Photoelectric effect — Einstein's explanation, work function: Worked examples

Everything here uses only two master relations, so let's pin them down before any symbol appears again.

Two conversion tools we will lean on constantly:

The scenario matrix

Every photoelectric problem is one of these cells. The examples below are tagged with the cell they hit, so together they cover all of them. Each cell also lives at a spot on the $V_{0}$ -vs- $ν$ line in the figure — read the table and the figure together.

#	Cell (case class)	What's given → what's asked	Example
A	Standard forward	$λ$ (or $ν$ ), $ϕ$ → $K E_{m a x}$ , $V_{0}$	Ex 1
B	Threshold / degenerate	$ϕ$ → $λ_{0}$ or $ν_{0}$ (the "just barely" edge, $K E_{m a x} = 0$ )	Ex 2
C	Below threshold (zero output)	$h ν < ϕ$ → prove nothing emitted	Ex 3
D	Reverse solve for $ϕ$	$λ$ , $V_{0}$ → $ϕ$	Ex 4
E	Two-point slope (measure $h$ )	$(ν_{1}, V_{0, 1}), (ν_{2}, V_{0, 2})$ → $h$ , $ϕ$	Ex 5
F	Ratio / scaling	halve $λ$ , what happens to $K E_{m a x}$ ? (non-proportional trap)	Ex 6
G	Intensity vs frequency (word problem)	double brightness — does $V_{0}$ move?	Ex 7
H	Limiting behaviour	$ν \to \infty$ and $ν \to ν_{0}^{+}$ — shape of the line	Ex 8
I	Exam twist (electron speed)	$K E_{m a x}$ → actual velocity $v_{m a x}$ in m/s	Ex 9

Look at the figure below — the whole matrix lives on one straight-line graph of $V_{0}$ vs $ν$ . The lavender line is $V_{0} = (h / e) ν - ϕ / e$ ; its slope (steepness) is $h / e$ , its $x$ -intercept (red dot on the axis) is the threshold $ν_{0}$ , and its $y$ -intercept is $- ϕ / e$ . The mint-shaded strip on the left is Cell C, where no line exists because no electron escapes. Each labelled dot ties a cell letter to its geometric home.

Figure — Photoelectric effect — Einstein's explanation, work function

The examples

Ex 1 — Cell A: standard forward

Worked example Sodium under violet light

$ϕ = 2.3$ eV, light of wavelength $λ = 400$ nm. Find $K E_{m a x}$ and $V_{0}$ .

Forecast: guess before solving — will the electron come out fast or barely crawl out? (Violet is fairly energetic; $ϕ$ is modest, so expect a smallish positive KE.) This is the grey dot in the figure, sitting on the line above threshold.

Step 1. Photon energy: $E = \frac{1240}{400} = 3.10$ eV. Why this step? We must know what one packet is worth before we can subtract the toll.

Step 2. $K E_{m a x} = E - ϕ = 3.10 - 2.3 = 0.80$ eV. Why this step? Einstein's equation — energy left after paying $ϕ$ .

Step 3. $V_{0} = K E_{m a x} / e = 0.80$ V. In full SI: $K E_{m a x} = 0.80 eV = 0.80 \times 1.6 \times 1 0^{- 19} = 1.28 \times 1 0^{- 19}$ J, and $V_{0} = \frac{K E _{m a x}}{e} = \frac{1.28 \times 1 0 ^{- 19} J}{1.6 \times 1 0 ^{- 19} C} = 0.80$ V. Why this step? Working the joules-and-coulombs division out in full shows why the eV number and the volt number come out identical — the two factors of $1.6 \times 1 0^{- 19}$ cancel.

Verify: SI division gives $0.80$ V, matching the eV shortcut ✓. Sanity: $K E_{m a x} = 0.80$ eV is positive and less than the photon energy, as it must be (you can't keep more than you were given). ✓

Ex 2 — Cell B: threshold (degenerate edge)

Worked example Longest wavelength that still works

A metal has $ϕ = 4.0$ eV. What is the longest wavelength $λ_{0}$ that ejects electrons?

Forecast: "Longest wavelength" = lowest energy = the just-barely case. Expect $K E_{m a x} = 0$ at the answer, and the wavelength should be in the UV (since 4 eV is more than any visible photon). In the figure this is the red dot on the $ν$ -axis where the line touches down.

Step 1. At threshold, $K E_{m a x} = 0$ , so all photon energy is spent escaping: $ϕ = \frac{h c}{λ _{0}}$ . Why this step? Longer wavelength = weaker photon; push $λ$ up until the photon has exactly enough and no leftover. That is the boundary of emission.

Step 2. $λ_{0} = \frac{1240}{4.0} = 310$ nm. Why this step? Rearrange the eV–nm shortcut for $λ$ .

Verify: $310$ nm is ultraviolet (below the $\sim 400$ nm violet edge) ✓ — so ordinary visible light fails on this metal, exactly what the "threshold" idea predicts. Plug back: $1240/310 = 4.0$ eV $= ϕ$ ✓.

Ex 3 — Cell C: below threshold (zero output)

Worked example Red light on the same metal

Same metal, $ϕ = 4.0$ eV, now lit with $λ = 620$ nm (red). What is $K E_{m a x}$ ? How many red photons per second are needed to eventually free an electron?

Forecast: Red is low energy. Compared with the $310$ nm threshold, $620$ nm is longer → weaker photon → probably nothing comes out. This is the mint-shaded strip in the figure — left of $ν_{0}$ , no line at all. And the "how many photons" question is a trap.

Step 1. Photon energy: $E = \frac{1240}{620} = 2.0$ eV. Why this step? We compare one packet against the toll.

Step 2. $E - ϕ = 2.0 - 4.0 = - 2.0$ eV. A negative "leftover" is impossible — physically it means no electron escapes, so $K E_{m a x} = 0$ (there simply are no ejected electrons). Why this step? One photon pays for one electron; if the single packet can't cover the toll, the electron stays put.

Step 3. Number of photons needed: irrelevant — no number works. Electrons de-excite instantly and cannot "save up" many weak photons. Why this step? This is the classical accumulation myth; each absorption is a one-shot event.

Verify: $2.0 eV < 4.0 eV = ϕ$ , so $h ν < ϕ$ ✓ → the sub-threshold case, emission = zero regardless of intensity. Consistent with Ex 2: $620 > 310$ nm means below threshold. ✓

Ex 4 — Cell D: reverse-solve for $ϕ$

Worked example Given the stopping voltage, find the metal's work function

Light of $λ = 300$ nm on an unknown metal gives a stopping potential $V_{0} = 1.13$ V. Find $ϕ$ .

Forecast: We run the equation backwards: photon energy minus the measured KE gives the toll. Expect $ϕ$ around 3 eV.

Step 1. Photon energy: $E = \frac{1240}{300} = 4.133$ eV. Why this step? Always establish the packet value first.

Step 2. $K E_{m a x} = e V_{0} = 1.13$ eV (volts → eV directly). Why this step? The stopping potential measures the fastest electron's KE.

Step 3. $ϕ = E - K E_{m a x} = 4.133 - 1.13 = 3.00$ eV. Why this step? Rearranged Einstein equation: toll = packet − leftover.

Verify: Forward-check: $K E_{m a x} = 4.133 - 3.00 = 1.13$ eV → $V_{0} = 1.13$ V ✓ matches the given. Threshold of this metal: $λ_{0} = 1240/3.00 = 413$ nm (violet), and $300 < 413$ nm so emission is expected ✓.

Ex 5 — Cell E: two data points → measure $h$ (Millikan style)

Worked example Extract Planck's constant from a voltage graph

For one metal: $ν_{1} = 7 \times 1 0^{14}$ Hz gives $V_{0} = 0.50$ V; $ν_{2} = 10 \times 1 0^{14}$ Hz gives $V_{0} = 1.74$ V. Find $h$ and $ϕ$ .

Forecast: Two points on the line $V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}$ . Slope gives $h / e$ ; extending back to $V_{0} = 0$ gives $ν_{0}$ , hence $ϕ$ . Expect $h \approx 6.6 \times 1 0^{- 34}$ . These are the two butter-yellow squares in the figure.

Step 1. Slope $= \frac{Δ V _{0}}{Δ ν} = \frac{1.74 - 0.50}{( 10 - 7 ) \times 1 0 ^{14}} = 4.133 \times 1 0^{- 15}$ V·s. Why this step? From $V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}$ , the coefficient of $ν$ is $h / e$ — geometry of the straight line encodes $h$ .

Step 2. $h = slope \times e = 4.133 \times 1 0^{- 15} \times 1.6 \times 1 0^{- 19} = 6.61 \times 1 0^{- 34}$ J·s. Why this step? Multiply out the $/ e$ to isolate $h$ .

Step 3. Find $ϕ$ using point 1: $e V_{0, 1} = h ν_{1} - ϕ$ , so $ϕ = h ν_{1} - e V_{0, 1} = (6.61 \times 1 0^{- 34}) (7 \times 1 0^{14}) - (1.6 \times 1 0^{- 19}) (0.50) .$ That is $ϕ = 4.63 \times 1 0^{- 19} - 0.80 \times 1 0^{- 19} = 3.83 \times 1 0^{- 19}$ J $= 2.39$ eV. Why this step? Once the slope is known, either point fixes the intercept, i.e. the toll.

Verify: $h = 6.61 \times 1 0^{- 34}$ J·s matches the accepted $6.63 \times 1 0^{- 34}$ within rounding ✓. Cross-check with point 2 by computing $h ν_{2}$ directly: $h ν_{2} = (6.61 \times 1 0^{- 34}) (10 \times 1 0^{14}) = 6.61 \times 1 0^{- 19}$ J. Then $K E_{m a x} = h ν_{2} - ϕ = 6.61 \times 1 0^{- 19} - 3.83 \times 1 0^{- 19} = 2.78 \times 1 0^{- 19}$ J $= 1.74$ eV → $V_{0} = 1.74$ V ✓, exactly the given second point.

Ex 6 — Cell F: ratio / scaling trap

Worked example Halving the wavelength — does

$K E_{m a x}$ double? A metal has $ϕ = 2.0$ eV. Light of $λ_{1} = 400$ nm gives some $K E_{m a x}$ . If we halve the wavelength to $λ_{2} = 200$ nm, does $K E_{m a x}$ double?

Forecast: Careful! Photon energy doubles when wavelength halves, but $K E_{m a x}$ has the $ϕ$ subtracted, so it will more than double. Common trap: people answer "just doubles."

Step 1. $E_{1} = 1240/400 = 3.10$ eV → $K E_{m a x, 1} = 3.10 - 2.0 = 1.10$ eV. Why this step? Baseline value.

Step 2. $E_{2} = 1240/200 = 6.20$ eV → $K E_{m a x, 2} = 6.20 - 2.0 = 4.20$ eV. Why this step? Halving $λ$ doubles photon energy ( $6.20 = 2 \times 3.10$ ✓), but the toll $ϕ$ is a fixed subtraction.

Step 3. Ratio $K E_{m a x, 2} / K E_{m a x, 1} = 4.20/1.10 = 3.82$ , not 2. Why this step? Because $K E_{m a x} = E - ϕ$ is affine, not proportional — subtracting a constant breaks simple doubling.

Verify: Photon energies did double ( $3.10 \to 6.20$ ) ✓. Yet KE grew by factor $3.82 > 2$ ✓, confirming the trap. Only the photon energy scales as $1/ λ$ , not $K E_{m a x}$ .

Ex 7 — Cell G: intensity vs frequency (word problem)

Worked example The brightness question

You shine $λ = 350$ nm light ( $>$ threshold) on caesium ( $ϕ = 2.1$ eV) and record $V_{0}$ . A friend doubles the brightness (same colour). Does $V_{0}$ change? Does the photocurrent change?

Forecast: Brightness = photons per second, not energy per photon. Predict: $V_{0}$ unchanged, current doubles.

Step 1. Photon energy: $E = 1240/350 = 3.543$ eV. Why this step? Establish per-packet energy — this is what sets $K E_{m a x}$ .

Step 2. $K E_{m a x} = 3.543 - 2.1 = 1.44$ eV → $V_{0} = 1.44$ V. Doubling intensity does not change $E$ , $ϕ$ , or $K E_{m a x}$ , so $V_{0}$ stays $1.44$ V. Why this step? Each electron eats exactly one photon; a photon's energy depends on colour ( $ν$ ), not on how many arrive.

Step 3. Photocurrent doubles: twice the photons per second → twice the ejected electrons per second. Why this step? Current counts electrons, and electron count tracks photon count.

Verify: $E = 3.543 > ϕ = 2.1$ so emission occurs ✓. $V_{0}$ depends only on $(ν, ϕ)$ , both unchanged, so $V_{0} = 1.44$ V before and after ✓ — matching the classical-vs-quantum table in the parent note.

Ex 8 — Cell H: limiting behaviour of the line

Worked example The two ends of the graph

On the $V_{0}$ vs $ν$ line for a metal with $ϕ = 2.0$ eV: (a) what is $V_{0}$ as $ν \to ν_{0}^{+}$ (just above threshold)? (b) how does $V_{0}$ behave as $ν \to \infty$ ?

Forecast: At the low end $V_{0} \to 0$ (electrons barely escape). At the high end $V_{0}$ grows without bound, linearly. Look at the figure's two flagged ends — the red dot (low) and the lavender arrow (high).

Step 1. Convert the work function to joules first: $ϕ = 2.0 eV \times 1.6 \times 1 0^{- 19} J/eV = 3.2 \times 1 0^{- 19}$ J. Then threshold: $ν_{0} = \frac{ϕ}{h} = \frac{3.2 \times 1 0 ^{- 19} J}{6.63 \times 1 0 ^{- 34} J\cdots} = 4.83 \times 1 0^{14}$ Hz. Why this step? $ν_{0}$ is where the line meets the $ν$ -axis; below it there is no line at all (no emission). We must convert eV→J because $h$ is in SI units.

Step 2. As $ν \to ν_{0}^{+}$ : $K E_{m a x} = h ν - ϕ \to 0$ , so $V_{0} \to 0$ . Why this step? At threshold every bit of the photon pays the toll; nothing left to stop.

Step 3. As $ν \to \infty$ : $V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}$ grows linearly and without bound; slope stays $h / e = 4.14 \times 1 0^{- 15}$ V·s. Why this step? The $ϕ$ term becomes negligible next to a huge $h ν$ ; the line's steepness is fixed forever.

Verify: $ν_{0} = 4.83 \times 1 0^{14}$ Hz corresponds to $λ_{0} = c / ν_{0} = 3 \times 1 0^{8} /4.83 \times 1 0^{14} = 621$ nm ✓ and $λ_{0} = 1240/2.0 = 620$ nm from the shortcut ✓ (agree within rounding). Slope $h / e = 6.63 \times 1 0^{- 34} /1.6 \times 1 0^{- 19} = 4.14 \times 1 0^{- 15}$ V·s ✓.

Ex 9 — Cell I: exam twist (actual electron speed)

Worked example From energy to velocity

An electron leaves a metal with $K E_{m a x} = 0.80$ eV. Find its maximum speed $v_{m a x}$ . (Electron mass $m_{e} = 9.11 \times 1 0^{- 31}$ kg.)

Forecast: We must first convert eV to joules, then invert $K E = \frac{1}{2} m v^{2}$ . Expect $v$ around $5 \times 1 0^{5}$ m/s (electrons are light, so even a small energy gives high speed).

Step 1. Convert energy to SI: $K E_{m a x} = 0.80 \times 1.6 \times 1 0^{- 19} = 1.28 \times 1 0^{- 19}$ J. Why this step? The kinetic-energy formula needs joules, kilograms, metres — you can't mix in eV.

Step 2. From $K E = \frac{1}{2} m v^{2}$ : $v_{m a x} = \frac{2 K E _{m a x}}{m _{e}} = \frac{2 ( 1.28 \times 1 0 ^{- 19} )}{9.11 \times 1 0 ^{- 31}}$ . Why this step? We know energy and mass and want speed — invert the classical KE definition (non-relativistic since $v ≪ c$ ).

Step 3. $v_{m a x} = 2.81 \times 1 0^{11} = 5.30 \times 1 0^{5}$ m/s. Why this step? Take the square root of the numeric ratio.

Verify: $v_{m a x} = 5.3 \times 1 0^{5}$ m/s is $\approx 0.0018 c$ , safely non-relativistic ✓ so our classical formula was valid. Forward-check: $\frac{1}{2} (9.11 \times 1 0^{- 31}) (5.30 \times 1 0^{5})^{2} = 1.28 \times 1 0^{- 19}$ J $= 0.80$ eV ✓.

Recall Which cell is this problem?

Given wavelength and work function, asked for stopping potential? ::: Cell A — standard forward (Ex 1). Given only $ϕ$ , asked for the longest wavelength? ::: Cell B — threshold, set $K E_{m a x} = 0$ (Ex 2). Photon energy less than $ϕ$ ? ::: Cell C — no emission at all, any intensity (Ex 3). Given $λ$ and $V_{0}$ , asked for $ϕ$ ? ::: Cell D — reverse solve (Ex 4). Two $(ν, V_{0})$ pairs, asked for $h$ ? ::: Cell E — slope $= h / e$ (Ex 5). Does halving $λ$ double $K E_{m a x}$ ? ::: No — Cell F, KE is affine not proportional (Ex 6). Does doubling brightness change $V_{0}$ ? ::: No — Cell G, only current changes (Ex 7).

Connections

Parent: Einstein's photoelectric equation — this page is its worked-example drill.
Energy conservation — every example is just $h ν = ϕ + K E_{m a x}$ bookkeeping.
Millikan oil drop & measurement of h — Ex 5 is the slope method that measured $h$ .
Planck's quantum hypothesis & blackbody radiation — where $E = h ν$ comes from.
Wave-particle duality of light — why "one photon, one electron" is legitimate.
de Broglie wavelength — Ex 9's electron speed leads into matter waves.
Compton effect — further photon-as-particle evidence.