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State Board · Class 12 · Physics

Lesson 14: Dual Nature of Radiation and Matter

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

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Experimental facts of the photoelectric effect and the failure of wave theory

From daily life

A solar calculator works only in light — light frees charges.

Simple definition

The photoelectric effect is the emission of electrons from a metal when light of suitable frequency falls on it.

Photoelectric effect : the set-upmetalcollectore⁻hf ≥ φ₀ ⇒ emission

In the figure : f ≥ f₀ ⇒ electrons are emitted.

Rule 1: Experimental facts — (1) No emission below the threshold frequency f₀. (2) E_k(max) depends on f, not on intensity. (3) Emission is instantaneous (< 10⁻⁹ s). (4) Current ∝ intensity.

  • Easy : f < f₀ ⇒ no emission however bright the light.
  • Medium : Intensity doubled ⇒ current doubles, E_k(max) unchanged.
  • Exam (CET) : E_k(max) depends on? (1) f (2) intensity (3) time (4) area → answer 1)

Rule 2: Failure of wave theory — Wave theory predicts more intensity ⇒ more E_k, no threshold and a delay in emission — all contradicting observation.

  • Easy : Wave theory : energy absorbed continuously ⇒ minutes of delay.
  • Medium : In reality emission is instantaneous.
  • Exam (CET) : Which fact does wave theory fail to explain? (1) threshold f (2) interference (3) diffraction (4) polarisation → answer 1)

Rule 3: Work function and threshold — φ₀ = h f₀ (the minimum energy to free an electron), unit eV; λ₀ = hc ÷ φ₀.

  • Easy : φ₀ = 2.3 eV ⇒ λ₀ = 1240 ÷ 2.3 = 539 nm.
  • Medium : f₀ = 5 × 10¹⁴ Hz ⇒ φ₀ = 2.07 eV.
  • Exam (CET) : hc = ? (1) 1240 eV nm (2) 12400 eV nm (3) 124 eV nm (4) 6.63 eV nm → answer 1)

Derivation : λ₀ = 1240 ÷ φ₀(eV) nm

  • Step 1 : φ₀ = h c ÷ λ₀ (J).
  • Step 2 : h c = 6.63 × 10⁻³⁴ × 3 × 10⁸ = 1.989 × 10⁻²⁵ J m.
  • Step 3 : In eV : ÷ 1.6 × 10⁻¹⁹ ⇒ 1.243 × 10⁻⁶ eV m = 1243 eV nm.
  • Step 4 : So λ(nm) = 1240 ÷ E(eV).
Remember

φ₀ = hf₀; λ₀(nm) = 1240/φ₀(eV).

Common mistake / CET trap

Raising intensity does not raise E_k; raising f does.

Shortcut

Memory clue : "f sets the energy, intensity sets the number".

Spot the difference

FactWave theoryPhoton theory
Threshold fnoyes
E_k ∝intensityf
Time lagyesinstant

Practice questions

Easy
  1. 1What is the threshold frequency?
  2. 2Unit of work function?
  3. 3State one experimental fact.
Medium
  1. 4φ₀ = 4.14 eV : λ₀?
  2. 5λ₀ = 620 nm : φ₀?
  3. 6f₀ = 8 × 10¹⁴ Hz : φ₀ (eV)?
Challenging
  1. 72000 Å light, φ₀ = 4.2 eV : will there be emission?
  2. 8Intensity halved : current and E_k?
Show answers (try them yourself first)
  1. 1Minimum frequency for emission
  2. 2eV
  3. 3A threshold f exists
  4. 4λ₀ = 300 nm
  5. 5φ₀ = 2 eV
  6. 6φ₀ = 3.31 eV
  7. 7E = 6.2 eV ⇒ 6.2 > 4.2 ⇒ yes
  8. 8Current halves; E_k same

Photon energy and Einstein's photoelectric equation

From daily life

A gift comes in a whole packet, not in halves — like a photon.

Simple definition

A photon is a packet of light energy : E = hf.

A photon is a packet of energyE = h f = hc ÷ λhf = φ₀ + E_k(max)one photon ⇒ one electron

In the figure : One photon ⇒ one electron.

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Photon energy and Einstein's photoelectric equation

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Rule 1: Photon energy — E = h f = h c ÷ λ; h = 6.63 × 10⁻³⁴ J s. The photon has zero rest mass.

  • Easy : f = 1.2 × 10¹⁵ Hz ⇒ E = 4.97 eV.
  • Medium : λ = 200 nm ⇒ E = 6.2 eV.
  • Exam (CET) : E ∝ ? (1) f (2) λ (3) 1/f (4) f² → answer 1)

Rule 2: Einstein's equation — h f = φ₀ + E_k(max) ⇒ E_k(max) = h f − φ₀ = ½ m v²(max) = e V₀.

  • Easy : E = 4.97 eV, φ₀ = 2.3 eV ⇒ E_k = 2.67 eV.
  • Medium : v(max) = 9.69 × 10⁵ m/s.
  • Exam (CET) : At f = f₀, E_k = ? (1) 0 (2) hf (3) φ₀ (4) ∞ → answer 1)

Rule 3: How it explains the facts — One photon gives its energy to one electron ⇒ instant; hf < φ₀ ⇒ no emission; intensity = number of photons.

  • Easy : Intensity doubled ⇒ twice the photons ⇒ twice the current.
  • Medium : 1 W of 500 nm light : photons/s = 1 ÷ (3.98 × 10⁻¹⁹) = 2.5 × 10¹⁸.
  • Exam (CET) : Intensity ∝ ? (1) number of photons (2) f (3) λ (4) φ₀ → answer 1)

Derivation : hf = φ₀ + ½mv²

  • Step 1 : Energy conservation : photon energy = work function + kinetic.
  • Step 2 : Most loosely bound electron : E_k maximum.
  • Step 3 : hf = φ₀ + E_k(max).
  • Step 4 : E_k(max) = h(f − f₀).
Remember

E = hf = hc/λ; hf = φ₀ + E_k.

Common mistake / CET trap

One photon gives energy to one electron, not several together.

Shortcut

Memory clue : "photon energy = work function + motion".

Spot the difference

QuantityFormulaUnit
Ehf = hc/λJ, eV
E_k(max)hf − φ₀eV
h6.63 × 10⁻³⁴J s; [ML²T⁻¹]

Practice questions

Easy
  1. 1What is a photon?
  2. 2Unit of h?
  3. 3Einstein's equation?
Medium
  1. 4λ = 400 nm : E (eV)?
  2. 5E = 5 eV, φ₀ = 2 eV : E_k?
  3. 6f = 10¹⁵ Hz : E (eV)?
Challenging
  1. 7E_k(max) = 1.5 eV, φ₀ = 2.5 eV : f?
  2. 8E_k(max) = 2 eV; f doubled, φ₀ = 1 eV : new E_k?
Show answers (try them yourself first)
  1. 1A packet of light energy
  2. 2J s
  3. 3hf = φ₀ + E_k
  4. 4E = 3.1 eV
  5. 5E_k = 3 eV
  6. 6E = 4.14 eV
  7. 7f = 9.65 × 10¹⁴ Hz
  8. 8E = 3 eV ⇒ E_k = 5 eV

Stopping potential, saturation current and graphs

From daily life

Going up a slope, the fastest climber stops last — the stopping potential does the same for electrons.

Simple definition

The stopping potential V₀ stops the fastest electrons.

Current versus potential graph−V₀higher intensitylower intensityVI

In the figure : Intensity ⇒ I_sat; f ⇒ V₀.

Rule 1: Stopping potential — The reverse potential V₀ that makes I = 0 : e V₀ = E_k(max) = h f − φ₀. V₀ depends on f, not intensity.

  • Easy : E_k = 2.67 eV ⇒ V₀ = 2.67 V.
  • Medium : E_k = 3 eV ⇒ V₀ = 3 V.
  • Exam (CET) : V₀ depends on? (1) f (2) intensity (3) area (4) distance → answer 1)

Rule 2: Saturation current — When all emitted electrons reach the collector, I stops rising; I_sat ∝ intensity.

  • Easy : Intensity doubled ⇒ I_sat doubles; V₀ unchanged.
  • Medium : Raise f ⇒ V₀ rises; I_sat (same photon number) unchanged.
  • Exam (CET) : I_sat ∝ ? (1) intensity (2) f (3) V₀ (4) φ₀ → answer 1)

Rule 3: V₀ versus f graph — A straight line : V₀ = (h/e) f − φ₀/e; slope h/e (the same for all metals); intercept on the f axis is f₀.

  • Easy : Slope = 4.14 × 10⁻¹⁵ V s ⇒ h = 4.14 × 10⁻¹⁵ × 1.6 × 10⁻¹⁹ = 6.62 × 10⁻³⁴ J s.
  • Medium : A different metal ⇒ a parallel line with a different f₀.
  • Exam (CET) : Slope of the graph? (1) h/e (2) h (3) e/h (4) φ₀ → answer 1)

Derivation : V₀ = (h/e)(f − f₀)

  • Step 1 : e V₀ = h f − φ₀.
  • Step 2 : V₀ = (h/e) f − φ₀/e.
  • Step 3 : y = m x + c; slope m = h/e.
  • Step 4 : At V₀ = 0, f = f₀ = φ₀/h.
Remember

eV₀ = hf − φ₀; I_sat ∝ intensity; slope h/e.

Common mistake / CET trap

V₀ does not depend on intensity; the slope does not change with the metal.

Shortcut

Memory clue : "intensity = number, f = speed".

Spot the difference

GraphWhat changesWhat is fixed
I–V (intensity)I_satV₀
I–V (f)V₀I_sat
V₀–fslope h/esame for all metals

Practice questions

Easy
  1. 1Stopping potential?
  2. 2I_sat depends on?
  3. 3Slope of V₀–f?
Medium
  1. 4E_k = 2.5 eV : V₀?
  2. 5V₀ = 1.8 V, φ₀ = 2 eV : hf (eV)?
  3. 6Slope 4.1 × 10⁻¹⁵ : h?
Challenging
  1. 7Does doubling f double V₀?
  2. 8Intensity tripled : I_sat and V₀?
Show answers (try them yourself first)
  1. 1Reverse potential that makes I = 0
  2. 2Intensity
  3. 3h/e
  4. 4V₀ = 2.5 V
  5. 5hf = 3.8 eV
  6. 6h = 6.56 × 10⁻³⁴ J s
  7. 7No; V₀ = (h/e)f − φ₀/e
  8. 8I_sat triples; V₀ same

Photon momentum, Compton scattering and the wave–particle duality of radiation

From daily life

A comet's tail points away from the Sun — pushed by light pressure.

Simple definition

A photon's momentum is p = h/λ; light has a dual wave–particle nature.

Photon momentum and Compton scatteringhfhf'e⁻p = h ÷ λ = E ÷ c

In the figure : Photon–electron collision; p = h/λ.

Rule 1: Momentum of a photon — p = E ÷ c = h ÷ λ = h f ÷ c; zero rest mass; direction of the light.

  • Easy : λ = 500 nm ⇒ p = 1.33 × 10⁻²⁷ kg m/s.
  • Medium : E = 3.1 eV ⇒ p = 3.1 × 1.6 × 10⁻¹⁹ ÷ 3 × 10⁸ = 1.65 × 10⁻²⁷.
  • Exam (CET) : p ∝ ? (1) 1/λ (2) λ (3) λ² (4) constant → answer 1)

Rule 2: Radiation pressure — Pressure on an absorber = I ÷ c; on a reflector = 2I ÷ c, from the photon's change of momentum.

  • Easy : I = 1000 W/m² ⇒ pressure = 3.3 × 10⁻⁶ Pa (absorber).
  • Medium : A reflector ⇒ double.
  • Exam (CET) : Pressure on a reflector? (1) 2I/c (2) I/c (3) I c (4) I/2c → answer 1)

Rule 3: Wave–particle duality — Light behaves as a wave in interference and diffraction and as particles in photoelectric and Compton effects; no single experiment shows both together.

  • Easy : Young's slits ⇒ wave.
  • Medium : Photoelectric effect ⇒ particle.
  • Exam (CET) : Nature of light? (1) dual (2) only wave (3) only particle (4) neither → answer 1)

Derivation : p = h ÷ λ

  • Step 1 : Photon : E = pc (m₀ = 0).
  • Step 2 : E = h f = h c ÷ λ.
  • Step 3 : p = E ÷ c = h ÷ λ.
  • Step 4 : λ = h ÷ p (the basis of de Broglie's relation).
Remember

p = h/λ = E/c; radiation pressure I/c; 2I/c.

Common mistake / CET trap

A photon has zero rest mass but non-zero momentum.

Shortcut

Memory clue : "long λ — small p".

Spot the difference

QuantityFormulaUnit
ph/λ = E/ckg m/s
Pressure (absorber)I/cPa
Pressure (reflector)2I/cPa

Practice questions

Easy
  1. 1Momentum of a photon?
  2. 2What is duality?
  3. 3Pressure on a reflector?
Medium
  1. 4λ = 600 nm : p?
  2. 5I = 500 W/m² : pressure (absorber)?
  3. 6E = 6.2 eV : p?
Challenging
  1. 7λ halved : p?
  2. 81 W fully reflected : force?
Show answers (try them yourself first)
  1. 1h/λ
  2. 2Both wave and particle properties
  3. 32I/c
  4. 4p = 1.1 × 10⁻²⁷ kg m/s
  5. 5Pressure = 1.67 × 10⁻⁶ Pa
  6. 6p = 3.31 × 10⁻²⁷ kg m/s
  7. 7Doubles
  8. 8F = 2P/c = 6.67 × 10⁻⁹ N

de Broglie wavelength of matter waves

From daily life

A squirrel shows no wave nature, but an electron does.

Simple definition

Matter waves are waves associated with moving particles; λ = h/p.

de Broglie wave : λ = h ÷ pvλ = h ÷ m v = h ÷ √(2 m E_k)λ = 12.27 ÷ √V Å (e⁻)

In the figure : A particle ⇒ a wave; λ = h/mv.

Rule 1: de Broglie wavelength — λ = h ÷ p = h ÷ m v; from kinetic energy λ = h ÷ √(2 m E_k).

  • Easy : Electron accelerated through 100 V : λ = 1.23 Å (= 12.27 ÷ √100).
  • Medium : 0.1 kg ball at 10 m/s : λ = 6.63 × 10⁻³⁴ m (undetectable).
  • Exam (CET) : λ ∝ ? (1) 1/p (2) p (3) p² (4) constant → answer 1)

Rule 2: Handy formula for electrons — An electron accelerated through V volts : λ = 12.27 ÷ √V Å; same with E in eV.

  • Easy : V = 150 ⇒ λ = 1.00 Å.
  • Medium : V four times ⇒ λ halves.
  • Exam (CET) : V = 49 ⇒ λ? (1) 1.75 Å (2) 12.27 Å (3) 0.5 Å (4) 7 Å → answer 1) Trick : 12.27 ÷ 7.

Rule 3: Comparing particles — For the same accelerating V : λ ∝ 1/√(m q). For the same E_k : λ ∝ 1/√m.

  • Easy : Proton through 100 V : λ = 2.87 × 10⁻¹² m.
  • Medium : Equal E_k : λ_e ÷ λ_p = √(m_p/m_e) = 43.
  • Exam (CET) : Same V : λ_p : λ_α = ? (1) 2√2 : 1 (2) 1 : 1 (3) 4 : 1 (4) 2 : 1 → answer 1)

Derivation : λ = 12.27 ÷ √V Å

  • Step 1 : E_k = e V = p² ÷ 2m ⇒ p = √(2 m e V).
  • Step 2 : λ = h ÷ √(2 m e V).
  • Step 3 : h = 6.63 × 10⁻³⁴; m = 9.11 × 10⁻³¹; e = 1.6 × 10⁻¹⁹.
  • Step 4 : λ = 12.28 ÷ √V Å ⇒ 12.27 ÷ √V Å.
Remember

λ = h/mv = h/√(2mE_k); electron : 12.27/√V Å.

Common mistake / CET trap

λ ∝ 1/p, 1/√E_k, 1/√V; take mass in kg.

Shortcut

Memory clue : "12.27 ÷ √V".

Spot the difference

ParticleSame VSame E_k
λ ∝1/√(mq)1/√m
e : p43 : 1 (e larger)43 : 1

Practice questions

Easy
  1. 1de Broglie's formula?
  2. 2The 12.27 formula for electrons?
  3. 3Why can't we see λ of large objects?
Medium
  1. 4V = 25 : λ (e)?
  2. 5m = 2 g, v = 5 m/s : λ?
  3. 6p = 3.3 × 10⁻²⁴ : λ?
Challenging
  1. 7Electron of 100 eV : λ?
  2. 8Same λ : ratio of kinetic energies of e and p?
Show answers (try them yourself first)
  1. 1λ = h/p
  2. 2λ = 12.27/√V Å
  3. 3h is tiny ⇒ λ negligible
  4. 4λ = 2.45 Å
  5. 5λ = 6.63 × 10⁻³² m
  6. 6λ = 2.01 × 10⁻¹⁰ m
  7. 71.227 Å
  8. 8E_e : E_p = m_p : m_e ≈ 1833 : 1

Davisson–Germer experiment and when the wave nature shows

From daily life

A CD shows rainbow colours by diffraction; electrons diffract likewise at a crystal.

Simple definition

The Davisson–Germer experiment experimentally confirmed the wave nature of electrons.

Davisson–Germer : electron diffractiongunNiφ = 50°2 d sin θ = n λ54 V ⇒ λ = 1.67 Å

In the figure : 54 V electrons; maximum at φ = 50°.

Rule 1: Davisson–Germer experiment — Electrons accelerated through 54 V strike a nickel crystal; a maximum at φ = 50° confirms their wave nature.

  • Easy : λ (theory) = 12.27 ÷ √54 = 1.67 Å.
  • Medium : λ (experiment) = d sin φ = 2.15 × sin 50° = 1.65 Å.
  • Exam (CET) : What did the experiment prove? (1) wave nature of electrons (2) photons (3) neutrons (4) α → answer 1)

Rule 2: Bragg condition — For lattice spacing d : 2 d sin θ = n λ; for the surface grating of nickel d sin φ = n λ.

  • Easy : d = 2.15 Å, φ = 50°, n = 1 ⇒ λ = 1.65 Å.
  • Medium : Theory and experiment nearly agree.
  • Exam (CET) : Bragg's condition? (1) 2d sinθ = nλ (2) d = nλ (3) sinθ = nλ (4) d sinθ = 2nλ → answer 1)

Rule 3: When does wave nature show — Only when λ is comparable to the obstacle size. For tiny λ everyday objects behave as particles.

  • Easy : An electron (λ ≈ 1 Å) diffracts at a crystal.
  • Medium : A cricket ball with λ ≈ 10⁻³⁴ m shows no diffraction.
  • Exam (CET) : Condition for diffraction? (1) λ ≈ size (2) λ ≫ size (3) λ ≪ size (4) any → answer 1)

Derivation : 2d sin θ = nλ

  • Step 1 : Path difference of rays from adjacent planes = 2 d sin θ.
  • Step 2 : Constructive interference : path difference = nλ.
  • Step 3 : 2 d sin θ = n λ.
  • Step 4 : Here θ is the glancing angle (from the plane).
Remember

2d sinθ = nλ; λ = 12.27/√V; diffraction needs λ ≈ size.

Common mistake / CET trap

θ is the glancing angle, not the angle of incidence.

Shortcut

Memory clue : "54 V — 50° — 1.65 Å".

Spot the difference

ExperimentProvesCondition
Young/diffractionlight is a wave—
Photoelectriclight is a particlehf ≥ φ₀
Davisson–Germerelectron waveλ ≈ d

Practice questions

Easy
  1. 1Who did the experiment?
  2. 2Which crystal was used?
  3. 3Significance of the experiment?
Medium
  1. 4V = 54 : λ?
  2. 5d = 2 Å, θ = 30°, n = 1 : λ?
  3. 6λ = 1.2 Å, d = 2.4 Å : sin θ (n = 1)?
Challenging
  1. 7Why does a large ball not diffract?
  2. 8V = 150 : λ and the consequence?
Show answers (try them yourself first)
  1. 1Davisson and Germer
  2. 2Nickel
  3. 3Wave nature of electrons
  4. 4λ = 1.67 Å
  5. 5λ = 2.0 Å
  6. 6sin θ = 0.25
  7. 7λ is far too small
  8. 8λ = 1 Å ⇒ diffraction at crystals

Electron microscope and methods of electron emission

From daily life

A virus can be imaged only with an electron microscope, not an optical one.

Simple definition

An electron microscope uses the small λ of electron waves to give high resolution.

Electron microscope : small λ ⇒ high resolutionguncondenserobjectobjectivescreena ≈ 0.61λ ÷ N.A.

In the figure : Gun → lenses → specimen → screen.

Rule 1: Electron microscope — Electrons with λ ≈ 0.05 Å (50 kV) give resolution about a thousand times that of a light microscope. They use magnetic or electrostatic lenses.

  • Easy : 50 kV : λ = 12.27 ÷ √50000 = 0.055 Å.
  • Medium : Light (500 nm) : a ≈ 300 nm; electrons : ≈ 0.03 Å.
  • Exam (CET) : Better resolution when? (1) λ is small (2) λ is large (3) V is low (4) no change → answer 1)

Rule 2: Limitations — Needs a vacuum; the specimen must be thin and conducting; living specimens do not survive.

  • Easy : Why a vacuum? Air molecules scatter the electrons.
  • Medium : TEM shows internal structure; SEM shows surfaces.
  • Exam (CET) : What does a TEM show? (1) internal structure (2) surface (3) colour (4) temperature → answer 1)

Rule 3: Ways of emitting electrons — (1) thermionic (heating), (2) field emission (strong E), (3) photoelectric, (4) secondary emission.

  • Easy : A hot filament : thermionic emission.
  • Medium : E ≈ 10⁸ V/m : field emission.
  • Exam (CET) : A photocell uses? (1) photoelectric (2) thermionic (3) field (4) secondary → answer 1)

Derivation : Resolution depends on λ

  • Step 1 : In a microscope a = 0.61 λ ÷ (n sin θ).
  • Step 2 : Electrons : λ = 12.27 ÷ √V Å.
  • Step 3 : Large V ⇒ small λ ⇒ small a.
  • Step 4 : Hence electrons can image down to atomic scale.
Remember

e λ = 12.27/√V Å; a = 0.61λ/N.A.; methods : thermionic, field, photoelectric, secondary.

Common mistake / CET trap

In an electron microscope a higher V means a smaller λ and better resolution.

Shortcut

Memory clue : "small λ = better resolution".

Spot the difference

MethodCauseExample
Thermionicheatfilament
Fieldstrong EFEM
Photoelectricphotonsphotocell
SecondaryimpactPMT

Practice questions

Easy
  1. 1Lenses used in an electron microscope?
  2. 2One limitation?
  3. 3Methods of emission?
Medium
  1. 4V = 10 kV : λ?
  2. 5λ halved : resolution?
  3. 6V four times : λ?
Challenging
  1. 7a = 0.61λ/N.A., λ = 0.1 Å, N.A. = 0.01 : a?
  2. 8Why a vacuum for electrons?
Show answers (try them yourself first)
  1. 1Magnetic/electrostatic
  2. 2Vacuum, thin specimen
  3. 3Thermionic, field, photoelectric, secondary
  4. 4λ = 0.123 Å
  5. 5Resolution doubles
  6. 6λ halves
  7. 7a = 6.1 Å
  8. 8Air scatters electrons

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
PhotonE = hf = hc/λ; λ(nm) = 1240/E(eV)J, eV
Einsteinhf = φ₀ + E_k(max)eV
V₀eV₀ = E_k(max)V
GraphV₀–f slope = h/eV s
Momentump = h/λ = E/ckg m/s
de Broglieλ = h/mv; e : 12.27/√V Åm
Davisson2d sinθ = nλÅ
Microscopea = 0.61λ/N.A.m

🎯 Most important points for the exam

  • f < f₀ ⇒ no emission; E_k independent of intensity.
  • hf = φ₀ + E_k; λ(nm) = 1240/E(eV).
  • V₀ independent of intensity; I_sat ∝ intensity.
  • Slope of the V₀–f graph is h/e.
  • p = h/λ; radiation pressure I/c, 2I/c.
  • λ = h/√(2mE_k); e : 12.27/√V Å.
  • Diffraction needs λ ≈ size; 2d sinθ = nλ.
Study mantra

Start with E = hf → subtract φ₀ → E_k → V₀ = E_k/e → λ = h/p.