Why does raising the intensity leave the maximum kinetic energy unchanged?

Intensity only adds photons per second, while each photon keeps the energy set by its frequency; one electron still absorbs one photon, so the maximum kinetic energy stays put and only the current grows.

How does the stopping potential relate to the maximum kinetic energy?

The stopping potential is the reverse voltage that just reduces the photocurrent to zero, so eVs equals the maximum kinetic energy, which makes Vs a direct reading of that energy in electronvolts.

S6 Physics: Photoelectric Effect Paper 1 (photon energy, work function and stopping potential)

2026-11-03

S · 6

This paper starts from photon energy: find E = hf, subtract the work function to get the maximum kinetic energy, then convert that into a stopping potential. Use h = 6.63 x 10^-34 J s, c = 3.00 x 10^8 m/s and 1 eV = 1.60 x 10^-19 J throughout.

Quick answer　3.315 x 10^-19 J (2.07 eV); 4.97 x 10^-19 J; 4.83 x 10^14 Hz; current rises, kinetic energy unchanged; 1.8 x 10^-19 J (1.125 eV, stopping potential 1.125 V); 501 nm; kinetic energy unchanged; 2.64 eV; 3.99 x 10^-19 J and 1.98 x 10^-19 J; the error is confusing intensity with frequency.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10 questions
Suggested time	25 minutes
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S6 Physics: Photoelectric Effect
Level	S6

02　Questions (10 items)

FoundationA photon has frequency 5.0 x 10^14 Hz. What is its energy in joules, and in electronvolts, using h = 6.63 x 10^-34 J s and 1 eV = 1.60 x 10^-19 J?
FoundationLight of wavelength 400 nm falls on a metal. What is its frequency, and what is the energy of one photon in joules?
FoundationA metal has a work function of 2.0 eV. What is that in joules, and what is the threshold frequency?
FoundationThe frequency is already above the threshold and only the intensity is doubled with the frequency fixed. What happens to the photocurrent and to the maximum kinetic energy?
IntermediateA metal of work function 2.0 eV is struck by photons of energy 5.0 x 10^-19 J. What are the maximum kinetic energy in joules and in electronvolts, and what is the stopping potential?
IntermediateA metal has a work function of 2.48 eV. What is its threshold wavelength in nanometres?
IntermediateFor the same metal and frequency the intensity is raised from I to 2I. How do the photocurrent and the maximum kinetic energy change, and why?
ChallengeLight of wavelength 300 nm gives a stopping potential of 1.5 V. Find the photon energy in electronvolts and the work function of the metal.
ChallengeA graph of maximum kinetic energy against frequency is a straight line of gradient h. Given 2.0 x 10^-19 J at 6.0 x 10^14 Hz, find the maximum kinetic energy at 9.0 x 10^14 Hz and the work function in joules and electronvolts.
Error redoA student writes that increasing the intensity raises the maximum kinetic energy until electrons come out. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	3.315 x 10^-19 J, that is 2.07 eV	Six point six three times ten to the minus thirty-four multiplied by five times ten to the fourteen gives 3.315 x 10^-19 J, which over 1.60 x 10^-19 is 2.07 eV.
2	7.50 x 10^14 Hz, with 4.97 x 10^-19 J per photon	Three times ten to the eight over 400 times ten to the minus nine gives 7.50 x 10^14 Hz, and multiplying by h gives 4.97 x 10^-19 J.
3	3.2 x 10^-19 J with a threshold frequency of 4.83 x 10^14 Hz	Two times 1.60 x 10^-19 gives 3.2 x 10^-19 J, which over h is 4.83 x 10^14 Hz; below that frequency no electrons come out however bright the light.
4	The photocurrent roughly doubles and the maximum kinetic energy stays the same	Twice the photons per second means twice the electrons emitted, while each photon carries the same energy, so the maximum kinetic energy of an electron is untouched.
5	1.8 x 10^-19 J, that is 1.125 eV, so the stopping potential is 1.125 V	The photon energy 5.0 x 10^-19 J minus the work function 3.2 x 10^-19 J leaves 1.8 x 10^-19 J, which is 1.125 eV and therefore a stopping potential of 1.125 V.
6	about 501 nm	The work function is 3.968 x 10^-19 J, and hc over W gives 5.01 x 10^-7 m, that is 501 nm.
7	The current doubles and the maximum kinetic energy is unchanged, since photon energy depends on frequency alone.	Doubling the intensity doubles the photons per second and the electron count with it, but each photon still carries hf, so the maximum kinetic energy is capped at the same value.
8	Photon energy 4.14 eV and work function 2.64 eV	hc over lambda gives 6.63 x 10^-19 J or 4.14 eV, and subtracting the 1.5 eV stopping potential leaves a work function of 2.64 eV.
9	Kinetic energy 3.99 x 10^-19 J with a work function of 1.98 x 10^-19 J, that is 1.24 eV	A rise of 3.0 x 10^14 Hz adds h times that, 1.989 x 10^-19 J, giving 3.99 x 10^-19 J, and the work function is hf minus the kinetic energy, 1.98 x 10^-19 J.
10	The mistake is confusing intensity with frequency: intensity adds photons, while emission depends on whether the frequency exceeds the threshold.	Below the threshold frequency one photon cannot supply the work function, so no intensity however great will release an electron, and intensity only sets the size of the current.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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