⁷ kg/m³.
  • Exam (CET) : R ∝ ? (1) A^(1/3) (2) A (3) A² (4) 1/A → answer 1)

Rule 3: Nuclear force — Strong, attractive, short-range (~ 2–3 fm), charge-independent (p–p, n–n, p–n similar) and saturated.

  • Easy : At 1 fm the nuclear force ≫ Coulomb force.
  • Medium : Nearly zero beyond ~ 3 fm.
  • Exam (CET) : The nuclear force is? (1) short-range (2) long-range (3) repulsive (4) charge-dependent → answer 1)

Derivation : Density independent of A

  • Step 1 : Mass ≈ A u.
  • Step 2 : Volume = ⁴⁄₃ π R₀³ A.
  • Step 3 : ρ = A u ÷ (⁴⁄₃ π R₀³ A) = u ÷ (⁴⁄₃ π R₀³).
  • Step 4 : A cancels ⇒ ρ = 2.29 × 10¹⁷ kg/m³.
Remember

R = 1.2 A^(1/3) fm; ρ ≈ 2.3 × 10¹⁷ kg/m³; A = Z + N.

Common mistake / CET trap

Density is constant; R ∝ A^(1/3), not A.

Shortcut

Memory clue : "R goes as the cube root".

Spot the difference

ForceRangeRelative strength
Nuclear~ 2 fm1
Electricinfinite~ 10⁻²
Gravityinfinite~ 10⁻³⁸

Practice questions

Easy
  1. 1Relation of A, Z, N?
  2. 2What are isobars?
  3. 3Range of the nuclear force?
Medium
  1. 4A = 27 : R?
  2. 5²³⁵₉₂U : N?
  3. 6A = 8 and 64 : ratio of R?
Challenging
  1. 7Why is nuclear density independent of A?
  2. 8A = 125 : R?
Show answers (try them yourself first)
  1. 1A = Z + N
  2. 2Same A
  3. 3~ 2–3 fm
  4. 4R = 3.6 fm
  5. 5N = 143
  6. 61 : 2
  7. 7V ∝ A and m ∝ A
  8. 8R = 6 fm

Mass defect, binding energy, nuclear fission and fusion

From daily life

A tightly packed bundle needs force to break — binding energy is like that.

Simple definition

Binding energy is the energy needed to break a nucleus into its nucleons.

Binding energy per nucleonFe (≈ 8.8)fusionfissionA →

In the figure : The BE/A curve peaks near Fe.

Rule 1: Mass defect — Δm = Z m_p + (A−Z) m_n − M; E = Δm c²; 1 u = 931.5 MeV.

  • Easy : Deuteron (₁²H) : Δm = 1.007825 + 1.008665 − 2.013553 = 0.002937 u ⇒ BE = 2.74 MeV.
  • Medium : 1 g of matter fully to energy : E = 9 × 10¹³ J.
  • Exam (CET) : 1 u = ? (1) 931.5 MeV (2) 9.315 MeV (3) 93 MeV (4) 1 MeV → answer 1)

Rule 2: Binding energy — BE = Δm c²; a larger BE/A means more stable. The maximum near Fe-56 is ≈ 8.8 MeV.

  • Easy : ₁²H : BE/A = 1.37 MeV.
  • Medium : ⁴₂He : BE ≈ 28.3 MeV ⇒ BE/A ≈ 7.1 MeV.
  • Exam (CET) : Most stable? (1) Fe-56 (2) H-2 (3) U-238 (4) He-4 → answer 1)

Rule 3: Fission and fusion — Fission : a heavy nucleus splits (U-235 + n), ≈ 200 MeV per fission. Fusion : light nuclei join (H → He), needs ~ 10⁷ K.

  • Easy : 1 kg of U-235 : 2.5 × 10²⁴ nuclei × 200 MeV = 8 × 10¹³ J.
  • Medium : The Sun's energy : fusion.
  • Exam (CET) : High temperature is needed for? (1) fusion (2) fission (3) neither (4) decay → answer 1)

Derivation : Fission energy ≈ 200 MeV

  • Step 1 : At A = 240, BE/A ≈ 7.6 MeV.
  • Step 2 : At A ≈ 120 (two fragments) : BE/A ≈ 8.5 MeV.
  • Step 3 : Gain ≈ 0.9 MeV × 240.
  • Step 4 : Q ≈ 200 MeV.
Remember

Δm = Zm_p + (A−Z)m_n − M; E = Δm × 931.5 MeV; Q ≈ 200 MeV.

Common mistake / CET trap

Do not mix atomic and nuclear masses; Δm in u, then × 931.5.

Shortcut

Memory clue : "Fe is the peak; fission and fusion both move toward it".

Spot the difference

PropertyFissionFusion
Nucleiheavylight
Conditionneutron~ 10⁷ K
ExamplereactorSun

Practice questions

Easy
  1. 1What is mass defect?
  2. 21 u in MeV?
  3. 3What is fusion?
Medium
  1. 4Δm = 0.03 u : BE?
  2. 5BE = 160 MeV, A = 20 : BE/A?
  3. 62 × 10¹⁰ fissions × 200 MeV : joules?
Challenging
  1. 7Why is Fe-56 most stable?
  2. 8Why does fusion need high temperature?
Show answers (try them yourself first)
  1. 1Mass less than the sum of constituents
  2. 2931.5 MeV
  3. 3Light nuclei joining
  4. 4BE = 27.9 MeV
  5. 58 MeV
  6. 60.64 J
  7. 7Maximum BE/A
  8. 8To overcome Coulomb repulsion

Radioactive decay and the law of radioactive decay

From daily life

Popcorn pops one at a time — which and when is random, yet the rate is fixed.

Simple definition

Radioactive decay is the spontaneous disintegration of an unstable nucleus.

Radioactive decay : N versus tT½N₀/2N = N₀ e^(−λt)T½ = 0.693 ÷ λ

In the figure : An exponential fall; half at T½.

Live figure

Radioactive decay and the law of radioactive decay

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Rule 1: Law of decay — dN/dt = −λ N ⇒ N = N₀ e^(−λt); activity A = λ N, unit Bq (1 decay/s); 1 Ci = 3.7 × 10¹⁰ Bq.

  • Easy : T½ = 10 days ⇒ λ = 0.0693/day.
  • Medium : After 30 days N/N₀ = e^(−λt) = 0.125 = 1/8.
  • Exam (CET) : 1 Bq = ? (1) 1 decay/s (2) 1 Ci (3) 1 J (4) 1 Hz only → answer 1)

Rule 2: Half-life and mean life — T½ = 0.693 ÷ λ; mean life τ = 1 ÷ λ = 1.44 T½; after n half-lives N = N₀/2ⁿ.

  • Easy : T½ = 5 years ⇒ τ = 7.2 years.
  • Medium : 20 years = 4 half-lives ⇒ N₀/16.
  • Exam (CET) : τ = ? (1) 1.44 T½ (2) 0.693 T½ (3) T½ (4) 2 T½ → answer 1) Trick : τ is longer.

Rule 3: Decay types and uses — α : A − 4, Z − 2; β⁻ : Z + 1; γ : releases excitation energy. Uses : carbon dating (C-14, T½ ≈ 5730 y), medical diagnosis.

  • Easy : ²³⁸₉₂U → α : ²³⁴₉₀Th.
  • Medium : ¹⁴₆C → β⁻ : ¹⁴₇N.
  • Exam (CET) : After α decay Z changes by? (1) −2 (2) −4 (3) +1 (4) 0 → answer 1)

Derivation : N = N₀ e^(−λt)

  • Step 1 : dN/dt = −λ N.
  • Step 2 : dN/N = −λ dt; integrating : ln N − ln N₀ = −λ t.
  • Step 3 : N = N₀ e^(−λt).
  • Step 4 : N = N₀/2 at t = T½ ⇒ T½ = ln 2 ÷ λ = 0.693 ÷ λ.
Remember

N = N₀e^(−λt); T½ = 0.693/λ; τ = 1/λ; A = λN.

Common mistake / CET trap

Half-life is statistical for a sample; after two half-lives ¼ remains, not zero.

Shortcut

Memory clue : "halve every T½".

Spot the difference

QuantityFormulaUnit
NN₀e^(−λt)—
AλNBq
T½0.693/λs
τ1/λs

Practice questions

Easy
  1. 1Law of decay?
  2. 2Unit of λ?
  3. 3What is half-life?
Medium
  1. 4T½ = 20 min : N/N₀ after 1 h?
  2. 5λ = 0.1 s⁻¹ : T½?
  3. 6N = 10²⁰, λ = 10⁻⁹ s⁻¹ : A?
Challenging
  1. 7C-14 sample with 25% left : age (T½ = 5730 y)?
  2. 8Ratio of τ to T½?
Show answers (try them yourself first)
  1. 1dN/dt = −λN
  2. 2s⁻¹
  3. 3Time for N to halve
  4. 41/8
  5. 5T½ = 6.93 s
  6. 6A = 10¹¹ Bq
  7. 711460 years
  8. 81.44

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Spectrum1/λ = R(1/n₁² − 1/n₂²)m⁻¹
Bohrmvr = nh/2π; r_n = 0.529n²/Z ÅJ s; m
EnergyE_n = −13.6Z²/n² eVeV
Transitionhf = E_i − E_f; λ(nm) = 1240/ΔEnm
NucleusR = 1.2A^(1/3) fmm
BEΔm c²; 1 u = 931.5 MeVMeV
DecayN = N₀e^(−λt); T½ = 0.693/λs
ActivityA = λNBq

🎯 Most important points for the exam

  • Rutherford : small, massive, positive nucleus; r₀ = 2kZe²/E.
  • Balmer is visible; n₂ > n₁.
  • r_n ∝ n²/Z; v_n ∝ Z/n; E_n ∝ −Z²/n².
  • ΔE positive; lines = n(n−1)/2.
  • R = 1.2A^(1/3) fm; density constant.
  • BE/A peaks near Fe; 1 u = 931.5 MeV.
  • N = N₀/2ⁿ; τ = 1.44T½.
Study mantra

Fix n → formula (r, v, E) → ΔE → λ = 1240/ΔE → units.