Photoelectric effect — Einstein's photon model: Question bank

This bank targets the parent note 2.1.2 Photoelectric effect. Before you start, here are the symbols every trap leans on — no letter appears below until it is named here.

Definition Every symbol used on this page

$h$ = Planck's constant, $6.626 \times 1 0^{- 34} J\cdots$ — a fixed number of nature that converts a light wave's frequency into the energy of one packet. Think of it as the "price per unit of frequency."
$ν$ (Greek "nu") = frequency of the light (in hertz, Hz) — how fast the wave wiggles; this is the colour dial.
$ϕ$ (Greek "phi") = work function — the minimum energy to pull one electron out of the metal, set by the metal.
$ν_{0}$ = threshold frequency — the exact frequency whose packet energy equals $ϕ$ , i.e. $ϕ = h ν_{0}$ .
$K E_{m a x}$ = maximum kinetic energy — the leftover energy a surface electron flies off with.
$e$ = elementary charge, $1.602 \times 1 0^{- 19} C$ — the (magnitude of the) charge on one electron.
$V_{0}$ = stopping potential — defined in the next callout.

Figure — Photoelectric effect — Einstein's photon model

Recall The three-way distinction (memorise before the traps)

Work function $ϕ$ — set by the metal, the minimum energy to escape. Threshold frequency $ν_{0}$ — the frequency whose photon energy exactly equals $ϕ$ , i.e. $ϕ = h ν_{0}$ . $K E_{m a x}$ — the leftover energy, $K E_{m a x} = h (ν - ν_{0})$ , set by the light's frequency. Metal fixes the first two; light fixes the last. The figure above draws all three: the line's slope hides $h$ , its intercept hides $ϕ / e$ , and its height above the axis is $V_{0}$ .

True or false — justify

Doubling the intensity doubles the maximum kinetic energy of the ejected electrons.

False. Intensity means more photons of the same energy

h ν

; each electron still absorbs exactly one photon, so

K E_{m a x} = h ν - ϕ

is unchanged — only the electron count doubles.

If light is below the threshold frequency, waiting longer will eventually eject an electron.

False. One electron absorbs one photon in one instant; energy does not accumulate. A million too-small packets never combine into one big enough packet, so emission never happens.

A photon of frequency exactly equal to

ν_{0}

ejects an electron with zero kinetic energy.

True. At

ν = ν_{0}

h ν = ϕ

exactly, so

K E_{m a x} = h (ν - ν_{0}) = 0

— the electron barely escapes with nothing to spare.

Two metals lit by the same light always emit electrons with the same

K E_{m a x}

False.

K E_{m a x} = h ν - ϕ

and

ϕ

differs per metal; the one with the smaller work function emits faster electrons for the identical photon.

Red light, if bright enough, can eject electrons from any metal.

False. If the red photon's energy

h ν < ϕ

, no single photon can pay the escape toll; brightness only adds more equally-weak photons, so nothing is ejected regardless of intensity.

The stopping potential

V_{0}

depends on how bright the light is.

False.

e V_{0} = K E_{m a x} = h ν - ϕ

has no intensity term;

V_{0}

tracks frequency only. More intensity raises the current before cutoff but not the cutoff voltage.

The graph of

V_{0}

against frequency

ν

has a slope that depends on the metal used.

False. The slope is

h / e

, made only of universal constants; it is identical for every metal. Different metals shift the intercept (

- ϕ / e

), not the slope.

Increasing frequency above threshold increases the number of photoelectrons.

False. Number of electrons is set by number of photons (intensity); frequency sets their energy. Raising

ν

makes each electron faster, not more numerous.

At threshold, the work function equals the photon energy.

True. By definition

ϕ = h ν_{0}

, and at threshold the incident frequency is exactly

ν_{0}

, so

h ν = ϕ

Spot the error

"Since

K E_{m a x} = h (ν - ν_{0})

came out to

- 0.4 eV

, the electron is emitted moving backwards."

Error: a negative result means

ν < ν_{0}

, i.e. no emission at all. The equation is only valid when

ν \geq ν_{0}

; negative "KE" is the algebra signalling "nothing comes out," never a real backward electron.

"The work function is the energy the ejected electron carries away."

Error:

ϕ

is the energy spent escaping (the toll), not carried away. The electron carries away the leftover,

K E_{m a x} = h ν - ϕ

"Longest wavelength that ejects electrons is

λ_{0} = ϕ / h c

Error: the formula is inverted. Threshold is

h c / λ_{0} = ϕ

, so

λ_{0} = h c / ϕ

(an energy over an energy gives length, once you keep

h c

on top).

"Doubling the frequency doubles

K E_{m a x}

Error:

K E_{m a x} = h ν - ϕ

is linear but has an offset

- ϕ

. Doubling

ν

gives

2 h ν - ϕ

, which is more than double the old

K E_{m a x} = h ν - ϕ

unless

ϕ = 0

"Because energy is

\frac{1}{2} m v^{2}

, the fastest electrons come from photons absorbed deepest in the metal."

Error: deep electrons lose extra energy fighting their way out, so they emerge slower.

K E_{m a x}

belongs to a surface electron that pays only the minimum toll

ϕ

"eV is a unit of voltage, so

K E_{m a x}

in eV and

V_{0}

in volts are literally the same number by coincidence."

Error: it is not coincidence —

e V_{0} = K E_{m a x}

, so dividing an energy in eV by the charge

e

leaves the number unchanged in volts. It is the definition of the electron-volt at work.

"Intensity is amplitude squared, so classical theory is right that intensity sets the electron energy."

Error: classical amplitude-squared reasoning is exactly what fails. Experiment shows energy tracks frequency; the photon model, not the wave-amplitude model, matches reality.

Why questions

Why does the ejected electron's energy depend on frequency but the current depend on intensity?

Each photon carries

h ν

(a frequency quantity) and each ejection is a one photon–one electron event, so energy per electron follows

ν

; the number of such events follows the number of photons, which is intensity.

Why is emission instantaneous even for very dim light above threshold?

An electron needs only one photon of sufficient energy, and absorption is a single instantaneous event. Dimness reduces how many electrons come out, not how fast each absorption happens.

Why did Einstein need photons rather than just Planck's quantised oscillators?

Planck quantised the emitting oscillators of a hot body; Einstein made the light itself granular, so that a single localised packet can hand its whole energy to one electron — the step that explains threshold and instantaneity.

Why does the

V_{0}

ν

line let us measure Planck's constant?

Rearranged,

V_{0} = (h / e) ν - ϕ / e

, a straight line whose slope is

h / e

. Measure the slope from the graph, multiply by the known charge

e

, and you have

h

with no need to know

ϕ

Why can't you get electrons out with a huge number of low-frequency photons?

Because there is no mechanism for two photons to combine their energy on one electron in a single event; the process is strictly one photon per electron, so each too-weak photon fails individually.

Why is

K E_{m a x}

written with "max" rather than just

K E

Electrons start at different depths and lose different amounts of energy escaping; the maximum is the special case of a surface electron paying only the minimum toll

ϕ

, giving the cleanest link to

h ν - ϕ

Edge cases

What is

K E_{m a x}

when

ν

is far below

ν_{0}

— is it a large negative number?

There is no

K E_{m a x}

at all — no electron is emitted. The negative value from the formula is not a physical energy; it is the equation's way of flagging "below threshold."

What happens exactly at

ν = ν_{0}

to the stopping potential?

K E_{m a x} = 0

, so

V_{0} = 0

. Electrons just barely escape with zero speed, and no reverse voltage is needed to stop them.

ϕ = 0

(an idealised free electron with no binding), what does the equation predict?

K E_{m a x} = h ν

— every photon of any frequency ejects an electron and hands over its entire energy, with no threshold. Real metals always have

ϕ > 0

, so a threshold always exists.

Two photons of half the threshold energy hit the same electron nearly simultaneously — does it escape?

In the standard (low-intensity) photon model, no: absorption is one-photon-per-electron, so two half-energy photons cannot pool their energy. (Two-photon absorption exists only at extreme laser intensities, outside this model.)

At the very brightest possible light below threshold, is the photocurrent tiny or exactly zero?

Exactly zero. Every photon is individually too weak, so no electron is ever freed no matter how many arrive per second.

If the frequency is doubled from just above

ν_{0}

, does the stopping potential more than double?

Yes.

V_{0} = (h ν - ϕ) / e

; at first

h ν

barely exceeds

ϕ

, so

K E_{m a x}

is small, and the extra

h ν_{0}

worth of energy makes the new

K E_{m a x}

(and

V_{0}

) grow faster than a straight doubling.

Connections

Work Function and Binding Energy — the escape toll $ϕ$ , central to nearly every trap above.
Planck's Quantum Theory — origin of $E = h ν$ , the seed of the one-photon-one-electron rule.
Wave-Particle Duality — why light must be granular here yet wavelike elsewhere.
Photoelectron Spectroscopy — the same $h ν = ϕ + K E$ bookkeeping turned into a measurement tool.
Electromagnetic Spectrum — why frequency (colour), not brightness, is the deciding dial.