Class 12 Photoelectric Effect: Energy, Intensity and Stopping Potential Practice
A brighter beam contains more photon energy arriving per unit time, but at a fixed frequency each photon still has the same energy. This distinction explains why increasing intensity cannot compensate for a photon energy below the work function in the basic photoelectric model.
Curriculum checked 2 October 2026 · original practice for selected skills.
PREDICT · CHANGE · EXPLAIN
Above threshold: change photon energy
For a metal with work function 2.1 eV, this above-threshold model gives Kmax=E−2.1. The slider starts at the threshold; it does not model negative kinetic energy.
The curve shows this simplified model over the labelled range. Read the exact point below; the sketch is not a measuring instrument.
Maximum KE: 1.5 eV
What happens to Kmax when photon energy rises by 1 eV?
Now close the explanation and try the independent retest. Moving a slider alone does not check your understanding.
Before you start
You will need: Energy units, electron charge and potential difference.
Attempt each task before opening a hint. If you need help, reveal the first hint, then the setup, then compare your written steps with the solution. Finish with the retests without referring back.
Three ideas to keep in view
For emission, photon energy must reach the work function. Above threshold, maximum kinetic energy is photon energy minus work function.
One electron-volt is the energy gained by one elementary charge through one volt. With e=1.6 × 10⁻¹⁹ C, 1 eV=1.6 × 10⁻¹⁹ J.
At fixed frequency above threshold, greater intensity can increase photoelectron emission rate. It does not increase the maximum kinetic energy or the stopping-potential magnitude in this model.
Try the worked example first
P12-PHOTO-W1Energy and potentialFocused practice
Photons of energy 3.6 eV strike a metal with work function 2.1 eV. Find Kmax in eV and J, and stopping-potential magnitude. Use e=1.6 × 10⁻¹⁹ C.
Show your working
Need a starting hint?
Subtract the work function first.
Still stuck? Reveal the setup
Kmax=1.5 eV; convert energy to joules before using K=eVs if needed.
Show worked solution+
1
Energy balance
Kmax=3.6−2.1=1.5 eV, equal to 2.4 × 10⁻¹⁹ J.
2
Stopping potential
Divide the energy in joules by e.
∣Vs∣=1.6×10−192.4×10−19=1.5 V
Answer: 1.5 eV; 2.4 × 10⁻¹⁹ J; 1.5 V
Before moving on: can you explain why your method works, as well as give the answer?
Your turn: three different checks
Write a method as well as an answer. The tasks change the reasoning, not just the numbers.
P12-PHOTO-Q1Below thresholdFocused practice
Photons of 1.8 eV strike a metal with work function 2.2 eV. Would doubling intensity cause photoemission in this model?
Show your working
Need a starting hint?
Compare energy per photon with the work function.
Still stuck? Reveal the setup
1.8 eV is below 2.2 eV.
Show worked solution+
1
Threshold test
Each photon lacks sufficient energy.
2
Intensity change
More photons of the same insufficient energy do not change the threshold result.
Answer: No photoemission in the basic model.
Before moving on: can you explain why your method works, as well as give the answer?
P12-PHOTO-Q2Double intensityFocused practice
At fixed frequency above threshold, intensity doubles. What happens to maximum kinetic energy and stopping potential?
Show your working
Need a starting hint?
Energy per photon is unchanged.
Still stuck? Reveal the setup
Both depend on hf−work function, not beam intensity.
Show worked solution+
1
Hold frequency fixed
Photon energy stays the same.
2
Predict
Kmax and stopping-potential magnitude stay unchanged; emission rate may increase.
Answer: Both remain unchanged.
Before moving on: can you explain why your method works, as well as give the answer?
P12-PHOTO-Q3Work backwardFocused practice
Photon energy is 5.0 eV and the stopping-potential magnitude is 1.8 V. Find the work function.
Show your working
Need a starting hint?
The maximum electron energy is 1.8 eV.
Still stuck? Reveal the setup
Work function = photon energy − Kmax.
Show worked solution+
1
Convert the meaning
For an electron, 1.8 V stopping magnitude corresponds to 1.8 eV.
2
Subtract
5.0−1.8=3.2 eV.
Answer: 3.2 eV
Before moving on: can you explain why your method works, as well as give the answer?
Catch a likely mistake
A tempting claim: “Brighter light always gives faster photoelectrons.”
At fixed frequency brightness changes photon arrival rate, not energy per photon.
Close the examples and try again
These changed questions test whether you can reconstruct the method. If you use a hint, record where you got stuck and retry later. Completing this small set does not establish full chapter mastery.
P12-PHOTO-R1Changed photonIndependent retest
Use photon energy 2.8 eV and work function 2.1 eV. Find Kmax in J and stopping-potential magnitude.
Show your working
Need a starting hint?
First subtract in eV.
Still stuck? Reveal the setup
Kmax=0.7 eV.
Show worked solution+
1
Convert
0.7 × 1.6 × 10⁻¹⁹=1.12 × 10⁻¹⁹ J.
2
Potential
Divide by the elementary charge.
Answer: 1.12 × 10⁻¹⁹ J; 0.7 V
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
P12-PHOTO-R2New threshold testIndependent retest
A 2.1 eV work-function metal receives 1.9 eV photons. State whether stopping potential can be inferred as −0.2 V.
Show your working
Need a starting hint?
Stopping potential assumes photoelectrons exist.
Still stuck? Reveal the setup
The photons are below threshold.
Show worked solution+
1
Threshold
No photoemission occurs in the model.
2
Interpretation
There is no emitted-electron stopping potential to calculate from a negative energy difference.
Answer: No; the proposed −0.2 V result is invalid.
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Choose the next useful step
If a retest exposed the same error, rewrite the first incorrect step and explain its correction aloud. If both were independent, return to a mixed exercise or a missed paper question. A parent can ask what changed in the method rather than only asking for the answer.
No. They are original targeted practice for selected photoelectric-effect skills, not copied board questions or a prediction of exam marks.
Can I print the questions and worked solutions separately?+
Yes. Use the two print buttons above. The question sheet leaves working space and hides solutions; the worked-solutions option includes the solution steps. Your browser can save either view as a PDF.
PERSONAL SUPPORT
Does the same mistake keep returning?
Bring one attempted question and the step that caused difficulty. Discuss suitable TutorMax support for the learner's current level.
