Photoelectric effect — Einstein's photon model: Worked examples

This is the drill page for Photoelectric Effect — Einstein's Photon Model. The parent note built the one equation everything hangs on.

First, the fixed constants that appear on every line — memorise their meaning, not their digits:

Next, the stopping potential $V_{0}$ — a lab trick we lean on heavily below:

Now the master equation everything hangs on:

Before we start: three symbols you must never mix up.

The scenario matrix

Every photoelectric problem falls into one of these cells. The examples below are labelled with the cell they cover, and together they hit all of them.

#	Cell (case class)	What's special	Example
A	Above threshold ( $h ν > ϕ$ )	electrons emit, positive $K E_{m a x}$	Ex 1
B	Exactly at threshold ( $h ν = ϕ$ )	$K E_{m a x} = 0$ , the boundary	Ex 2
C	Below threshold ( $h ν < ϕ$ )	formula gives "negative" → no emission	Ex 3
D	Stopping potential	measuring $K E_{m a x}$ via voltage	Ex 4
E	Graph / slope ( $V_{0}$ vs $ν$ )	extracting $h$ and $ϕ$ from a line	Ex 5
F	Intensity vs frequency	which knob changes count vs speed	Ex 6
G	Real-world word problem	UV lamp, counting photons	Ex 7
H	Exam twist ( $λ$ ratios, no numbers)	algebra only, limiting behaviour	Ex 8

The three energy regimes (A, B, C) are the heart of it — here they are on one picture:

Figure — Photoelectric effect — Einstein's photon model

Cell A — Above threshold

Worked example Example 1 — find

$K E_{m a x}$ and electron speed A metal has $ϕ = 2.5 eV$ . Light of $λ = 350 nm$ strikes it. Find $K E_{m a x}$ in eV and in joules, then the maximum electron speed.

Forecast: 350 nm is near-UV. Is the photon energy above or below 2.5 eV? Guess whether electrons come out at all.

Step 1 — Photon energy. $E = \frac{1240}{350} = 3.543 eV$ . Why this step? We must know what each photon delivers before asking what survives the toll. The eV·nm shortcut skips unit conversion.

Step 2 — Check the regime. $E = 3.543 eV > ϕ = 2.5 eV$ , so we are in cell A: electrons emit. Why this step? Never apply $K E_{m a x} = E - ϕ$ without confirming it's positive — otherwise you're in cell C.

Step 3 — Leftover energy. $K E_{m a x} = 3.543 - 2.5 = 1.043 eV$ . Why this step? The electron keeps the change after paying $ϕ$ .

Step 4 — Convert to joules. $1.043 eV \times 1.602 \times 1 0^{- 19} = 1.671 \times 1 0^{- 19} J$ . Why this step? Speed needs SI energy, not eV.

Step 5 — Speed from $K E = \frac{1}{2} m v^{2}$ . With $m_{e} = 9.11 \times 1 0^{- 31} kg$ : $v_{m a x} = \frac{2 K E}{m _{e}} = \frac{2 ( 1.671 \times 1 0 ^{- 19} )}{9.11 \times 1 0 ^{- 31}} = 6.06 \times 1 0^{5} m/s .$ Why this step? $K E_{m a x}$ is energy; to get speed we invert the definition of kinetic energy.

Verify: $v \approx 6 \times 1 0^{5}$ m/s is about 0.2% of light speed — non-relativistic, so $\frac{1}{2} m v^{2}$ is valid. Units: $J / kg = m^{2} / s^{2} = m/s$ . ✓

Cell B — Exactly at threshold

Cell C — Below threshold (the "no emission" case)

Worked example Example 3 — what the formula does when it "fails"

Same metal, $ϕ = 2.5 eV$ . Now shine green light of $λ = 550 nm$ at high intensity. Find $K E_{m a x}$ .

Forecast: 550 nm is longer than the $λ_{0} = 496$ nm from Ex 2. Predict what happens.

Step 1 — Photon energy. $E = \frac{1240}{550} = 2.255 eV$ .

Step 2 — Apply the equation blindly. $E - ϕ = 2.255 - 2.5 = - 0.245 eV$ . Why this step? To show the trap: the algebra spits out a negative number.

Step 3 — Interpret. A negative $K E_{m a x}$ is physically impossible. It means the photon can't even cover the toll, so no electron is ejected. The correct answer is $K E_{m a x} = 0$ and photocurrent $= 0$ . Why this step? The equation is only valid when $h ν \geq ϕ$ ; below that, the answer is "no emission", not "negative energy".

Verify: $λ = 550 nm > λ_{0} = 496 nm$ , so below threshold — matches "no emission". Cranking intensity to a million photons per second changes nothing: each photon is still 2.255 eV, still short of 2.5 eV. ✓

Cell D — Stopping potential

Worked example Example 4 — measure

$K E_{m a x}$ with a voltage For the setup of Ex 1 ( $ϕ = 2.5$ eV, $λ = 350$ nm, $K E_{m a x} = 1.043$ eV), find the stopping potential $V_{0}$ .

Forecast: We apply a reverse voltage to brake the electrons. Guess the value of $V_{0}$ before computing.

Step 1 — Energy balance at stopping. The electron climbs the reverse voltage and loses all its KE: $e V_{0} = K E_{m a x}$ (this is the definition of $V_{0}$ from the top of the page). Why this step? "Stopping" means the fastest electron just fails to reach the collector — all its KE spent.

Step 2 — Solve. $V_{0} = \frac{K E _{m a x}}{e} = \frac{1.043 eV}{e}$ . Because $K E$ measured in eV numerically equals volts (the $e$ cancels), $V_{0} = 1.043 V$ . Why this step? The whole point of eV units: an electron with $1.043$ eV is stopped by exactly $1.043$ V.

Verify: Dimensionally $e V_{0}$ = (coulomb)(volt) = joule = energy. ✓ And $1.043 eV = 1.671 \times 1 0^{- 19} J$ matches Ex 1. ✓

Cell E — Graph: extract $h$ and $ϕ$ from a line

The stopping-potential equation rearranges into a straight line: $V_{0} = \frac{h}{e} ν - \frac{ϕ}{e} .$

Compare to $y = m x + c$ : plotting $V_{0}$ (vertical) against $ν$ (horizontal) gives a line with slope $h / e$ and vertical intercept $- ϕ / e$ . This is how Millikan measured Planck's constant.

Worked example Example 5 — read

$h$ and $ϕ$ off two points An experiment gives: at $ν_{1} = 7.0 \times 1 0^{14} Hz$ , $V_{0} = 1.00 V$ ; at $ν_{2} = 10.0 \times 1 0^{14} Hz$ , $V_{0} = 2.24 V$ . Find Planck's constant $h$ and the work function $ϕ$ .

Forecast: The slope should come out near $4 \times 1 0^{- 15} V\cdots$ (that's $h / e$ ). Watch for it.

Step 1 — Slope = $h / e$ . $\frac{Δ V _{0}}{Δ ν} = \frac{2.24 - 1.00}{( 10.0 - 7.0 ) \times 1 0 ^{14}} = \frac{1.24}{3.0 \times 1 0 ^{14}} = 4.133 \times 1 0^{- 15} V\cdots .$ Why this step? Slope of the line is $h / e$ by the equation above — two points fix it.

Step 2 — Recover $h$ . $h = slope \times e = 4.133 \times 1 0^{- 15} \times 1.602 \times 1 0^{- 19} = 6.62 \times 1 0^{- 34} J\cdots$ . Why this step? Multiply out the $e$ we divided by to isolate $h$ .

Step 3 — Find $ϕ$ using one point. From $ϕ = h ν_{1} - e V_{0}$ (i.e. $ϕ / e = (h / e) ν_{1} - V_{0}$ ): $\frac{ϕ}{e} = 4.133 \times 1 0^{- 15} \times 7.0 \times 1 0^{14} - 1.00 = 2.893 - 1.00 = 1.893 V \Rightarrow ϕ = 1.893 eV .$ Why this step? With the slope known, any single data point pins down the intercept, hence $ϕ$ .

Verify: Recomputed $h = 6.62 \times 1 0^{- 34}$ J·s matches the known $6.626 \times 1 0^{- 34}$ to 3 sig figs. ✓ Using point 2 as a cross-check: $ϕ / e = 4.133 \times 1 0^{- 15} \times 1 0^{15} - 2.24 = 4.133 - 2.24 = 1.893$ V — same $ϕ$ . ✓ Threshold: $ν_{0} = ϕ / h = 1.893 \times 1.602 \times 1 0^{- 19} /6.626 \times 1 0^{- 34} = 4.58 \times 1 0^{14}$ Hz, exactly where the line would cross the horizontal axis. ✓

Cell F — Intensity vs frequency (which knob does what)

Worked example Example 6 — three knobs, three effects

Start from Ex 1 ( $ϕ = 2.5$ eV, $λ = 350$ nm, giving $K E_{m a x} = 1.043$ eV, photocurrent $I$ ). Keeping the definition above, predict $K E_{m a x}$ and photocurrent for each change: (a) double the power, same $λ$ ; (b) halve the wavelength to 175 nm, same power; (c) switch to 700 nm, same power.

Forecast: Which of these changes the electron speed? Which changes the count? Which switches it off?

Step (a) — double power, same $λ$ . Each photon still carries $3.543$ eV, so $K E_{m a x}$ unchanged at $1.043$ eV. Double power at the same photon energy = double the photon rate = double the electrons ejected, so photocurrent doubles to $2 I$ . Why this step? At fixed $λ$ , power and photon rate move together; power multiplies photon count, never photon size. "Intensity Increases the count."

Step (b) — halve $λ$ , same power. New energy $E = \frac{1240}{175} = 7.086 eV$ , so $K E_{m a x} = 7.086 - 2.5 = 4.586 eV$ — electrons emerge far faster. But at the same power each photon now carries twice the energy, so the photon rate is halved; with one electron per photon the photocurrent drops to about $I /2$ . Why this step? Wavelength sets photon size ( $K E_{m a x}$ ), while fixed power with bigger photons means fewer of them — count goes down even as speed goes up. "Frequency Frees (and speeds) them."

Step (c) — 700 nm, same power. $E = \frac{1240}{700} = 1.771 eV < 2.5 eV$ → cell C, no emission. Photocurrent $= 0$ , no matter how high the power. Why this step? Same trap as Ex 3 — always check against $ϕ$ first.

Verify: (a) $K E_{m a x} = 1.043$ eV unchanged, current $2 I$ ✓; (b) $K E_{m a x} = 4.586$ eV up, current $\approx I /2$ down (photon energy doubled) ✓; (c) $1.771 < 2.5$ → zero ✓. The pattern "frequency sets speed, intensity/power sets count" holds — with the honest twist in (b) that raising frequency at fixed power lowers the count.

Cell G — Real-world word problem

Worked example Example 7 — a UV lamp counting photons

A $λ = 200 nm$ UV lamp delivers $P = 1.0 mW$ of power onto a sodium plate ( $ϕ = 2.28 eV$ ). (a) Find $K E_{m a x}$ of photoelectrons. (b) If every photon ejects one electron, how many electrons per second, and what photocurrent?

Forecast: 200 nm is deep UV — energetic. Guess whether $K E_{m a x}$ is bigger or smaller than 3 eV.

Step 1 — Photon energy. $E = \frac{1240}{200} = 6.20 eV$ .

Step 2 — $K E_{m a x}$ . $6.20 - 2.28 = 3.92 eV$ (well above threshold, cell A). Why this step? Answer to (a); confirms emission.

Step 3 — Energy per photon in joules. $6.20 eV \times 1.602 \times 1 0^{- 19} = 9.932 \times 1 0^{- 19} J$ . Why this step? Power is in watts (J/s), so we need photon energy in J to count them.

Step 4 — Photons per second. $N = \frac{P}{E} = \frac{1.0 \times 1 0 ^{- 3} J/s}{9.932 \times 1 0 ^{- 19} J} = 1.007 \times 1 0^{15} s^{- 1} .$ Why this step? Total power ÷ energy per packet = number of packets per second.

Step 5 — Photocurrent. Each ejected electron carries charge $e$ : $I = N e = 1.007 \times 1 0^{15} \times 1.602 \times 1 0^{- 19} = 1.61 \times 1 0^{- 4} A = 0.161 mA .$ Why this step? Current = charge per second = (electrons/s)(charge each).

Verify: $0.161$ mA from a $1$ mW lamp is plausibly small. Units: (J/s)/(J) = 1/s ✓; (1/s)(C) = C/s = A ✓. In reality only a fraction of photons eject electrons, so real current is lower — this is the ideal ceiling.

Cell H — Exam twist (algebra only, limiting behaviour)

Worked example Example 8 — two wavelengths, no numbers given for

$ϕ$ When light of wavelength $λ_{1}$ hits a metal, the stopping potential is $V_{1}$ . With $λ_{2}$ (shorter) it is $V_{2}$ . Show that Planck's constant can be found without knowing $ϕ$ , and give the formula.

Forecast: Two equations, two unknowns ( $h$ and $ϕ$ ). Which unknown cancels when we subtract?

Step 1 — Write both equations. Using $e V_{0} = \frac{h c}{λ} - ϕ$ : $e V_{1} = \frac{h c}{λ _{1}} - ϕ, e V_{2} = \frac{h c}{λ _{2}} - ϕ .$ Why this step? The stopping-potential form is linear in the two unknowns.

Step 2 — Subtract to kill $ϕ$ . $e (V_{2} - V_{1}) = h c (\frac{1}{λ _{2}} - \frac{1}{λ _{1}}) .$ Why this step? $ϕ$ is the same metal in both, so it cancels — that's the whole trick.

Step 3 — Solve for $h$ . $h = \frac{e ( V _{2} - V _{1} )}{c ( \frac{1}{λ _{2}} - \frac{1}{λ _{1}} )}$ Why this step? Everything on the right is measurable; $ϕ$ never appears.

Step 4 — Sanity numbers. Let $λ_{1} = 400$ nm, $λ_{2} = 300$ nm. The two photon energies are $1240/400 = 3.10$ eV and $1240/300 = 4.133$ eV. Because the same toll $ϕ$ is subtracted from both, the difference in stopping potentials is just the difference in photon energies: $V_{2} - V_{1} = 4.133 - 3.10 = 1.033$ V. Why this step? Check that the " $ϕ$ cancels" logic gives the plain photon-energy difference.

Verify: $V_{2} - V_{1} = 1240 (1/300 - 1/400) = 1240 (0.003333 - 0.0025) = 1240 \times 8.333 \times 1 0^{- 4} = 1.033$ V. ✓ Independent of $ϕ$ , exactly as the derivation promised. Limiting check: if $λ_{1} = λ_{2}$ the denominator → 0 and $V_{2} = V_{1}$ , giving $0/0$ — you need two different colours to measure $h$ this way, which makes physical sense.

Active Recall

Recall Which cell? Quick classification drill

Photon 4.0 eV, metal $ϕ = 4.0$ eV — which cell? ::: Cell B (exactly at threshold, $K E_{m a x} = 0$ ). Photon 1.8 eV, $ϕ = 2.5$ eV — emission or not? ::: Cell C — no emission (formula would give negative $K E$ ). You double the power (intensity) at fixed $ν$ — does $V_{0}$ change? ::: No (Cell F): $V_{0}$ depends only on $ν$ , not power. Slope of $V_{0}$ vs $ν$ ? ::: $h / e$ (Cell E). Where does the $V_{0}$ vs $ν$ line cross the horizontal axis? ::: At the threshold frequency $ν_{0} = ϕ / h$ (Cell E). To find $h$ from two wavelengths, what cancels? ::: The work function $ϕ$ (Cell H).

Connections

Work Function and Binding Energy — the toll $ϕ$ ; extended by Photoelectron Spectroscopy.
Planck's Quantum Theory — origin of $E = h ν$ used in every cell here.
Wave-Particle Duality — why light behaves as counted packets (photons).
Electromagnetic Spectrum — Cells B/C are exactly "which colours are energetic enough".
Bohr Model of the Atom · Compton Effect — same photon accounting elsewhere.