A Level Edexcel Physics Paper 2 Exam Questions and
Answers A+ Graded (2025)
1. Electron .Gun .Model .Answer: .- .Low .voltage .supply .heats .filament .causes
.thermionic .emission/ .electrons .have .enough .energy .to .leave
- Electrons .are .repulsed .from .the .cathode .(filament) .and .attracted .towards .the .anode
- This .electric .field .causes .electrons .to .accelerate
- There .is .a .hole .in .the .anode .which .creates .a .beam .for .electrons .that .pass .through
- Electrons .are .then .fired .through .a .vacuum .(to .stop .interfering .particles) .onto .a
.fluorescent .screen
2. Photo .Electric .Effect .Model .Answer: .- .1 .Electron .near .the .surface .of .a .metal
.gains .1 .photon .and .gains .enough .energy .to .be .liberated. .1-2-1 .interaction
- The .work .function .is .the .minimum .energy .needed .for .the .electron .to .escape .the
.surface
- increasing .intensity .will .increase .the .number .of .photons
- increasing .frequency .will .increase .the .energy .of .the .photons
- These .electrons .released .are .called .photoelectrons
- This .effect .can .be .explained .through .E .= .hf
- Extra .energy .is .then .stored .as .the .kinetic .energy .store .in .the .photoelectron
- hf .= .¦ .+EK
3. PhotoElectric .Equation: .Ekmax .= .hf .- .work .fucntion
,Ekmax .= .max .kinetic .energy .of .freed .electrons
.
hf .= .energy .of .photon
work .function .= .the .minimum .energy .to .release .an .electron
4. Why .does .the .photoelectric .effect .equation .refer .to .kinetic .energy .as .maxi-
.mum .kinetic .energy?: .It .will .not .always .be .an .equal .energy .transfer:
- Some .energy .may .have .been .transferred .to .electrons .below .the .surface
, - Therefore .for .these .electrons .to .liberate .to .the .surface .there .will .require .more
.energy
- therefore .the .electrons .leaves .the .surface .will .less .than .max
5. Energy .Frequency .equation: .E .= .hf
E .= .energy
h .= .plank's .constant
.f .= .frequency
6. Work .function .equation: .Work .Function .= .h .x .threshold .frequency
7. Work .Function: .the .minimum .energy .needed .to .liberate .an .electron .from .a .metal
.atom
8. Threshold .Frequency: .minimum .light .frequency .necessary .to .liberate .an .elec-
.tron .from .a .given .metal
Energy .is .proportional .to .frequency
9. Wave .Particle .Duality .Model .Answer:
10. Nuclear .Binding .Energy: .The .energy .needed .to .separate .a .nucleus .into .indi-
.vidual .protons .and .neutrons
E .= .mc^2
Binding .Energy .= .mass .difference .x .the .speed .of .light^2
11. How .to .calculate .nuclear .binding .energy .from .a .nuclear .decay .equation?-
: .Th .-> .Ra .+ .a
Mass .of .Th .nuclear-(Mass .of .Ra .nuclear .+ .mass .of .alpha .particle)
. Convert .u .to .kg
Multiply .by .speed .of .light^2 .E .= .mc^2
Convert .to .Mev
12. How .to .convert .u .to .kg: .kg ./ .1.6605x10^-27
13. How .to .convert .kg .to .u: .kg ./ .1.6605x10^-27
14. What .is .1 .u?: .1.6605x10^-27kg
15. Converting .u .to .Mev: .u .x .931.5
16. Converting .Mev .to .u: .Mev ./ .931.5
Answers A+ Graded (2025)
1. Electron .Gun .Model .Answer: .- .Low .voltage .supply .heats .filament .causes
.thermionic .emission/ .electrons .have .enough .energy .to .leave
- Electrons .are .repulsed .from .the .cathode .(filament) .and .attracted .towards .the .anode
- This .electric .field .causes .electrons .to .accelerate
- There .is .a .hole .in .the .anode .which .creates .a .beam .for .electrons .that .pass .through
- Electrons .are .then .fired .through .a .vacuum .(to .stop .interfering .particles) .onto .a
.fluorescent .screen
2. Photo .Electric .Effect .Model .Answer: .- .1 .Electron .near .the .surface .of .a .metal
.gains .1 .photon .and .gains .enough .energy .to .be .liberated. .1-2-1 .interaction
- The .work .function .is .the .minimum .energy .needed .for .the .electron .to .escape .the
.surface
- increasing .intensity .will .increase .the .number .of .photons
- increasing .frequency .will .increase .the .energy .of .the .photons
- These .electrons .released .are .called .photoelectrons
- This .effect .can .be .explained .through .E .= .hf
- Extra .energy .is .then .stored .as .the .kinetic .energy .store .in .the .photoelectron
- hf .= .¦ .+EK
3. PhotoElectric .Equation: .Ekmax .= .hf .- .work .fucntion
,Ekmax .= .max .kinetic .energy .of .freed .electrons
.
hf .= .energy .of .photon
work .function .= .the .minimum .energy .to .release .an .electron
4. Why .does .the .photoelectric .effect .equation .refer .to .kinetic .energy .as .maxi-
.mum .kinetic .energy?: .It .will .not .always .be .an .equal .energy .transfer:
- Some .energy .may .have .been .transferred .to .electrons .below .the .surface
, - Therefore .for .these .electrons .to .liberate .to .the .surface .there .will .require .more
.energy
- therefore .the .electrons .leaves .the .surface .will .less .than .max
5. Energy .Frequency .equation: .E .= .hf
E .= .energy
h .= .plank's .constant
.f .= .frequency
6. Work .function .equation: .Work .Function .= .h .x .threshold .frequency
7. Work .Function: .the .minimum .energy .needed .to .liberate .an .electron .from .a .metal
.atom
8. Threshold .Frequency: .minimum .light .frequency .necessary .to .liberate .an .elec-
.tron .from .a .given .metal
Energy .is .proportional .to .frequency
9. Wave .Particle .Duality .Model .Answer:
10. Nuclear .Binding .Energy: .The .energy .needed .to .separate .a .nucleus .into .indi-
.vidual .protons .and .neutrons
E .= .mc^2
Binding .Energy .= .mass .difference .x .the .speed .of .light^2
11. How .to .calculate .nuclear .binding .energy .from .a .nuclear .decay .equation?-
: .Th .-> .Ra .+ .a
Mass .of .Th .nuclear-(Mass .of .Ra .nuclear .+ .mass .of .alpha .particle)
. Convert .u .to .kg
Multiply .by .speed .of .light^2 .E .= .mc^2
Convert .to .Mev
12. How .to .convert .u .to .kg: .kg ./ .1.6605x10^-27
13. How .to .convert .kg .to .u: .kg ./ .1.6605x10^-27
14. What .is .1 .u?: .1.6605x10^-27kg
15. Converting .u .to .Mev: .u .x .931.5
16. Converting .Mev .to .u: .Mev ./ .931.5
