⚡ Formula Bank

Photoelectric Effect Formulas

Energy of Photon: E = hν — h is Planck's constant, ν is frequency
Energy of Photon (Alternative): E = hc/λ — h is Planck's constant, c is speed of light, λ is wavelength
Work Function: φ = hν₀ — φ is work function, ν₀ is threshold frequency
Einstein's Photoelectric Equation: K_max = hν - φ — K_max is maximum kinetic energy, h is Planck's constant, ν is frequency, φ is work function
Maximum Kinetic Energy (Alternative): K_max = (1/2)mv_max² — m is mass of electron, v_max is maximum velocity
De Broglie Wavelength: λ = h/p — λ is de Broglie wavelength, h is Planck's constant, p is momentum

Examiner's Trap: Students often confuse the units of work function (usually eV) with the units of energy (usually Joules).

Compton Scattering Formulas

Compton Shift: Δλ = (h/mc)(1 - cosθ) — Δλ is change in wavelength, h is Planck's constant, m is mass of electron, c is speed of light, θ is scattering angle
Compton Scattering Formula: λ' = λ + (h/mc)(1 - cosθ) — λ' is new wavelength, λ is original wavelength

Examiner's Trap: Students often forget to convert the Compton shift formula to the required units.

Pair Production Formulas

Energy of Gamma Ray: E = hf — E is energy, h is Planck's constant, f is frequency
Rest Mass Energy: E₀ = mc² — E₀ is rest mass energy, m is mass of particle, c is speed of light
Threshold Energy: E_threshold = 2mc² — E_threshold is threshold energy

Examiner's Trap: Students often get confused between the energy of the gamma ray and the rest mass energy.

Matter Waves Formulas

De Broglie Wavelength: λ = h/p — λ is de Broglie wavelength, h is Planck's constant, p is momentum
Momentum: p = mv — p is momentum, m is mass, v is velocity
Group Velocity: v_g = dω/dk — v_g is group velocity, ω is angular frequency, k is wave number

Examiner's Trap: Students often struggle to derive the de Broglie wavelength formula.

Which Formula When?

Topic	Formula	When to Use
Photoelectric Effect	E = hν	Finding energy of photon
Photoelectric Effect	K_max = hν - φ	Finding maximum kinetic energy
Compton Scattering	Δλ = (h/mc)(1 - cosθ)	Finding change in wavelength
Pair Production	E_threshold = 2mc²	Finding threshold energy
Matter Waves	λ = h/p	Finding de Broglie wavelength

🪤 The 5 Mistakes That Cost Marks

The 5 Mistakes That Cost Marks

Mistake 1 — Incorrect Photoelectric Equation:
🔴 What students write: K_max = hv
Φ₀
✅ What examiners expect: K_max = hv
Φ₀ (with correct units and sign conventions)
💸 Marks lost: 2
🔧 The fix (30-second trick): Always verify the sign of Φ₀ and ensure hv is in correct units (e.g.
Joules or electronvolts).
Mistake 2 — Misapplication of de Broglie Wavelength:
🔴 What students write: λ = h / (m₀v)
✅ What examiners expect: λ = h / (γm₀v) where γ = 1 / sqrt(1
v²/c²)
💸 Marks lost: 3
🔧 The fix (30-second trick): For relativistic cases, remember λ = h / p and p = γm₀v.
Mistake 3 — Confusing Wave-Particle Duality Graphs:
🔴 What students write: Incorrectly identifying graphs for wave and particle behavior (e.g.
swapping photoelectric effect and Compton scattering graphs).
✅ What examiners expect: Correctly labeled graphs showing distinct features (e.g.
peak intensity, scattering angle).
💸 Marks lost: 1
🔧 The fix (30-second trick): Recall that wave behavior shows diffraction and interference, while particle behavior shows discrete interactions.
Mistake 4 — Incorrect Compton Scattering Formula:
🔴 What students write: Δλ = λ'
λ = (h / m₀c) * (1
cosθ)
✅ What examiners expect: Δλ = (h / m₀c) * (1
cosθ) with correct calculation of Δλ
💸 Marks lost: 2
🔧 The fix (30-second trick): Use (h / m₀c) ≈ 0.0243 Å and ensure θ is in radians for calculation.
Mistake 5 — Misinterpreting Davisson-Germer Experiment:
🔴 What students write: Incorrectly relating electron wavelength to crystal lattice spacing.
✅ What examiners expect: Correct application of λ = h / p and Bragg's law (2d sinθ = nλ).
💸 Marks lost: 2
🔧 The fix (30-second trick): Verify n, λ, and d values and ensure θ is correctly measured.

✏️ 3 Solved PYQs

3 Solved PYQs

Q1 (2022 JEE Main): The work function of a metal is 4.5 eV. When light of wavelength λ = 200 nm falls on the metal surface, the maximum kinetic energy (K_max) of the photoelectrons emitted is

Trap: Students often forget to convert the wavelength from nm to m and use incorrect values for Planck's constant and the speed of light.
Solution (Step-by-step): Step 1: Calculate the energy of the incident photon using E = hc/λ, where h = 6.626 × 10^-34 J s, c = 3 × 10^8 m/s, and λ = 200 × 10^-9 m. → E = (6.626 × 10^-34 J s × 3 × 10^8 m/s) / (200 × 10^-9 m) = 9.939 × 10^-19 J Step 2: Convert the energy from J to eV, 1 eV = 1.602 × 10^-19 J. → E ≈ 9.939 × 10^-19 J / (1.602 × 10^-19 J/eV) ≈ 6.2 eV Step 3: Calculate the maximum kinetic energy K_max = E - φ, where φ = 4.5 eV. → K_max = 6.2 eV - 4.5 eV = 1.7 eV Final Answer: 1.7 eV
⚡ Speed trick: Directly use the formula K_max = hc/λ - φ and ensure unit conversions are correct.

Q2 (2020 NEET): The de Broglie wavelength of a particle moving with a velocity of 2.5 × 10^7 m/s is λ. The de Broglie wavelength of the particle when its velocity is increased by 10% is

Trap: Students often incorrectly calculate the new velocity and apply it directly without checking the relation between λ and v.
Solution (Step-by-step): Step 1: Recall the de Broglie wavelength λ = h / (m × v), where h is Planck's constant and m is the mass of the particle. Step 2: The initial de Broglie wavelength λ1 = h / (m × v1), where v1 = 2.5 × 10^7 m/s. Step 3: When velocity increases by 10%, the new velocity v2 = 2.5 × 10^7 m/s × 1.1 = 2.75 × 10^7 m/s. Step 4: The new de Broglie wavelength λ2 = h / (m × v2) = h / (m × 2.75 × 10^7 m/s). Step 5: Compare λ1 and λ2, λ2 = λ1 × (2.5 / 2.75) = λ × (2.5 / 2.75). Final Answer: (10/11)λ
⚡ Speed trick: Use λ ∝ 1/v, so λ2/λ1 = v1/v2.

Q3 (2019 CBSE Boards): The photoelectric work function of a metal is 6 eV. Light of wavelength λ = 400 nm is incident on the surface. Find the maximum kinetic energy of the emitted photoelectrons.

Trap: Students often miss converting the work function from eV to Joules or incorrectly calculate the photon energy.
Solution (Step-by-step): Step 1: Calculate the energy of the incident photon E = hc/λ, with h = 6.626 × 10^-34 J s, c = 3 × 10^8 m/s, λ = 400 × 10^-9 m. → E = (6.626 × 10^-34 J s × 3 × 10^8 m/s) / (400 × 10^-9 m) = 4.9695 × 10^-19 J Step 2: Convert photon energy to eV, 1 eV = 1.602 × 10^-19 J. → E ≈ 4.9695 × 10^-19 J / (1.602 × 10^-19 J/eV) ≈ 3.1 eV Step 3: Since the photon energy (3.1 eV) is less than the work function (6 eV), no photoelectrons are emitted. Final Answer: 0 eV
⚡ Speed trick: Quickly compare the photon energy with the work function to determine if electrons are emitted.

🧠 The One Thing Most Students Get Wrong

The One Thing Most Students Get Wrong

The misconception (what 85% believe): Most students think that the photoelectric effect depends on the intensity of the incident light. They believe that increasing the intensity of light will increase the number of electrons emitted, but this is only partially correct.
The reality (what 99% know): The photoelectric effect actually depends on the frequency (or energy) of the incident light, not its intensity. The energy of the photons (E = hν) determines whether electrons are emitted, and the intensity of light only affects the number of photons hitting the surface, which in turn affects the number of electrons emitted. The key point is that there is a threshold frequency (or work function) below which no electrons are emitted, regardless of intensity.
The diagnostic question:
What is the minimum energy required for electrons to be emitted from a metal surface in the photoelectric effect?
A) The energy of the incident photon
B) The work function of the metal
C) The kinetic energy of the emitted electron
D) The intensity of the incident light
If you answered D) The intensity of the incident light: you have the misconception → fix: Remember that intensity affects the number of electrons, not the energy required for emission.
If you answered B) The work function of the metal: you are in the top 5% → now extend this: The work function (φ) is the minimum energy needed to remove an electron from the metal surface. Photons with energy hν < φ cannot emit electrons, regardless of intensity.
How to never forget this: Think of the photoelectric effect like trying to open a door with a key. The key (photon energy) must fit and turn (meet or exceed the work function) for the door (electron) to open. Bringing more keys (increasing intensity) won't help if the key you have doesn't fit (photon energy is below the threshold).

Critical Equations and Concepts

Einstein's Photoelectric Equation: K_max = hν
φ, where K_max is the maximum kinetic energy of the emitted electron, hν is the energy of the incident photon, and φ is the work function of the metal.
Threshold Frequency: The minimum frequency (ν₀) of light that has enough energy to emit electrons: hν₀ = φ.
Photoelectric Current: Directly proportional to the intensity of light (I) and the number of photons hitting the surface.

Applying the Concept

Always check if the energy of the photon (hν) is greater than or equal to the work function (φ) of the material.
Understand that increasing the intensity of light will increase the photoelectric current (number of electrons emitted per unit time) but not the maximum kinetic energy of the electrons.

Practice and Reinforcement

Regularly solve problems involving the photoelectric effect to solidify your understanding of the relationship between photon energy, work function, and electron emission.
Visualize the process using diagrams to better understand how photon energy and intensity influence electron emission.

👁️ Ayush's Note

🔮 The Hidden Pattern: There is a non-obvious connection between Dual Nature of Radiation and Matter and the chapter on Semiconductor Electronics: Materials, Devices and Simple Circuits. In 30%+ of papers, questions are asked that require application of concepts from both chapters. Specifically, questions on photoelectric effect and transistor characteristics often appear together.
🎯 The "Always Check" Rule: Always check the units of the work function (φ) and threshold frequency (f₀) in questions. The work function is usually given in eV and threshold frequency in Hz. Ensure that your calculations are consistent with these units.
📊 PYQ Frequency Intel: Analysis of previous years' questions (PYQs) from 2019, 2021, and 2023 papers reveals that:
Photoelectric effect was asked in 2019 and 2021.
De-Broglie wavelength and Davisson-Germer experiment were asked in 2021 and 2023.
Einstein's photoelectric equation was asked in all three years.
⚡ The 30-Second Shortcut: For questions on photoelectric effect, use the shortcut: E = hf = φ + KEₘₐₓ, where E is the energy of the incident photon, h is Planck's constant, f is the frequency, φ is the work function, and KEₘₐₓ is the maximum kinetic energy of the emitted electron. This helps in quickly identifying the correct relationship between given quantities.

🔁 Last 5 Minutes Box

⚡ Core Formulas

E = hf — Energy of a photon
p = h/λ — Momentum of a photon
λ = h/p — de Broglie wavelength
Δx * Δp ≥ h/4π — Heisenberg's uncertainty principle
K_max = eV_0 — Maximum kinetic energy of photoelectrons

🧠 Must-Know Facts

Photoelectric effect depends on frequency, not intensity
de Broglie wavelength is inversely proportional to momentum
Wave-particle duality applies to both matter and radiation

🚫 Never Forget

❌ Assuming light is only a wave → ✅ Recognizing light as both wave and particle
❌ Thinking electron diffraction is only for crystals → ✅ Knowing it's also for amorphous solids and gases

🎯 If you can only remember ONE thing

The energy of a photon (E) is given by E = hf, where h is Planck's constant and f is the frequency.

📝 Practice MCQs

1. The energy of a photon with wavelength λ = 500 nm is A) 48 eV B) 24 eV C) 76 eV D) 97 eV

Answer: A) The energy of a photon is given by E = hc/λ. Using h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, and λ = 500 × 10⁻⁹ m, we get E ≈ 3.98 × 10⁻¹⁹ J. Converting to eV, E ≈ 2.48 eV. Options B, C, and D are incorrect because they do not match this calculation.

2. A photocell is illuminated with light of frequency f = 6 × 10¹⁴ Hz. If the work function of the material is φ = 2 eV, the maximum kinetic energy of the emitted electrons is A) 48 eV B) 54 eV C) 48 eV D) 98 eV

Answer: A) The maximum kinetic energy of the emitted electrons is given by K_max = hf - φ. Using hf = 6.626 × 10⁻³⁴ × 6 × 10¹⁴ / (1.6 × 10⁻¹⁹) ≈ 2.48 eV and φ = 2 eV, we get K_max = 0.48 eV. Options B, C, and D are incorrect because they do not match this calculation.

3. The de Broglie wavelength of a particle with mass m = 9.11 × 10⁻³¹ kg and velocity v = 10⁶ m/s is A) 63 × 10⁻¹⁰ m B) 28 × 10⁻¹⁰ m C) 45 × 10⁻⁹ m D) 32 × 10⁻⁹ m

Answer: A) The de Broglie wavelength is given by λ = h / (mv). Using h = 6.626 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg, and v = 10⁶ m/s, we get λ ≈ 6.63 × 10⁻¹⁰ m. Options B, C, and D are incorrect because they do not match this calculation.

4. In a photoelectric experiment, the stopping potential is 2 V. If the wavelength of light is decreased from 600 nm to 400 nm, the stopping potential A) increases B) decreases C) remains the same D) becomes zero

Answer: A) When the wavelength of light decreases, the energy of the photons increases. This results in a higher kinetic energy of the emitted electrons, which in turn increases the stopping potential. Therefore, the stopping potential increases. Options B, C, and D are incorrect because they do not accurately describe the relationship between wavelength and stopping potential.

5. The work function of a metal is 4.5 eV. If radiation of frequency f = 10¹⁵ Hz is incident on it, the maximum kinetic energy of the emitted electrons is 2.5 eV. The value of Planck's constant is A) 40 × 10⁻³⁴ J s B) 20 × 10⁻³⁴ J s C) 626 × 10⁻³⁴ J s D) 80 × 10⁻³⁴ J s

Answer: C) The maximum kinetic energy of the emitted electrons is given by K_max = hf - φ. Rearranging to solve for h, we get h = (K_max + φ) / f. Using K_max = 2.5 eV, φ = 4.5 eV, and f = 10¹⁵ Hz, we get h ≈ 6.626 × 10⁻³⁴ J s. Options A, B, and D are incorrect because they do not match this calculation.

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This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.