A full lesson: a cartoon movie to start, then why the wave picture fails, Einstein's energy packets, how K_max = hν − φ₀ explains every observation, Millikan's straight line with slope h/e, a numerical calculator, NEET shortcuts, problems to solve yourself, and 12 MCQs.
A thousand peas cannot knock the coconut down, one cricket ball can. Light behaves the same way: what matters is the energy in one packet, hν, not how many packets arrive. Energy left after paying the work function becomes kinetic energy.
At the village mela, the shopkeeper's challenge is simple: knock a coconut off its stand and take it home. Arjun empties a whole bag of dried peas at it, hundreds of hits, and the coconut only wobbles. Meera throws one cricket ball. The coconut flies off. One strong hit beats any number of weak ones, because the energies of separate hits do not add up on the coconut.
In Lesson 2 you saw light behave exactly like this: very bright red light frees no electrons, while dim ultraviolet frees them at once. In 1905 Einstein explained it: light energy comes in packets (quanta) of hν, and one electron absorbs one whole packet. That one idea gives the equation K_max = hν − φ₀, which NEET uses in almost every paper.
"The coconut is the electron, the stand's grip is the work function, the cricket ball is a high-frequency photon, and the peas are low-frequency photons. More peas, brighter light, still no coconut."
"If Meera throws the ball harder than needed, where does the extra energy go?"
Into the coconut's flight: it flies off faster. For the electron, energy left over after paying the work function becomes its kinetic energy: K_max = hν − φ₀.
In the wave picture, light energy is spread smoothly over the wavefront and every electron soaks up a little of it. That makes three predictions. Pick a test, move the sliders, and compare what wave theory says with what the experiment shows. The metal is sodium (φ₀ = 2.75 eV).
| Test | Wave theory predicts | Experiment shows |
|---|---|---|
| Brighter | faster electrons: K_max grows with intensity | K_max does not depend on intensity |
| Low ν | any frequency works if bright enough or long enough | no emission below ν₀, whatever the intensity |
| Dim | a wait of hours or more while energy builds up | instantaneous (≲ 10⁻⁹ s) |
"NCERT's line: the wave picture is unable to explain the most basic features of photoelectric emission. Three predictions, three failures. The problem is not the experiment; it is the idea that energy is spread out smoothly."
Light of frequency ν is a stream of quanta, each of energy E = hν. One electron absorbs one whole quantum in a single event. It spends at least φ₀ to escape; the rest is kinetic energy. Drag the top of the photon arrow (or use the slider) to change hν, and watch electrons from different depths.
h = 6.63 × 10⁻³⁴ J s. A 400 nm photon carries 3.1 eV.
The least tightly bound electrons pay only φ₀ and leave with the most energy.
Electrons bound more firmly lose more energy on the way out, so they leave with less than K_max.
"One photon, one electron, one event. Never add the energies of two photons for one electron. This is the cricket ball, not the bag of peas."
Photons rain on a sodium plate. Change the number of photons per second (intensity) and the energy of each (frequency). The graph on the right plots K_max against ν: watch which slider moves the point.
Intensity does not appear in the equation. Brighter light means more photons, not bigger ones.
Write the equation as K_max = h(ν − ν₀). Equivalently, λ₀ = hc/φ₀.
Above threshold, each absorbed photon can free one electron, so current ∝ intensity. One absorption is a single quick event: no build-up, no time lag.
"Weak light reduces how many electrons come out, not how soon. That one sentence answers the time-lag question every time."
Put K_max = eV₀ into the equation: V₀ = (h/e) ν − φ₀/e. A graph of V₀ against ν must be a straight line of slope h/e, the same for every metal. Millikan worked from 1906 to 1916 hoping to prove Einstein wrong. Be Millikan: drag the two handles to fit a line through his kind of data (sodium) and read off h.
Millikan's slope gave h ≈ 6.6 × 10⁻³⁴ J s, agreeing with Planck. For a K_max–ν graph (in joule) the slope is h itself.
Extend the line backwards: it cuts the V₀-axis at −φ₀/e. In eV the intercept is simply −φ₀.
The data supported Einstein. Einstein got the 1921 Nobel prize for the photoelectric law; Millikan got his in 1923.
"A scientist who tried hard to disprove a theory, and ended up proving it. That is how good physics is done. And for you: slope h by e, never just h, on a V₀–ν graph."
Pick a real light source and a metal. The calculator does every step the way you should write it in the exam: photon energy, compare with φ₀, K_max, stopping potential and maximum speed.
Low work function: visible light is above threshold, so ordinary light gives a current.
Enough to free electrons even from copper. The same high photon energy damages the DNA of germs.
Below sodium's own 2.75 eV and just below caesium's 2.14 eV: a street lamp frees electrons from neither. Check the numbers, never guess.
300 nm → 4.13 eV, 400 nm → 3.1 eV, 620 nm → 2.0 eV. Saves the whole hc/λ calculation.
300 nm on 2.5 eV: V₀ = 4.13 − 2.5 = 1.63 V.
φ₀ = 2.0 eV → λ₀ = 620 nm. Emission needs λ < λ₀ (or ν > ν₀).
ν = 2ν₀ and 3ν₀ → K_max ratio 1 : 2 (not 2 : 3).
Subtract the two equations: φ₀ cancels. Use it to find h or V₂.
Both straight lines, parallel for all metals. ν-intercept ν₀; V-intercept −φ₀/e.
"Every numerical here has the same three lines: photon energy with the 1240 rule, subtract the work function, then convert to whatever is asked: volts, joules or speed."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute. Use hc = 1240 eV nm, e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg.
Ultraviolet light of wavelength 250 nm falls on sodium (φ₀ = 2.75 eV). Find the maximum kinetic energy, the stopping potential and the maximum speed of the photoelectrons.
E = 1240/λ, then K_max = E − φ₀, V₀ = K_max/e.
The work function of caesium is 2.14 eV. Find its threshold frequency and threshold wavelength. Will red light of 650 nm free electrons from it?
ν₀ = φ₀/h, λ₀ = 1240/φ₀.
For a metal, V₀ = 0.20 V at ν = 6.0 × 10¹⁴ Hz and V₀ = 1.03 V at ν = 8.0 × 10¹⁴ Hz. Find Planck's constant and the work function.
Subtract: e(V₂ − V₁) = h(ν₂ − ν₁). Then φ₀ = hν₁ − eV₁.
Light of 400 nm gives photoelectrons with K_max = 1.0 eV. What is K_max for 200 nm light on the same metal? Is it simply double?
Find φ₀ first; it does not change.
Light of frequency 1.5ν₀ (ν₀ = threshold) gives photoelectrons of maximum kinetic energy K. The frequency is raised to 2ν₀. Find the new K_max in terms of K.
K_max = h(ν − ν₀).
Use hc = 1240 eV nm, h = 6.6 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C. Write the energy balance first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "light of wavelength λ falls on a metal of work function φ₀" | K_max = 1240/λ − φ₀ (eV) | Forgetting to subtract φ₀, or mixing J and eV |
| "stopping potential" | eV₀ = hν − φ₀ | Writing V₀ = hν/e |
| "threshold wavelength / frequency" | λ₀ = hc/φ₀, ν₀ = φ₀/h | Taking λ > λ₀ as "enough" |
| "wavelength halved" | find φ₀, recompute | Doubling K_max |
| "slope of V₀–ν graph" | h/e | Answering h (that is the K_max–ν slope) |
| "intercept on the V₀-axis" | −φ₀/e | Answering ν₀ (that is the ν-axis intercept) |
| "two wavelengths, two stopping potentials" | subtract: e(V₁ − V₂) = hc(1/λ₁ − 1/λ₂) | Solving each one for φ₀ separately and rounding |
| "which prediction of wave theory fails" | K_max with intensity, threshold, time lag | Choosing interference or "light carries energy" |
Lesson 4 takes Einstein's packets seriously as particles: the photon, with energy hν and momentum h/λ. Then light behaves as a wave in interference and as a particle in the photoelectric effect: the dual nature of radiation.
Values used: h = 6.63 × 10⁻³⁴ J s = 4.14 × 10⁻¹⁵ eV s, c = 3 × 10⁸ m/s, hc ≈ 1240 eV nm, e = 1.602 × 10⁻¹⁹ C, m = 9.11 × 10⁻³¹ kg. Work functions from NCERT Table 11.1. The Millikan-style data points in Stage 5 are generated for sodium-like behaviour with small scatter, not his original readings.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.