Photoelectric effect — Einstein's explanation, work function: Exercises

L1 — Recognition

Goal: read the equation, plug numbers, respect units.

Problem 1.1

A photon has frequency $ν = 6.0 \times 1 0^{14}$ Hz. Find its energy in joules and in eV.

Recall Solution 1.1

WHAT: apply $E = h ν$ . WHY: frequency was given, so use the $ν$ form directly. $E = h ν = (6.63 \times 1 0^{- 34}) (6.0 \times 1 0^{14}) = 3.98 \times 1 0^{- 19} J .$ Convert to eV by dividing by $e$ (because $1 eV = 1.6 \times 1 0^{- 19}$ J): $E = \frac{3.98 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 2.49 eV .$

Problem 1.2

Light of wavelength $λ = 620$ nm strikes a metal. Find the photon energy in eV.

Recall Solution 1.2

WHAT: use the wavelength shortcut. WHY: wavelength is given in nm, so $h c = 1240$ eV·nm converts in one step. $E = \frac{h c}{λ} = \frac{1240 eV\cdotnm}{620 nm} = 2.0 eV .$

Problem 1.3

A metal has work function $ϕ = 2.0$ eV. Its threshold frequency $ν_{0}$ is asked.

Recall Solution 1.3

WHAT: rearrange $ϕ = h ν_{0}$ . WHY: at threshold the photon energy exactly equals the escape toll — nothing left over. First convert $ϕ$ to joules: $ϕ = 2.0 \times 1.6 \times 1 0^{- 19} = 3.2 \times 1 0^{- 19}$ J. $ν_{0} = \frac{ϕ}{h} = \frac{3.2 \times 1 0 ^{- 19}}{6.63 \times 1 0 ^{- 34}} = 4.83 \times 1 0^{14} Hz .$

L2 — Application

Goal: chain two equations together.

Problem 2.1

Sodium has $ϕ = 2.3$ eV. Light of $λ = 450$ nm hits it. Find $K E_{m a x}$ (eV) and the stopping potential $V_{0}$ (V).

Recall Solution 2.1

Step 1 — photon energy. $E = \frac{1240}{450} = 2.756 eV$ . Step 2 — Einstein's equation. WHY: whatever the photon brings minus the toll is the leftover kinetic energy. $K E_{m a x} = E - ϕ = 2.756 - 2.3 = 0.456 eV .$ Step 3 — stopping potential. Since $e V_{0} = K E_{m a x}$ , and working in eV the charge $e$ cancels numerically: $V_{0} = 0.456 V .$

Problem 2.2

A metal's threshold wavelength is $λ_{0} = 300$ nm. Find its work function $ϕ$ in eV, and decide whether green light ( $λ = 550$ nm) can eject electrons.

Recall Solution 2.2

Step 1 — work function. $ϕ = \frac{h c}{λ _{0}} = \frac{1240}{300} = 4.13 eV .$ Step 2 — test green light. Its photon energy is $\frac{1240}{550} = 2.25 eV$ . Compare: $2.25 eV < 4.13 eV$ , so a green photon cannot pay the toll. No emission, no matter how bright. The rule: longer wavelength = lower energy, and $λ > λ_{0}$ always fails.

Problem 2.3

When $λ = 200$ nm illuminates a metal, the stopping potential is $V_{0} = 2.2$ V. Find the work function $ϕ$ .

Recall Solution 2.3

Step 1 — photon energy. $E = \frac{1240}{200} = 6.2 eV$ . Step 2 — invert Einstein's equation. WHY: $e V_{0} = K E_{m a x}$ , and $K E_{m a x} = E - ϕ$ , so $ϕ = E - e V_{0}$ . In eV, $e V_{0}$ is just $2.2$ eV numerically: $ϕ = 6.2 - 2.2 = 4.0 eV .$

L3 — Analysis

Goal: reason about the graph and how quantities move together.

Problem 3.1

The stopping potential $V_{0}$ is plotted against frequency $ν$ and gives a straight line. From two data points, $(ν_{1}, V_{01}) = (5.0 \times 1 0^{14} Hz, 0.30 V)$ and $(ν_{2}, V_{02}) = (9.0 \times 1 0^{14} Hz, 1.96 V)$ , find Planck's constant $h$ and the work function $ϕ$ (in eV).

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

Recall Solution 3.1

Step 1 — slope. From $V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}$ , the slope is $h / e$ . Look at the red line in the figure — its rise over run is: $slope = \frac{1.96 - 0.30}{( 9.0 - 5.0 ) \times 1 0 ^{14}} = \frac{1.66}{4.0 \times 1 0 ^{14}} = 4.15 \times 1 0^{- 15} V\cdots .$ Step 2 — Planck's constant. $h = slope \times e = 4.15 \times 1 0^{- 15} \times 1.6 \times 1 0^{- 19} = 6.64 \times 1 0^{- 34} J\cdots .$ ✓ Step 3 — work function from a point. Use point 1: $V_{01} = \frac{h}{e} ν_{1} - \frac{ϕ}{e}$ , so $\frac{ϕ}{e} = \frac{h}{e} ν_{1} - V_{01} = (4.15 \times 1 0^{- 15}) (5.0 \times 1 0^{14}) - 0.30 = 2.075 - 0.30 = 1.775 V .$ Therefore $ϕ = 1.775 eV \approx 1.78 eV$ (multiplying $ϕ / e$ in volts by $e$ gives $ϕ$ in eV numerically unchanged).

Problem 3.2

Two metals A and B have $V_{0}$ – $ν$ lines drawn on the same axes. Metal B's line is shifted to the right of A's. Both lines are parallel. What does this tell you about (a) their work functions, (b) their Planck-constant slopes?

Recall Solution 3.2

(a) Work functions. The x-intercept is the threshold frequency $ν_{0} = ϕ / h$ . B's line crosses further right ⇒ larger $ν_{0}$ ⇒ larger $ϕ$ (harder to free B's electrons). (b) Slopes. Both lines are parallel and the slope is $h / e$ — a universal ratio. So the slopes are identical; the metal changes only the intercepts, never the tilt. This universality is exactly what let Millikan extract $h$ regardless of which metal he used.

L4 — Synthesis

Goal: combine photoelectric ideas with neighbouring concepts.

Problem 4.1

A photon of wavelength $λ = 300$ nm ejects an electron from a metal with $ϕ = 2.0$ eV. Find the de Broglie wavelength of the emitted electron. (Electron mass $m = 9.1 \times 1 0^{- 31}$ kg.) See de Broglie wavelength.

Recall Solution 4.1

Step 1 — kinetic energy of the electron. $E_{photon} = \frac{1240}{300} = 4.133 eV$ , so $K E_{m a x} = 4.133 - 2.0 = 2.133 eV .$ Convert to joules: $K E = 2.133 \times 1.6 \times 1 0^{- 19} = 3.41 \times 1 0^{- 19} J$ . Step 2 — momentum from KE. WHY: de Broglie needs momentum $p$ , and for a slow (non-relativistic) electron $K E = \frac{p ^{2}}{2 m}$ , so $p = 2 m K E$ . $p = 2 (9.1 \times 1 0^{- 31}) (3.41 \times 1 0^{- 19}) = 6.21 \times 1 0^{- 49} = 7.88 \times 1 0^{- 25} kg\cdotm/s .$ Step 3 — de Broglie wavelength. $λ_{d B} = \frac{h}{p}$ : $λ_{d B} = \frac{6.63 \times 1 0 ^{- 34}}{7.88 \times 1 0 ^{- 25}} = 8.41 \times 1 0^{- 10} m = 0.84 nm .$

Problem 4.2

A metal is lit by two wavelengths simultaneously: $λ_{1} = 400$ nm and $λ_{2} = 250$ nm. The work function is $ϕ = 3.0$ eV. Which wavelength ejects electrons, and what is the maximum kinetic energy of the fastest electron produced?

Recall Solution 4.2

Step 1 — test each photon. $E_{1} = \frac{1240}{400} = 3.10 eV$ ; $E_{2} = \frac{1240}{250} = 4.96 eV$ . Both exceed $ϕ = 3.0$ eV, so both eject electrons (each photon acts independently — one photon, one electron). Step 2 — the fastest electron comes from the more energetic photon. WHY: $K E_{m a x}$ grows with photon energy, so $λ_{2}$ (bluer, shorter) gives the fastest electron. $K E_{m a x} = E_{2} - ϕ = 4.96 - 3.0 = 1.96 eV .$ The 400 nm light also contributes electrons, but slower ones ( $3.10 - 3.0 = 0.10$ eV). The stopping potential of the whole setup is set by the fastest: $V_{0} = 1.96$ V.

L5 — Mastery

Goal: multi-step reasoning, degenerate cases, and a twist.

Problem 5.1

For a certain metal, doubling the incident frequency (from $ν$ to $2 ν$ ) changes the stopping potential from $V_{0} = 0.5$ V to $V_{0}^{'} = 2.3$ V. Find the work function $ϕ$ (eV) and the threshold frequency $ν_{0}$ in terms of the original $ν$ .

Recall Solution 5.1

Step 1 — write both equations. WHY: two unknowns ( $h ν$ scale and $ϕ$ ) need two equations. $e V_{0} = h ν - ϕ (1)$ $e V_{0}^{'} = h (2 ν) - ϕ (2)$ Step 2 — subtract (1) from (2). The $ϕ$ cancels, isolating $h ν$ : $e (V_{0}^{'} - V_{0}) = h ν ⟹ h ν = e (2.3 - 0.5) = 1.8 eV .$ (In eV, $e (V_{0}^{'} - V_{0})$ is numerically $1.8$ eV.) Step 3 — back-substitute into (1). $ϕ = h ν - e V_{0} = 1.8 - 0.5 = 1.3 eV .$ Step 4 — threshold frequency. $ν_{0} = \frac{ϕ}{h} = \frac{1.3}{h}$ . Since $h ν = 1.8$ eV, we have $h = 1.8/ ν$ (eV per Hz-unit), so $ν_{0} = \frac{1.3}{1.8/ ν} = \frac{1.3}{1.8} ν = 0.722 ν .$ So the threshold sits at about $72%$ of the original frequency — the original light was already above threshold, as it must be to give a positive $V_{0}$ .

Problem 5.2

A student claims that if you shine light of exactly the threshold frequency $ν_{0}$ , the emitted electrons will have a small but nonzero speed. Is this true? Support your answer with the equation, and describe the limiting behaviour as $ν \to ν_{0}^{+}$ .

Recall Solution 5.2

The claim is false. At $ν = ν_{0}$ : $K E_{m a x} = h ν_{0} - ϕ = ϕ - ϕ = 0.$ The photon's energy is spent entirely on the escape toll; there is nothing left to become motion. The electron just barely reaches the surface with zero speed. Limiting behaviour: as $ν \to ν_{0}^{+}$ (approaching threshold from above), $K E_{m a x} = h (ν - ν_{0}) \to 0^{+}$ — the electrons emerge slower and slower, and the stopping potential $V_{0} \to 0$ . Below $ν_{0}$ , $K E_{m a x}$ would be negative, which is physically impossible — meaning no electron escapes at all (the equation stops applying; emission simply switches off).

Problem 5.3

A metal surface receives light of intensity $I = 1.0 \times 1 0^{- 3} W/m^{2}$ over an area $A = 1.0 cm^{2}$ , at wavelength $λ = 400$ nm. If every photon that lands ejects one electron, what is the resulting current? ( $ϕ = 2.0$ eV, area in m $^{2}$ .)

Recall Solution 5.3

Step 1 — power on the surface. $A = 1.0 cm^{2} = 1.0 \times 1 0^{- 4} m^{2}$ . $P = I A = (1.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 4}) = 1.0 \times 1 0^{- 7} W .$ Step 2 — energy per photon (in joules). $E = \frac{h c}{λ} = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{400 \times 1 0 ^{- 9}} = 4.97 \times 1 0^{- 19} J .$ Step 3 — photons per second. WHY: power is energy/second, so dividing by energy-per-photon gives photons/second. $N = \frac{P}{E} = \frac{1.0 \times 1 0 ^{- 7}}{4.97 \times 1 0 ^{- 19}} = 2.01 \times 1 0^{11} photons/s .$ Step 4 — current. One electron per photon, each carrying charge $e$ : $I_{current} = N e = (2.01 \times 1 0^{11}) (1.6 \times 1 0^{- 19}) = 3.22 \times 1 0^{- 8} A \approx 32 nA .$ Note: $ϕ = 2.0$ eV was a distractor — it only decides whether emission happens (here $E = 3.10$ eV $> ϕ$ , so yes), not the current magnitude once every photon succeeds.

Recall Self-test summary (cloze)

The universal slope of the $V_{0}$ vs $ν$ graph is == $h / e$ ==. At exactly threshold frequency, $K E_{m a x}$ equals zero. To find the fastest electron under two wavelengths, use the shorter (more energetic) wavelength. The de Broglie wavelength of the ejected electron is $λ_{d B} =$ == $h / p$ == where $p = 2 m K E$ .

Which quantity is universal across all metals in the

V_{0}

–

ν

graph?

The slope

h / e

h ν < ϕ

, what is

K E_{m a x}

There is no emission; the equation gives a negative value that is physically meaningless.

Photons per second from power

P

at photon energy

E

N = P / E

Connections

Photoelectric effect — Einstein's explanation, work function (index 2.3.2) — the parent theory these exercises drill.
Energy conservation — the backbone of every $h ν = ϕ + K E_{m a x}$ line.
de Broglie wavelength — used in L4 to convert electron KE into a wavelength.
Millikan oil drop & measurement of h — the graph method of L3 mirrors Millikan's measurement.
Planck's quantum hypothesis & blackbody radiation — source of $E = h ν$ .
Wave-particle duality of light — why one photon feeds one electron.