Take a breath. This chapter has a reputation for being strange, and the strangeness is real — but the arithmetic is some of the friendliest in all of Class 12 Physics. Almost every question you will ever be asked here is one subtraction and one division in disguise. What trips students up is never the algebra. It is the units, and a handful of stubborn wrong pictures we carry in from Class 11.
This chapter is the sequel to a fight. In Wave Optics for Class 12 Physics you spent every worked example proving light is a wave; here the photoelectric effect refuses to fit that picture. Keep both chapters open side by side — the contrast is the whole point. If you also want the geometry side of light, Ray Optics and Optical Instruments covers it.
So here is the promise for this session. We are going to treat every single photoelectric problem as a one-line bank statement: one photon walks in with a fixed amount of energy, it pays a compulsory entry fee to get an electron out of the metal, and whatever is left over becomes the electron’s kinetic energy. Money in, money out, nothing borrowed, nothing shared. Once that ledger is in your head, the whole chapter collapses into a single habit.
These are your complete dual nature of radiation and matter class 12 physics notes with solved examples — built for the 2026-27 CBSE syllabus, with sixteen worked example cards, a ten-question practice worksheet, a full formula list, and three computed diagrams. If you have been searching for dual nature of radiation and matter class 12 physics important questions or a clean explanation of why bright light still cannot eject an electron, you are in the right place. Read it slowly the first time. Speed comes later.
Meet Your Tutor
Dual Nature becomes reliable when you separate what intensity changes from what frequency changes. I will help you read photoelectric graphs, track units through Einstein’s equation and de Broglie’s relation, and use quick physical checks so every numerical tells the same particle-and-wave story.
What You’ll Learn
- Dual Nature of Radiation: Why Light Needs Two Descriptions
- The Photon Ledger — One Photon, One Electron
- Hertz and Lenard’s Observations
- Work Function, Threshold Frequency and Threshold Wavelength
- Experimental Study of the Photoelectric Effect
- Stopping Potential and the Straight Line of Slope h/e
- Photocurrent vs Collector Potential: Reading the Graphs
- Einstein’s Photoelectric Equation — Particle Nature of Light
- eV vs Joule: The Unit Discipline Drill
- What Changes When You Vary Intensity, Frequency or Potential
- Matter Waves — The Wave Nature of Particles
- The de Broglie Relation and Its Three Faces
- de Broglie Wavelength from Accelerating Voltage
- Three Conceptual Traps Examiners Love
- What’s Been Trimmed for 2026-27
- Dual Nature of Radiation and Matter Formula List
Your Game Plan
- Learn the ledger sentence first: photon energy in = work function + maximum kinetic energy out. Say it aloud until it is boring.
- Fix your units before you touch a calculator. Decide, on every question, whether you are working in joules or in electronvolts — never both at once.
- Memorise exactly two shortcuts: E(eV) = 1240 / λ(nm), and λ = 12.27/√V ångström for an accelerated electron.
- Practise reading the two standard graphs until you can sketch them from memory with the axes labelled.
- Finish with the ten-question worksheet at the bottom. Do it with a pen, not by scrolling.
Study Notes
Dual Nature of Radiation: Why Light Needs Two Descriptions
For most of the nineteenth century, light was settled business. It was a wave. It diffracted around edges, it interfered to make bright and dark fringes, and Maxwell had shown it was a self-sustaining ripple of electric and magnetic fields travelling at c. If you have already worked through Electromagnetic Waves for Class 12 Physics, you have seen exactly how tidy that picture is.
Then a small, awkward experiment refused to behave. Shine light on a clean metal surface and electrons come flying off. That much is fine. But how they came off broke every wave prediction at once — the timing was wrong, the energy dependence was wrong, and the effect of brightness was wrong. Not slightly wrong. Wrong in kind.
The resolution was not “light is really a particle after all.” It was something more careful: light carries energy in indivisible lumps called photons, while still propagating as a wave. Which face you see depends on which question you ask. Ask about interference, you get waves. Ask about energy exchange with a single electron, you get particles. That is the dual nature of radiation.
And then, in 1924, Louis de Broglie asked the obvious follow-up that nobody had dared to ask: if a wave can behave like a particle, can a particle behave like a wave? It could. Electrons, protons, atoms — every piece of matter carries a wavelength. That is the dual nature of matter, and it is the second half of this chapter.
The Photon Ledger — One Photon, One Electron
This is the heart of the chapter, so we will build it slowly.
Imagine a small shop. A photon walks in carrying exactly hν joules — not a rupee more, not a rupee less. Inside the metal there is an electron who wants to leave. To get out through the surface, the electron must pay a compulsory exit fee. That fee is the work function, written φ0, and it is a fixed property of the metal, like a fixed toll.
The photon hands over its entire energy to one electron. Not half. Not spread over three electrons. All of it, to one. The electron pays the toll, and whatever is left is its kinetic energy as it leaves.
hν = φ0 + Kmax
Every photoelectric numerical in this chapter is this one line, rearranged. If you can identify which two of the three quantities a question gives you, you already have the answer.
Two consequences fall out immediately, and both of them are the classic exam traps.
- If the photon cannot afford the toll, nothing happens. If hν < φ0, the electron cannot leave. It cannot save up. It cannot combine its photon with the next photon’s. One photon, one electron — the accounts never merge.
- Brightness adds customers, not richer customers. Doubling the intensity sends in twice as many photons per second, each with the same hν. So twice as many electrons come out, but each one leaves with exactly the same maximum kinetic energy.
Notice that Kmax is the maximum. Electrons buried deeper inside the metal need to spend extra energy just reaching the surface, so they come out with less. The work function is the minimum possible toll, paid only by the electrons sitting right at the surface. That is why the equation gives a ceiling, not a fixed value.
Ledger step: we only need “energy in”. Use E = hc/λ.
E = (6.626 × 10−34 × 3.00 × 108) / (500 × 10−9)
hc = 1.9878 × 10−25 J·m, so E = 1.9878 × 10−25 / 5.00 × 10−7 = 3.98 × 10−19 J (3 s.f.).
In electronvolts: E = 3.9756 × 10−19 / 1.602 × 10−19 = 2.48 eV.
Cross-check with the shortcut: E(eV) = 1240.8 / λ(nm) = 1240.8 / 500 = 2.48 eV. Same answer, one line.
Energy of one photon: E = 1240.8 / 600 = 2.068 eV = 2.068 × 1.602 × 10−19 = 3.313 × 10−19 J.
Photons per second: n = P/E = 60 / (3.313 × 10−19) = 1.81 × 1020 photons per second (3 s.f.).
Why this matters: that number is so enormous that the lumpiness of light is completely invisible to your eye. The graininess only shows up when a single photon has to deal with a single electron — which is exactly the photoelectric situation.
Hertz and Lenard’s Observations
Physics rarely arrives fully formed, and this effect was stumbled into rather than hunted down. Heinrich Hertz, in 1887, was busy proving Maxwell right — making and detecting electromagnetic waves with a spark gap. He noticed something irritating on the side: the spark in his detector jumped more readily when ultraviolet light fell on the metal electrodes. He recorded it and moved on. It was, to him, a nuisance.
Wilhelm Hallwachs and Philipp Lenard picked up the nuisance and turned it into a proper investigation over the following years. Lenard’s setup is the one you should picture: two metal plates sealed in an evacuated glass tube, light let in through a quartz window onto one plate (the emitter), and the other plate (the collector) held at a controllable potential, with a sensitive ammeter in the circuit.
What they established is worth listing precisely, because CBSE loves asking for “the observations” as a straight three-mark question.
- Light ejects electrons from metal surfaces. The particles emitted were shown to be electrons — same charge-to-mass ratio, whatever the metal.
- Each metal has a threshold frequency. Below a certain frequency ν0, no electrons are emitted at all, no matter how intense the light or how long you wait.
- Above threshold, the maximum electron energy depends on frequency, not intensity. Making the light brighter did not make the electrons faster.
- The number of electrons depends on intensity. Brighter light at the same frequency gave a proportionally larger current.
- The emission is effectively instantaneous. Electrons appeared within about 10−9 s of the light arriving, even for very feeble light.
Notice that this whole experiment is a current-measurement problem at heart. If the idea of a controlled potential difference driving a measurable current in a circuit feels shaky, a quick revision of Current Electricity will make the apparatus much easier to picture.
Work Function, Threshold Frequency and Threshold Wavelength
The work function φ0 is the minimum energy needed to pull the least tightly bound electron out through a metal’s surface. It is a property of the metal and its surface condition, and it is almost always quoted in electronvolts because the numbers are conveniently small — roughly 2 eV for the alkali metals, 4 to 5 eV for most common metals.
Because the ledger says the photon must at minimum cover the toll, there is a lowest frequency that can do the job at all. Set Kmax = 0:
λ0 = c / ν0 = hc / φ0
Light with ν < ν0 (equivalently λ > λ0) produces no emission whatsoever.
Convert the toll to joules first: φ0 = 2.14 × 1.602 × 10−19 = 3.4283 × 10−19 J.
Threshold frequency: ν0 = φ0/h = 3.4283 × 10−19 / 6.626 × 10−34 = 5.17 × 1014 Hz (3 s.f.).
Threshold wavelength: λ0 = c/ν0 = 3.00 × 108 / 5.174 × 1014 = 5.80 × 10−7 m = 580 nm.
One-line check: λ0(nm) = 1240.8 / 2.14 = 579.8 nm. Agrees.
Physical reading: 580 nm is yellow-green. So caesium responds to visible light — which is exactly why caesium is used in photocells.
Energy in: E = 1240.8 / 650 = 1.909 eV.
Compare with the toll: 1.909 eV < 2.14 eV.
Answer: there are no emitted electrons. The question has no kinetic energy to report.
Do not write “Kmax = −0.23 eV”. A negative kinetic energy is your signal that emission does not occur. Say so in words — that is where the mark is. Increasing the laser power sends in more photons, each still worth only 1.909 eV, and none of them can afford a 2.14 eV toll.
Experimental Study of the Photoelectric Effect
Now the apparatus, because every graph question in this chapter is really a question about this circuit.
An evacuated tube holds two electrodes. Monochromatic light of chosen frequency and intensity falls on the emitter plate through a quartz window (ordinary glass absorbs ultraviolet, which is why quartz is specified). The collector plate faces it. A battery with a sliding contact lets you make the collector either positive or negative with respect to the emitter, and a microammeter reads the photocurrent.
You have three knobs. Turn each one and watch what moves — that is the entire experimental study.
- Knob 1: intensity (at fixed frequency and fixed positive collector potential). The photocurrent rises in direct proportion. More photons per second means more electrons per second.
- Knob 2: frequency (at fixed intensity). Below ν0, zero current at any intensity. Above ν0, electrons emerge and their maximum energy grows linearly with ν.
- Knob 3: collector potential. Make the collector more positive and the current rises, then flattens into a saturation current — every emitted electron is now being collected, so there are simply no more to gather. Make the collector negative and you push electrons back; at a particular negative value the current falls to exactly zero.
Stopping Potential and the Straight Line of Slope h/e
Turn the collector negative. Now the electric field between the plates pushes emitted electrons back towards the emitter. A slow electron gets turned around; a fast one still makes it across. As you make the collector more and more negative, you knock out the slower electrons one group at a time and the current falls.
At one precise value — call it the stopping potential V0 — even the fastest electron is turned back just before arriving. The current becomes exactly zero. That fastest electron had kinetic energy Kmax, and it spent all of it climbing the potential hill:
This is a beautiful measuring trick: you cannot weigh an electron’s speed directly, but you can read a voltmeter. The stopping potential converts an energy you cannot see into a voltage you can.
Substitute this into the ledger and divide through by e, and the whole chapter becomes a straight line:
Compare with y = mx + c:
• slope = h/e = 4.136 × 10−15 V·s — the same for every metal
• x-intercept = ν0 (the threshold frequency)
• y-intercept = −φ0/e (numerically the work function in eV, made negative)
Look at what that graph is telling you. Repeat the experiment with sodium instead of caesium and you get a parallel line — shifted right and down, but with identical slope. The slope belongs to nature, not to the metal. That is why this graph is the standard textbook method for measuring Planck’s constant.
Ledger, in eV: energy in = 1240.8 / 400 = 3.1021 eV.
Kmax = 3.1021 − 2.14 = 0.962 eV (3 s.f.).
Stopping potential: V0 = Kmax/e. Because Kmax is already in eV, the number transfers straight across: V0 = 0.962 V.
Speed — now you must convert: Kmax = 0.9621 × 1.602 × 10−19 = 1.5412 × 10−19 J.
vmax = √(2K/m) = √(2 × 1.5412 × 10−19 / 9.11 × 10−31) = 5.82 × 105 m s−1.
The habit to notice: stopping potential in volts equals Kmax in eV, always. Speed forces you into joules, always.
Use the slope — the work function cancels out:
slope = (2.823 − 1.169) / ((12.0 − 8.0) × 1014) = 1.654 / (4.0 × 1014) = 4.136 × 10−15 V·s.
h = slope × e = 4.136 × 10−15 × 1.602 × 10−19 = 6.63 × 10−34 J s (3 s.f.).
Bonus — recover the work function: φ0 = hν − eV0 = (4.136 × 10−15 × 8.0 × 1014) − 1.169 = 3.309 − 1.169 = 2.14 eV, in eV directly because we used h/e. Caesium, confirmed.
Photocurrent vs Collector Potential: Reading the Graphs
This is the second graph, and it is the one students most often draw wrong under exam pressure. Fix the frequency. Vary only the intensity. Plot photocurrent against collector potential for three different brightnesses.
Two features carry all the marks. First, the three saturation levels are different — they scale exactly with intensity, because intensity sets how many electrons per second leave the surface. Second, the three curves hit zero at exactly the same point. That single shared intercept is the experimental proof that intensity has nothing to do with electron energy.
Einstein’s Photoelectric Equation — Particle Nature of Light
We have been using it all along; now let us state it formally and see why it deserved a Nobel Prize.
In 1905 Einstein proposed that light is not merely emitted in quanta (Planck had already suggested that for the walls of a hot cavity) but travels and is absorbed in quanta. A beam of frequency ν is a stream of photons, each of energy hν, and the absorption is all-or-nothing by a single electron.
Equivalent everyday forms you should be able to write instantly:
• Kmax = h(ν − ν0)
• Kmax = hc(1/λ − 1/λ0)
• eV0 = hν − φ0
• ½mv2max = hν − φ0
Now check it against each stubborn observation, and watch them all fall into place:
- Threshold frequency — because Kmax cannot be negative, emission needs hν ≥ φ0. A sharp cut-off, exactly as observed.
- Instantaneous emission — the transaction is between one photon and one electron, so there is no waiting to accumulate energy. Even one photon per second gives immediate emission (just very rarely).
- Intensity controls current, not energy — intensity is photons per second, so it controls how many electrons leave, while hν alone fixes how energetic each one is.
- Linear V0 vs ν graph — a direct algebraic rearrangement, with a universal slope h/e.
The other half of the particle picture is momentum. A photon has zero rest mass but carries momentum p = h/λ = hν/c = E/c. You will not be asked to derive this, but you will be asked to compute it, and it is the bridge to the second half of the chapter.
(a) Energy in = 1240.8 / 300 = 4.1361 eV. Kmax = 4.1361 − 2.28 = 1.86 eV (3 s.f.).
(b) V0 = 1.86 V.
(c) Kmax = 1.8561 × 1.602 × 10−19 = 2.9735 × 10−19 J; vmax = √(2 × 2.9735 × 10−19 / 9.11 × 10−31) = 8.08 × 105 m s−1.
(d) λ0 = 1240.8 / 2.28 = 544 nm, so ν0 = 3.00 × 108 / 5.442 × 10−7 = 5.51 × 1014 Hz.
Photon: p = h/λ = 6.626 × 10−34 / 5.00 × 10−7 = 1.33 × 10−27 kg m s−1 (3 s.f.).
Electron: p = mv = 9.11 × 10−31 × 1.00 × 106 = 9.11 × 10−25 kg m s−1.
Ratio: the electron carries about 687 times more momentum, yet its energy (2.84 eV, if you work it out) is comparable to the photon’s 2.48 eV. Photons are energetic but momentum-light — a direct consequence of E = pc versus E = p2/2m.
eV vs Joule: The Unit Discipline Drill
Time for the drill that saves more marks than any concept in this chapter. If you skim one section, do not let it be this one.
An electronvolt is the energy gained by a charge e falling through 1 volt. So 1 eV = 1.602 × 10−19 J, exactly by definition. It is a unit of energy, not of voltage — the word “volt” hiding inside it is the entire problem.
eV lane: work function in eV, photon energy from E(eV) = 1240.8/λ(nm), Kmax in eV, and V0 in volts is the same number as Kmax in eV. Fast, and safe for most questions.
Joule lane: everything in SI. Compulsory the moment a question asks for speed, momentum, or a de Broglie wavelength, because those formulas contain the mass in kilograms.
Cross lanes only at a clearly written conversion step.
Here is the short conversion table to keep in your memory, all verified against h = 6.626 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.602 × 10−19 C:
| Conversion | Value | Use it when |
|---|---|---|
| 1 eV → joule | 1.602 × 10−19 J | Before any speed or de Broglie calculation |
| 1 J → eV | 6.242 × 1018 eV | Reporting an answer in the units the question used |
| hc | 1.988 × 10−25 J m | Photon energy in joules from λ in metres |
| hc/e | 1240.8 eV·nm (use 1240) | Photon energy in eV from λ in nm — the fastest line in the chapter |
| h/e | 4.136 × 10−15 V s | Slope of the V0 vs ν graph |
| 1 ångström | 10−10 m = 0.1 nm | de Broglie wavelengths of electrons |
(a) Stopping potential? Stay in the eV lane. V0 = 2.50 V. No conversion needed — and no multiplying by 1.602 × 10−19 either. That is the classic wasted step.
(b) Kinetic energy in joules? Cross lanes deliberately: 2.50 × 1.602 × 10−19 = 4.01 × 10−19 J.
(c) Speed? Joule lane is compulsory. v = √(2 × 4.005 × 10−19 / 9.11 × 10−31) = 9.38 × 105 m s−1.
(d) Momentum? p = mv = 9.11 × 10−31 × 9.376 × 105 = 8.54 × 10−25 kg m s−1.
Self-audit: in (a) the answer is a voltage; in (b) a joule; in (c) m s−1; in (d) kg m s−1. If your written answer does not carry a unit, it is not finished.
What Changes When You Vary Intensity, Frequency or Potential
Here is the whole experimental study compressed into one table. If you can reproduce this from memory, the graph questions cannot hurt you.
| You increase… | Saturation current | Kmax / stopping potential | Threshold frequency | Why |
|---|---|---|---|---|
| Intensity (ν fixed, above threshold) | Increases, in direct proportion | No change at all | No change | More photons per second, each still worth hν |
| Frequency (intensity fixed) | Roughly unchanged (same photon count for equal power only if photon number is held fixed) | Increases linearly with ν | No change | Each photon is richer, so more is left after the toll |
| Collector potential (positive) | Rises, then flattens at saturation | No change | No change | You are collecting electrons, not creating them |
| Collector potential (negative) | Falls to zero at V0 | V0 is the measurement, not a variable | No change | Retarding field turns back progressively faster electrons |
| Work function (change the metal) | Unchanged if still above threshold | Decreases as φ0 rises | Increases with φ0 | A bigger toll leaves less change |
Reading these graphs is a skill you have been building since Electric Charges and Fields, where potential and field first appeared, and through Electrostatic Potential and Capacitance where the idea of an electron gaining or losing eV of energy across a potential difference was introduced. This chapter simply cashes in that groundwork.
Matter Waves — The Wave Nature of Particles
Now we turn the whole chapter around.
By 1924 physicists had accepted, grudgingly, that light — the textbook wave — also behaves as particles. Louis de Broglie, in his doctoral thesis, made a leap of pure symmetry. Nature, he argued, ought to be even-handed. If waves carry particle properties, then particles should carry wave properties.
His reasoning ran through momentum. For a photon we already had p = h/λ. De Broglie simply refused to treat that as a photon-only rule and read it backwards for anything with momentum:
Every moving object — electron, proton, cricket ball, you walking to school — has an associated wavelength given by Planck’s constant divided by its momentum. The wavelength is real, not a metaphor.
The obvious objection is the right one: why have we never seen a cricket ball diffract? The answer is in the size of h. Planck’s constant is about 6.6 × 10−34 J s, which is unimaginably small. Divide it by the momentum of anything you can see and you get a wavelength so tiny that no aperture, no crystal, no obstacle in the universe is small enough to diffract it. Wave behaviour only shows up when the wavelength is comparable to the size of the thing the wave meets.
One more piece of quiet elegance: the de Broglie relation independently explains Bohr’s quantisation rule. If an electron in a circular orbit is a standing wave, the circumference must fit a whole number of wavelengths: 2πr = nλ = nh/mv, which rearranges to mvr = nh/2π. A postulate becomes a consequence.
The de Broglie Relation and Its Three Faces
In practice a question will hand you the particle’s state in one of three ways, so learn all three faces of the same relation.
Face 2 — given kinetic energy K: since p = √(2mK), λ = h/√(2mK)
Face 3 — given accelerating voltage V for charge q: K = qV, so λ = h/√(2mqV)
Face 2 and Face 3 are the same formula; Face 3 just substitutes where the energy came from.
Face 1: p = mv = 9.11 × 10−31 × 1.00 × 106 = 9.11 × 10−25 kg m s−1.
λ = h/p = 6.626 × 10−34 / 9.11 × 10−25 = 7.27 × 10−10 m = 0.727 nm = 7.27 ångström (3 s.f.).
Sanity check: that is a few atomic diameters — comparable to crystal plane spacings. This is precisely why electrons are diffracted by crystals while cricket balls are not.
Cross to the joule lane first: K = 120 × 1.602 × 10−19 = 1.9224 × 10−17 J.
Face 2: p = √(2mK) = √(2 × 9.11 × 10−31 × 1.9224 × 10−17) = √(3.5026 × 10−47) = 5.918 × 10−24 kg m s−1.
λ = 6.626 × 10−34 / 5.918 × 10−24 = 1.12 × 10−10 m = 1.12 ångström (3 s.f.).
Notice: higher energy gives a shorter wavelength. λ and K are inversely related through a square root, so quadrupling the energy halves the wavelength.
p = mv = 0.16 × 30 = 4.8 kg m s−1.
λ = 6.626 × 10−34 / 4.8 = 1.38 × 10−34 m (3 s.f.).
Comment (this is where the marks are): a nucleus is about 10−15 m across. This wavelength is roughly 1019 times smaller than a nucleus. No slit, crystal or obstacle in existence is that small, so the wave nature of the ball can never be detected. The relation still holds — it is simply unobservable, which is exactly why classical mechanics works so well for everyday objects.
de Broglie Wavelength from Accelerating Voltage
Electron diffraction experiments accelerate electrons through a known potential difference, so the accelerating voltage version deserves its own treatment — and it comes with a shortcut worth memorising.
An electron starting from rest and accelerated through a potential difference V gains kinetic energy K = eV. Substituting into Face 2:
Compute the bracket once, for all time:
√(2 × 9.11 × 10−31 × 1.602 × 10−19) = √(2.9188 × 10−49) = 5.4026 × 10−25
h / 5.4026 × 10−25 = 1.2264 × 10−9 m · V1/2
λ = 12.26 / √V ångström (V in volts)
Textbooks usually quote 12.27/√V å. The tiny difference is only because they use more precise constants (h = 6.62607 × 10−34, me = 9.10938 × 10−31, e = 1.602177 × 10−19). Either value is accepted; the difference is under 0.05%.
(a) Long way: K = eV = 1.602 × 10−19 × 100 = 1.602 × 10−17 J.
p = √(2 × 9.11 × 10−31 × 1.602 × 10−17) = √(2.9188 × 10−47) = 5.4026 × 10−24.
λ = 6.626 × 10−34 / 5.4026 × 10−24 = 1.2264 × 10−10 m = 1.23 å (3 s.f.).
(b) Shortcut: λ = 12.26/√100 = 12.26/10 = 1.226 å. Using the textbook 12.27 gives 1.227 å.
They agree. Two more for practice: V = 400 V → 12.26/20 = 0.613 å; V = 1000 V → 12.26/31.62 = 0.388 å.
Kinetic energy — note the 2: K = qV = 2 × 1.602 × 10−19 × 1000 = 3.204 × 10−16 J (that is 2000 eV, not 1000 eV).
Momentum: p = √(2 × 6.64 × 10−27 × 3.204 × 10−16) = √(4.2549 × 10−42) = 2.0627 × 10−21 kg m s−1.
λ = 6.626 × 10−34 / 2.0627 × 10−21 = 3.21 × 10−13 m = 0.00321 å (3 s.f.).
Why so short: the alpha is about 7300 times heavier than an electron and carries twice the charge, so its momentum at the same voltage is far larger. Heavier and more charged means shorter wavelength, every time.
Reason before you compute: at fixed K, λ = h/√(2mK), so λ ∝ 1/√m. The lighter particle has the longer wavelength — the electron.
Ratio: λe/λp = √(mp/me) = √(1.67 × 10−27 / 9.11 × 10−31) = √1833.2 = 42.8.
Check with actual values: λe = 1.226 å (from Example 13, since 100 eV is the same as 100 V for an electron) and λp = 6.626 × 10−34/√(2 × 1.67 × 10−27 × 1.602 × 10−17) = 2.86 × 10−12 m = 0.0286 å.
Ratio = 1.2264/0.028645 = 42.8. Confirmed.
Photon: E = hc/λ = 1240.8/1.00 nm = 1241 eV = 1.99 × 10−16 J.
Electron: p = h/λ = 6.626 × 10−34/1.00 × 10−9 = 6.626 × 10−25 kg m s−1.
K = p2/2m = (6.626 × 10−25)2 / (2 × 9.11 × 10−31) = 4.3904 × 10−49 / 1.822 × 10−30 = 2.410 × 10−19 J = 1.50 eV.
Ratio Ephoton/Kelectron = 825 (3 s.f.).
The lesson: equal wavelength does not mean equal energy. The photon obeys E = pc (linear in p); the slow electron obeys K = p2/2m (quadratic). Different energy–momentum relations, wildly different answers. This comparison is a favourite three-mark question.
Three Conceptual Traps Examiners Love
These three appear year after year, usually as short-answer reasoning questions. Learn the explanation, not just the conclusion — examiners award the reason, not the verdict.
The one-line answer: “Intensity increases the number of photons per second, not the energy per photon; since one photon is absorbed by one electron, Kmax = hν − φ0 is unchanged.”
The one-line answer: “Emission requires hν ≥ φ0 for a single photon; energy from separate photons cannot be accumulated by one electron.”
The one-line answer: “λ = h/mv applies universally, but for macroscopic masses the momentum is so large that λ is many orders of magnitude below any measurable length, so no diffraction can be observed.”
What’s Been Trimmed for 2026-27
A short housekeeping note, so you do not waste revision time.
Older question banks, YouTube playlists and second-hand notes will still contain it in full. Skip those sections. If it appears in a practice paper you have downloaded, that paper is following an earlier syllabus.
What has not been removed is the physics that experiment demonstrated. Electron diffraction as a concept — the fact that electrons show wave behaviour, and that their wavelength matches the de Broglie prediction — is still the reason matter waves are in the course at all, and it is fair game as a one-mark conceptual question. Keep the idea; drop the apparatus.
Everything else in this chapter is in scope: the dual nature of radiation, the photoelectric effect, Hertz and Lenard’s observations, Einstein’s photoelectric equation and the particle nature of light, the experimental study of the photoelectric effect (effects of intensity, frequency and potential; stopping potential; threshold frequency and work function), matter waves, and the de Broglie relation.
Dual Nature of Radiation and Matter Formula List
Your one-page revision sheet. Print it, stick it above your desk, and cover it up to test yourself. This is the dual nature of radiation and matter formula list in the order the chapter builds it.
| Quantity | Formula | Fast form / value |
|---|---|---|
| Photon energy | E = hν = hc/λ | E(eV) = 1240 / λ(nm) |
| Photon momentum | p = h/λ = E/c | Zero rest mass, non-zero momentum |
| Photons per second | n = P/(hν) | P in watt |
| Einstein’s equation | hν = φ0 + Kmax | The ledger — in = toll + out |
| Threshold frequency | ν0 = φ0/h | Emission needs ν ≥ ν0 |
| Threshold wavelength | λ0 = hc/φ0 | λ0(nm) = 1240 / φ0(eV); need λ ≤ λ0 |
| Stopping potential | eV0 = Kmax | V0(volt) = Kmax(eV), same number |
| V0 vs ν graph | V0 = (h/e)ν − φ0/e | slope h/e = 4.136 × 10−15 V s |
| Max speed | vmax = √(2Kmax/me) | K must be in joule |
| de Broglie (speed) | λ = h/mv | Face 1 |
| de Broglie (energy) | λ = h/√(2mK) | Face 2 — λ ∝ 1/√m at fixed K |
| de Broglie (voltage) | λ = h/√(2mqV) | Electron only: λ = 12.27/√V å |
| Constants | h, e, c, me | 6.626 × 10−34 J s · 1.602 × 10−19 C · 3.00 × 108 m/s · 9.11 × 10−31 kg |
If you want to see where the electromagnetic-wave description of light that this chapter complicates comes from, revisit Electromagnetic Waves; and for the induced-EMF circuits that share this chapter’s measurement style, Electromagnetic Induction is the natural companion.
Practice Worksheet
Ten original questions. Work each one on paper before you tap Show Answer — reading a solution feels like learning, but writing one is learning. Use h = 6.626 × 10−34 J s, e = 1.602 × 10−19 C, c = 3.00 × 108 m s−1, me = 9.11 × 10−31 kg, mp = 1.67 × 10−27 kg, mα = 6.64 × 10−27 kg. Give answers to 3 significant figures.
Q1. Calculate the energy of a photon of wavelength 250 nm, in both electronvolts and joules.
E(J) = 4.9633 × 1.602 × 10−19 = 7.95 × 10−19 J.
Direct check: hc/λ = 1.9878 × 10−25/2.50 × 10−7 = 7.951 × 10−19 J. Agrees.
Q2. A metal has work function 4.2 eV. Find its threshold wavelength and threshold frequency.
ν0 = φ0/h = (4.2 × 1.602 × 10−19)/(6.626 × 10−34) = 1.02 × 1015 Hz.
Check: c/λ0 = 3.00 × 108/2.954 × 10−7 = 1.015 × 1015 Hz. Agrees.
Q3. Light of wavelength 200 nm falls on the metal of Q2. Find the maximum kinetic energy of the photoelectrons and the stopping potential.
Kmax = 6.2041 − 4.2 = 2.00 eV.
V0 = 2.00 V (same number, because eV0 = Kmax).
Sanity: 200 nm is shorter than the 295 nm threshold, so emission does occur. Good.
Q4. The stopping potential in a photoelectric experiment is measured as 1.50 V. Express the maximum kinetic energy of the photoelectrons in joules.
Trap avoided: the answer is an energy in joules, not a voltage.
Q5. An electron starts from rest and is accelerated through a potential difference of 200 V. Find its de Broglie wavelength in ångström.
Long way: K = 1.602 × 10−19 × 200 = 3.204 × 10−17 J; p = √(2 × 9.11 × 10−31 × 3.204 × 10−17) = 7.641 × 10−24; λ = 6.626 × 10−34/7.641 × 10−24 = 8.672 × 10−11 m. Agrees.
Q6. Find the de Broglie wavelength of a proton moving at 2.0 × 105 m s−1.
λ = 6.626 × 10−34/3.34 × 10−22 = 1.98 × 10−12 m (0.0198 å).
Do not use 12.27/√V here — no accelerating voltage was given, and that constant is electron-only anyway.
Q7. In a photoelectric experiment the light intensity is doubled while the frequency is kept fixed and above threshold. State what happens to (a) the saturation current, (b) the stopping potential, (c) the maximum kinetic energy of the photoelectrons. Give a reason.
(b) The stopping potential is unchanged.
(c) Kmax is unchanged.
Reason: intensity determines the number of photons arriving per second, not the energy of each photon. Since one photon is absorbed entirely by one electron, twice as many photons free twice as many electrons (doubling the current), but each electron still receives energy hν and so leaves with the same Kmax = hν − φ0.
Q8. A graph of stopping potential against frequency for a certain metal is a straight line that cuts the frequency axis at 6.0 × 1014 Hz. Find (a) the work function in eV, (b) the intercept on the stopping-potential axis, (c) the slope.
(b) The y-intercept is −φ0/e = −2.48 V.
(c) Slope = h/e = 4.14 × 10−15 V s — the same for every metal, which is why this graph measures Planck’s constant.
Q9. Calculate the momentum of a photon of wavelength 600 nm. Would a photon of shorter wavelength carry more or less momentum?
Since p = h/λ, a shorter wavelength means larger momentum (and larger energy). Gamma-ray photons carry enormous momentum for this reason.
Q10. An electron and an alpha particle have the same kinetic energy. Find the ratio of their de Broglie wavelengths, and say which is longer.
λe/λα = √(mα/me) = √(6.64 × 10−27/9.11 × 10−31) = √7288.7 = 85.4.
The electron’s wavelength is longer, by a factor of about 85. Note that the charge does not appear here — charge only matters if the energy is supplied by an accelerating voltage.
And whenever a photoelectric question looks confusing, go back to the ledger and say it out loud: one photon in, one toll paid, one electron out. That sentence has never let a student down.