Photoelectric effect — Einstein's explanation, work function: Question bank

Two symbols before we start, so nothing is used unexplained:

$ν$ (Greek "nu") = frequency of the light = how many wave-cycles pass per second. Bigger $ν$ = bluer light = more energy per photon.
$ϕ$ (Greek "phi") = work function = the fixed "toll" (in energy) each metal charges an electron to escape its surface.

True or false — justify

Every item here has a hidden trap. The justification is what earns the mark.

T/F: Doubling the intensity of the incident light doubles the maximum kinetic energy of the ejected electrons.

False. Intensity is the number of photons per second, not the energy per photon; each electron absorbs exactly one photon, so

K E_{m a x} = h ν - ϕ

is untouched. Doubling intensity doubles the number of electrons, not their speed.

T/F: If the light frequency is below the threshold frequency

ν_{0}

, shining it for a much longer time will eventually eject electrons.

False. Energy arrives in indivisible packets; one sub-threshold photon (

h ν < ϕ

) can't pay the toll and electrons de-excite instantly rather than "saving up". No emission ever occurs, regardless of duration.

T/F: The stopping potential

V_{0}

depends on the intensity of the light.

False.

e V_{0} = h ν - ϕ

has no intensity term.

V_{0}

is fixed by the fastest electron's energy, which is set by

ν

and

ϕ

only. Brighter light gives more current but the same

V_{0}

T/F: The slope of the stopping-potential-versus-frequency graph is larger for a metal with a larger work function.

False. From

V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}

, the slope is

h / e

— a universal constant, identical for every metal. A larger

ϕ

only shifts the line right (bigger

ν_{0}

) and down (more negative intercept).

T/F: A photon whose energy exactly equals

ϕ

ejects an electron moving at high speed.

False. At

h ν = ϕ

we have

K E_{m a x} = 0

: the electron just barely escapes with essentially zero speed. This is precisely the threshold case.

T/F: Red light of very high intensity can eject electrons from a metal that green light of low intensity ejects easily.

False (in general). If red light is below

ν_{0}

, no amount of intensity works, while low-intensity green (above

ν_{0}

) ejects electrons every time. Frequency, not brightness, decides whether emission happens.

T/F: The work function of a metal equals the ionization energy of a free atom of that metal.

False.

ϕ

is the surface escape energy for the least-bound electron leaving the bulk solid; it is usually smaller than the energy to strip an electron from an isolated atom.

T/F: Increasing the frequency of light (above threshold) at fixed intensity increases the number of ejected electrons per second.

False (and subtly so). At fixed intensity (fixed energy delivered per second), higher-frequency photons each carry more energy, so there are fewer photons per second — hence fewer electrons, though each is faster. Number of electrons tracks photon count, not frequency.

Spot the error

Each statement below contains one flawed step. Name it and repair it.

"Since

K E_{m a x} = h ν - ϕ

, a metal with

ϕ = 0

would emit electrons for any light, so

ϕ = 0

metals are the best photodetectors."

Error: no real metal has

ϕ = 0

; electrons are always bound to some degree, so

ϕ > 0

always. The equation is right, the premise is physically impossible.

"The photoelectric equation

h ν = ϕ + K E_{m a x}

shows energy is created when the photon frees the electron."

Error: nothing is created — this is energy conservation. The photon's

h ν

is entirely spent:

ϕ

to escape and

K E_{m a x}

as leftover motion. See Energy conservation.

"Because emission is instantaneous, the electron must absorb energy faster than light can travel, violating relativity."

Error: instantaneous means a single absorption event, not superluminal energy transfer. One photon hands over

h ν

in one interaction; no accumulation, no speed limit broken.

"The stopping potential is the voltage that speeds up the fastest electron until it reaches the collector."

Error: it's a reverse (retarding) voltage that stops the fastest electron just short of the collector. When

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

, even the fastest electron is turned back.

K E_{m a x} = h ν - ϕ

, so plotting

K E_{m a x}

against intensity gives a straight line with slope

h

Error:

K E_{m a x}

does not depend on intensity at all — its graph vs intensity is a flat horizontal line. The straight line with slope

h / e

V_{0}

vs frequency, a different plot.

"All electrons ejected by a single frequency of light come out with the same speed, namely

K E_{m a x}

Error: electrons emerge with a range of speeds from

0

up to

K E_{m a x}

. Only the least-bound (surface) electron gets the full leftover; deeper electrons spend extra energy reaching the surface and come out slower.

Why questions

Why does the

V_{0}

-versus-

ν

graph have the same slope for every metal, even though metals differ?

Because the slope is

h / e

— built only from Planck's constant and the electron charge, both universal. The metal enters solely through

ϕ

, which shifts the intercept, not the tilt. This universality is what let Millikan measure $h$.

Why does classical wave theory predict a time delay for emission, and why is that prediction wrong?

A spread-out wave would deposit energy gradually, so a dim wave would need time to accumulate enough at one electron. Reality: a photon delivers

h ν

in one instant event, so emission begins within

\sim 1 0^{- 9}

s regardless of brightness.

Why is it

K E_{m a x}

in the equation and not just

K E

Only the surface (least-bound) electron pays exactly

ϕ

; deeper electrons pay more to reach the surface and emerge slower. The maximum leftover energy belongs to the surface electron, giving the ceiling

K E_{m a x} = h ν - ϕ

Why does a bigger work function push the threshold to higher frequency (shorter wavelength)?

From

ν_{0} = ϕ / h

, a larger toll

ϕ

demands a more energetic photon to just cover it, so

ν_{0}

rises. Since

λ_{0} = h c / ϕ

, the threshold wavelength shrinks — often into the UV.

Why does intensity control the photocurrent but not the stopping potential?

Intensity = photons per second = electrons per second = current. But each electron's energy is fixed by one photon's

h ν

, so the stopping voltage (which counters the fastest electron's energy) is untouched by how many arrive.

Why did the photoelectric effect force physicists to accept the particle nature of light?

The threshold, the instant emission, and the intensity-independence of

K E_{m a x}

are all inexplicable if light is a continuous wave, but fall out immediately if light comes in energy packets

h ν

. The data demanded quanta.

Edge cases

Edge: What is

K E_{m a x}

when

h ν = ϕ

exactly?

Exactly zero — the electron is liberated but with no spare energy to move. This defines the threshold:

ν = ν_{0}

λ = λ_{0}

Edge: What happens when

h ν < ϕ

(photon below threshold)?

No electron is emitted at all — the single packet can't pay the toll, and packets don't add up. The photocurrent is exactly zero.

Edge: What if two low-energy photons arrive at the same electron nearly together (below-threshold light)?

In the ordinary photoelectric regime this doesn't liberate the electron — absorption is one-photon-at-a-time and the electron de-excites before a second helps. (Multi-photon emission needs enormous laser intensities, outside this topic.)

Edge: What does the

V_{0}

-versus-

ν

line predict at

ν = 0

The y-intercept is

- ϕ / e

, a negative stopping potential — a purely mathematical extrapolation, since no real light has

ν = 0

and no emission occurs below

ν_{0}

anyway.

Edge: As the frequency

ν \to \infty

, what happens to

K E_{m a x}

It grows without bound as

K E_{m a x} = h ν - ϕ \to \infty

(in the non-relativistic idealization). Very high

ν

(X-rays) does eject very fast electrons — and eventually the Compton effect regime, where the photon behaves as a momentum-carrying particle, takes over.

Edge: Can the stopping potential ever be negative for light above threshold?

No. Above threshold

K E_{m a x} = h ν - ϕ > 0

, so

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

. A positive retarding voltage is always needed to stop a genuinely emitted electron.

Recall One-line self-test

If a friend says "just make the light brighter and dim red light will finally eject electrons," what's your reply? ::: Below threshold, brightness is irrelevant — each red photon individually fails to pay $ϕ$ , and photons don't pool their energy; only a bluer (higher- $ν$ ) photon can do it.

Connections

Photoelectric effect — Einstein's explanation, work function — the parent note these traps stress-test.
Energy conservation — every "spot the error" here reduces to honest energy bookkeeping.
Wave-particle duality of light — why quanta, not waves, explain the data.
Compton effect — the high-frequency limiting behaviour extends into here.
Millikan oil drop & measurement of h — the universal slope $h / e$ that several items probe.