A Level Physics - How to Solve: Photoelectric Effect & Energy Levels (Stopping Potential, de Broglie Wavelength)

How to Solve: Photoelectric Effect & Energy Levels (Stopping Potential, de Broglie Wavelength)

Complete Guide (For GCSE/A-Level Physics, Chemistry, Biology – Exam-Ready!)

Introduction

Master this, and you’ll crack 10–15% of your A-Level Physics paper—and even boost your Chemistry/Biology grades by understanding electron transitions. One question on stopping potential or de Broglie wavelength could be the difference between a B and an A—so let’s make sure you never lose marks here.

WHAT YOU NEED TO KNOW FIRST

Before diving in, you must already understand:
1. Wave-particle duality – Light behaves as both a wave (wavelength, frequency) and a particle (photons).
2. Energy conservation – Energy can’t be created or destroyed, only transferred.
3. Basic algebra – Rearranging equations like ( E = hf ) to solve for unknowns.

If any of these feel shaky, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas You Need

1. Photon Energy

[ E = hf ] - ( E ) = Energy of photon (J) - ( h ) = Planck’s constant (( 6.63 \times 10^{-34} ) Js) – given on exam sheet - ( f ) = Frequency of light (Hz)

2. Work Function & Threshold Frequency

[ \Phi = hf_0 ] - ( \Phi ) = Work function (J) - ( f_0 ) = Threshold frequency (Hz)

3. Maximum Kinetic Energy of Ejected Electrons

[ KE_{max} = hf - \Phi ] - ( KE_{max} ) = Maximum kinetic energy of ejected electron (J) - MEMORISE THIS – It’s the Einstein photoelectric equation.

4. Stopping Potential

[ eV_s = KE_{max} ] - ( e ) = Charge of electron (( 1.6 \times 10^{-19} ) C) – given on exam sheet - ( V_s ) = Stopping potential (V) - MEMORISE THIS – It links voltage to kinetic energy.

5. de Broglie Wavelength

[ \lambda = \frac{h}{p} ] - ( \lambda ) = de Broglie wavelength (m) - ( p ) = Momentum of particle (( p = mv )) (kg m/s) - MEMORISE THIS – Used for electrons, protons, etc.

6. Momentum (for de Broglie)

[ p = mv ] - ( m ) = Mass of particle (kg) - ( v ) = Velocity (m/s)

STEP-BY-STEP METHOD

How to Solve Photoelectric Effect Problems (Stopping Potential)

Follow these steps for any question involving stopping potential or kinetic energy of electrons.

1. Identify given values – Write down:
2. Frequency of light (( f ))
3. Work function (( \Phi )) or threshold frequency (( f_0 ))

4. Stopping potential (( V_s )) (if given)

5. Calculate photon energy – Use ( E = hf ).

6. If frequency isn’t given, use ( f = \frac{c}{\lambda} ) (where ( c = 3 \times 10^8 ) m/s).

7. Find maximum kinetic energy – Use ( KE_{max} = hf - \Phi ).

8. If ( \Phi ) isn’t given, use ( \Phi = hf_0 ).

9. Link kinetic energy to stopping potential – Use ( eV_s = KE_{max} ).

10. Rearrange to find ( V_s ) if needed.

11. Check units – Ensure all energies are in joules (J) before plugging into equations.

12. If given in eV, convert to joules: ( 1 \text{ eV} = 1.6 \times 10^{-19} ) J.

13. Answer the question – Does it ask for ( KE_{max} ), ( V_s ), or ( \Phi )? Write the final answer with units.

How to Solve de Broglie Wavelength Problems

Follow these steps for any question involving the wavelength of particles (e.g., electrons).

1. Identify given values – Write down:
2. Mass of particle (( m ))
3. Velocity (( v )) or kinetic energy (( KE ))

4. Planck’s constant (( h ))

5. Find momentum (( p )) – Use ( p = mv ).

6. If velocity isn’t given but ( KE ) is, use ( KE = \frac{1}{2}mv^2 ) to find ( v ), then ( p ).

7. Calculate de Broglie wavelength – Use ( \lambda = \frac{h}{p} ).

8. Check units – Ensure mass is in kg, velocity in m/s, and ( h ) in Js.

9. If mass is in u (atomic mass units), convert to kg: ( 1 \text{ u} = 1.66 \times 10^{-27} ) kg.

10. Answer the question – Write the final wavelength in metres (m).

WORKED EXAMPLES

Example 1 – Basic (Stopping Potential)

Question: Light of frequency ( 8.0 \times 10^{14} ) Hz shines on a metal with a work function of ( 3.2 \times 10^{-19} ) J. Calculate the stopping potential.

Solution:
1. Given: - ( f = 8.0 \times 10^{14} ) Hz - ( \Phi = 3.2 \times 10^{-19} ) J - ( h = 6.63 \times 10^{-34} ) Js (given) - ( e = 1.6 \times 10^{-19} ) C (given)

1. Calculate photon energy: ( E = hf = (6.63 \times 10^{-34}) \times (8.0 \times 10^{14}) ) ( E = 5.304 \times 10^{-19} ) J

2. Find ( KE_{max} ): ( KE_{max} = hf - \Phi = 5.304 \times 10^{-19} - 3.2 \times 10^{-19} ) ( KE_{max} = 2.104 \times 10^{-19} ) J

3. Link to stopping potential: ( eV_s = KE_{max} ) ( V_s = \frac{KE_{max}}{e} = \frac{2.104 \times 10^{-19}}{1.6 \times 10^{-19}} ) ( V_s = 1.315 ) V

Answer: The stopping potential is 1.32 V (to 3 s.f.).

What we did and why: - We used the photoelectric equation to find ( KE_{max} ). - Then linked ( KE_{max} ) to stopping potential using ( eV_s = KE_{max} ). - Key takeaway: Stopping potential depends on the excess energy after overcoming the work function.

Example 2 – Medium (Threshold Frequency & Stopping Potential)

Question: A metal has a work function of ( 4.2 \times 10^{-19} ) J. Light of wavelength ( 400 ) nm is shone on it. a) Calculate the threshold frequency. b) Determine the stopping potential.

Solution: Part a) Threshold frequency
1. Given: - ( \Phi = 4.2 \times 10^{-19} ) J - ( h = 6.63 \times 10^{-34} ) Js

1. Use ( \Phi = hf_0 ): ( f_0 = \frac{\Phi}{h} = \frac{4.2 \times 10^{-19}}{6.63 \times 10^{-34}} ) ( f_0 = 6.33 \times 10^{14} ) Hz

Answer (a): Threshold frequency = 6.33 × 10¹⁴ Hz

Part b) Stopping potential
1. Given: - ( \lambda = 400 ) nm = ( 400 \times 10^{-9} ) m - ( c = 3 \times 10^8 ) m/s

1. Find frequency of light: ( f = \frac{c}{\lambda} = \frac{3 \times 10^8}{400 \times 10^{-9}} ) ( f = 7.5 \times 10^{14} ) Hz

2. Calculate photon energy: ( E = hf = (6.63 \times 10^{-34}) \times (7.5 \times 10^{14}) ) ( E = 4.9725 \times 10^{-19} ) J

3. Find ( KE_{max} ): ( KE_{max} = hf - \Phi = 4.9725 \times 10^{-19} - 4.2 \times 10^{-19} ) ( KE_{max} = 0.7725 \times 10^{-19} ) J

4. Link to stopping potential: ( V_s = \frac{KE_{max}}{e} = \frac{0.7725 \times 10^{-19}}{1.6 \times 10^{-19}} ) ( V_s = 0.483 ) V

Answer (b): Stopping potential = 0.483 V

What we did and why: - For part (a), we used the work function to find the minimum frequency needed to eject electrons. - For part (b), we converted wavelength to frequency, then used the photoelectric equation to find ( KE_{max} ). - Key takeaway: If the light’s frequency is below ( f_0 ), no electrons are ejected—stopping potential would be 0 V.

Example 3 – Exam-Style (de Broglie Wavelength)

Question: An electron is accelerated through a potential difference of 500 V. a) Calculate its kinetic energy in joules. b) Determine its de Broglie wavelength.

Solution: Part a) Kinetic energy
1. Given: - ( V = 500 ) V - ( e = 1.6 \times 10^{-19} ) C - ( KE = eV ) (since ( KE = \frac{1}{2}mv^2 ) and ( eV = \frac{1}{2}mv^2 ))

1. Calculate ( KE ): ( KE = eV = (1.6 \times 10^{-19}) \times 500 ) ( KE = 8.0 \times 10^{-17} ) J

Answer (a): Kinetic energy = 8.0 × 10⁻¹⁷ J

Part b) de Broglie wavelength
1. Find velocity (( v )): ( KE = \frac{1}{2}mv^2 ) ( 8.0 \times 10^{-17} = \frac{1}{2} \times (9.11 \times 10^{-31}) \times v^2 ) ( v^2 = \frac{2 \times 8.0 \times 10^{-17}}{9.11 \times 10^{-31}} ) ( v^2 = 1.756 \times 10^{14} ) ( v = 1.325 \times 10^7 ) m/s

1. Find momentum (( p )): ( p = mv = (9.11 \times 10^{-31}) \times (1.325 \times 10^7) ) ( p = 1.207 \times 10^{-23} ) kg m/s

2. Calculate de Broglie wavelength: ( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.207 \times 10^{-23}} ) ( \lambda = 5.50 \times 10^{-11} ) m

Answer (b): de Broglie wavelength = 5.50 × 10⁻¹¹ m

What we did and why: - We used ( KE = eV ) to find the electron’s kinetic energy. - Then found velocity using ( KE = \frac{1}{2}mv^2 ). - Finally, used ( \lambda = \frac{h}{p} ) to find the wavelength. - Key takeaway: The higher the voltage, the shorter the de Broglie wavelength (since ( v ) increases).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

Listen up—this is your last-minute lifesaver.

1. Photoelectric effect:
2. Light hits metal → electrons ejected only if ( f > f_0 ).
3. Work function (( \Phi )) = minimum energy needed. ( \Phi = hf_0 ).
4. Maximum KE = ( hf - \Phi ).

5. Stopping potential = ( V_s = \frac{KE_{max}}{e} ).

6. de Broglie wavelength:

7. All particles have a wavelength: ( \lambda = \frac{h}{p} ).

8. Momentum (( p )) = ( mv ). If given ( KE ), find ( v ) first.

9. Units matter:

10. Joules (J) for energy, volts (V) for stopping potential.

11. Metres (m) for wavelength, kg for mass.

12. Exam tricks:

13. If frequency < ( f_0 ), no electrons → stopping potential = 0 V.
14. If given wavelength, convert to frequency first (( f = \frac{c}{\lambda} )).
15. Never use ( \frac{1}{2}mv^2 ) for photoelectric effect—only for de Broglie.

You’ve got this. Go smash that exam! ?