Skip to content

Photons and Atomic Physics

Photoelectric effect, photons, atomic models, energy levels, nuclear physics

Written and reviewed by the Study Mondo Education TeamLast updated
🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

⚛️ Photons and Atomic Physics

Wave-Particle Duality

Light exhibits both wave and particle properties!

Wave properties:

  • Interference
  • Diffraction
  • Polarization

Particle properties:

  • Photoelectric effect
  • Compton scattering
  • Discrete energy packets

💡 Quantum mechanics: Nature is fundamentally probabilistic, not deterministic!

Photons

Photon = particle of light (quantum of electromagnetic energy)

Energy of photon: E=hf=hcλE = hf = \frac{hc}{\lambda}

where:

  • h = Planck's constant = 6.626×10−346.626 \times 10^{-34} J·s
  • f = frequency (Hz)
  • c = speed of light
  • λ = wavelength

Higher frequency → higher energy

  • Gamma rays: Very high E
  • Radio waves: Very low E

Photoelectric Effect

Light shining on metal can eject electrons!

Key Observations:

  1. Threshold frequency f0f_0: Below this, NO electrons (even intense light!)
  2. Instantaneous: Electrons ejected immediately
  3. KE depends on f, not intensity
  4. Intensity affects number of electrons, not their energy

Cannot be explained by classical wave theory!

Einstein's Photoelectric Equation

KEmax=hf−ϕKE_{max} = hf - \phi

where:

  • KEmaxKE_{max} = maximum kinetic energy of ejected electron
  • hfhf = photon energy
  • ϕ\phi = work function (minimum energy to eject electron)

At threshold: hf0=ϕhf_0 = \phi (just enough to eject, KE = 0)

Stopping potential VsV_s: eVs=KEmaxeV_s = KE_{max}

Voltage needed to stop most energetic electrons.

💡 Won Nobel Prize 1921! Not for relativity, but photoelectric effect.

De Broglie Wavelength

Particles can behave like waves!

λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}

Matter waves confirmed by electron diffraction experiments!

Larger mass → smaller wavelength

  • Electron: λ ~ nm (measurable!)
  • Baseball: λ ~ 10⁻³⁴ m (unmeasurable)

Atomic Models

Rutherford Model (1911):

  • Nucleus: tiny, massive, positive
  • Electrons orbit (like planets)
  • Problem: Accelerating charges radiate → atom should collapse!

Bohr Model (1913):

  • Electrons in discrete energy levels (orbits)
  • Only certain radii allowed: rn=n2a0r_n = n^2 a_0 where a0=0.529a_0 = 0.529 Å
  • Quantized angular momentum: L=nℏL = n\hbar where ℏ=h/2π\hbar = h/2\pi

Energy levels (hydrogen): En=−13.6 eVn2E_n = -\frac{13.6 \text{ eV}}{n^2}

where n = 1, 2, 3, ... (principal quantum number)

  • n = 1: Ground state, E = -13.6 eV
  • n = ∞: Ionization, E = 0

Negative energy means bound (need energy to remove electron).

Atomic Transitions

Electron jumps between levels → photon absorbed or emitted!

Energy of photon: Ephoton=∣Ef−Ei∣=hfE_{photon} = |E_f - E_i| = hf

Emission (high → low): Photon out Absorption (low → high): Photon in

Emission spectrum: Discrete lines (fingerprint of element!)

Hydrogen Series:

  • Lyman (UV): n → 1
  • Balmer (visible): n → 2
  • Paschen (IR): n → 3

Heisenberg Uncertainty Principle

Cannot know position and momentum simultaneously with perfect precision!

ΔxΔp≥ℏ2\Delta x \Delta p \geq \frac{\hbar}{2}

Also for energy and time: ΔEΔt≥ℏ2\Delta E \Delta t \geq \frac{\hbar}{2}

Not limitation of measurement, but fundamental nature of reality!

💡 Key: More certain about position → less certain about momentum

Nuclear Physics

Nucleus Composition:

  • Protons: Z (atomic number), charge +e
  • Neutrons: N, charge 0
  • Mass number: A = Z + N

Isotopes: Same Z, different N (different A)

Example: 12C^{12}\text{C} vs 14C^{14}\text{C}

Mass-Energy Equivalence

Einstein's most famous equation:

E=mc2E = mc^2

Mass and energy are interchangeable!

Atomic mass unit: 1 u = 1.66×10−271.66 \times 10^{-27} kg = 931.5 MeV/c²

Binding energy: Mass of separated nucleons > mass of nucleus Δm=(mass of parts)−(mass of whole)\Delta m = \text{(mass of parts)} - \text{(mass of whole)} BE=Δmc2BE = \Delta m c^2

More binding energy → more stable

Nuclear Reactions

Fission:

  • Heavy nucleus splits → lighter nuclei
  • Releases energy (BE/nucleon increases)
  • Used in nuclear power plants
  • Example: 235U^{235}\text{U} + neutron → fission fragments + neutrons + energy

Fusion:

  • Light nuclei combine → heavier nucleus
  • Releases enormous energy
  • Powers the sun!
  • Example: 2H^2\text{H} + 3H^3\text{H} → 4He^4\text{He} + neutron + 17.6 MeV

Fusion releases MORE energy per nucleon than fission!

Radioactive Decay

Unstable nuclei decay spontaneously:

Types:

  • Alpha (α): 4He^4\text{He} nucleus, A↓4, Z↓2
  • Beta (β⁻): Electron, neutron→proton, A same, Z↑1
  • Gamma (γ): High-energy photon, A and Z same

Half-life t1/2t_{1/2}: N(t)=N0(12)t/t1/2N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}

After one half-life: 50% remain After two: 25% remain After three: 12.5% remain

Conservation Laws

All nuclear reactions must conserve:

  1. Mass-energy: Total E (including mc²) conserved
  2. Charge: Total Z conserved
  3. Mass number: Total A conserved
  4. Momentum: Total p conserved

Problem-Solving Strategy

Photoelectric:

  1. Find photon energy: E=hfE = hf or E=hc/λE = hc/\lambda
  2. Apply: KEmax=hf−ϕKE_{max} = hf - \phi
  3. Check threshold: if f<f0f < f_0, no electrons!

Atomic transitions:

  1. Find energy levels: En=−13.6/n2E_n = -13.6/n^2 eV
  2. Energy difference: ΔE=∣Ef−Ei∣\Delta E = |E_f - E_i|
  3. Photon: λ=hc/ΔE\lambda = hc/\Delta E

Nuclear:

  1. Check conservation (A and Z)
  2. Calculate mass defect: Δm
  3. Energy: E=Δmc2E = \Delta m c^2

Common Mistakes

❌ Using wavelength in meters with h in J·s (watch units!) ❌ Thinking intensity affects electron KE (only frequency does!) ❌ Forgetting negative sign in atomic energy levels ❌ Not converting eV to Joules (or vice versa): 1 eV = 1.60×10⁻¹⁹ J ❌ Confusing emission (high→low) with absorption (low→high) ❌ Using half-life formula incorrectly (power is t/t_1/2, not just t!)

📚 Practice Problems

1Problem 1easy

❓ Question:

Find the energy of a photon with wavelength 500 nm. Express in both Joules and eV.

💡 Show Solution

Given:

  • Wavelength: λ=500\lambda = 500 nm =5.00×10−7= 5.00 \times 10^{-7} m
  • Planck's constant: h=6.626×10−34h = 6.626 \times 10^{-34} J·s
  • Speed of light: c=3.00×108c = 3.00 \times 10^8 m/s

Solution:

Photon energy: E=hcλ=(6.626×10−34)(3.00×108)5.00×10−7E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{5.00 \times 10^{-7}} E=1.988×10−255.00×10−7=3.98×10−19 JE = \frac{1.988 \times 10^{-25}}{5.00 \times 10^{-7}} = 3.98 \times 10^{-19} \text{ J}

Convert to eV: E=3.98×10−191.60×10−19=2.49 eVE = \frac{3.98 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.49 \text{ eV}

Answer:

  • E = 3.98 × 10⁻¹⁹ J
  • E = 2.49 eV (green light)

2Problem 2easy

❓ Question:

Find the energy of a photon with wavelength 500 nm. Express in both Joules and eV.

💡 Show Solution

Given:

  • Wavelength: λ=500\lambda = 500 nm =5.00×10−7= 5.00 \times 10^{-7} m
  • Planck's constant: h=6.626×10−34h = 6.626 \times 10^{-34} J·s
  • Speed of light: c=3.00×108c = 3.00 \times 10^8 m/s

Solution:

Photon energy: E=hcλ=(6.626×10−34)(3.00×108)5.00×10−7E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{5.00 \times 10^{-7}} E=1.988×10−255.00×10−7=3.98×10−19 JE = \frac{1.988 \times 10^{-25}}{5.00 \times 10^{-7}} = 3.98 \times 10^{-19} \text{ J}

Convert to eV: E=3.98×10−191.60×10−19=2.49 eVE = \frac{3.98 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.49 \text{ eV}

Answer:

  • E = 3.98 × 10⁻¹⁹ J
  • E = 2.49 eV (green light)

3Problem 3medium

❓ Question:

Light with wavelength 400 nm strikes a metal surface with work function 2.0 eV. Find (a) maximum kinetic energy of ejected electrons, (b) stopping potential.

💡 Show Solution

Given:

  • Wavelength: λ=400\lambda = 400 nm =4.00×10−7= 4.00 \times 10^{-7} m
  • Work function: ϕ=2.0\phi = 2.0 eV =3.20×10−19= 3.20 \times 10^{-19} J

Part (a): Maximum kinetic energy

First, find photon energy: E=hcλ=(6.626×10−34)(3.00×108)4.00×10−7E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{4.00 \times 10^{-7}} E=4.97×10−19 J=3.11 eVE = 4.97 \times 10^{-19} \text{ J} = 3.11 \text{ eV}

Einstein's photoelectric equation: KEmax=hf−ϕ=3.11−2.0=1.11 eVKE_{max} = hf - \phi = 3.11 - 2.0 = 1.11 \text{ eV}

In Joules: KEmax=(1.11)(1.60×10−19)=1.78×10−19 JKE_{max} = (1.11)(1.60 \times 10^{-19}) = 1.78 \times 10^{-19} \text{ J}

Part (b): Stopping potential

eVs=KEmaxeV_s = KE_{max} Vs=KEmaxe=1.11 VV_s = \frac{KE_{max}}{e} = 1.11 \text{ V}

Answer:

  • (a) KE_max = 1.11 eV = 1.78 × 10⁻¹⁹ J
  • (b) V_s = 1.11 V

4Problem 4medium

❓ Question:

Light with wavelength 400 nm strikes a metal surface with work function 2.0 eV. Find (a) maximum kinetic energy of ejected electrons, (b) stopping potential.

💡 Show Solution

Given:

  • Wavelength: λ=400\lambda = 400 nm =4.00×10−7= 4.00 \times 10^{-7} m
  • Work function: ϕ=2.0\phi = 2.0 eV =3.20×10−19= 3.20 \times 10^{-19} J

Part (a): Maximum kinetic energy

First, find photon energy: E=hcλ=(6.626×10−34)(3.00×108)4.00×10−7E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{4.00 \times 10^{-7}} E=4.97×10−19 J=3.11 eVE = 4.97 \times 10^{-19} \text{ J} = 3.11 \text{ eV}

Einstein's photoelectric equation: KEmax=hf−ϕ=3.11−2.0=1.11 eVKE_{max} = hf - \phi = 3.11 - 2.0 = 1.11 \text{ eV}

In Joules: KEmax=(1.11)(1.60×10−19)=1.78×10−19 JKE_{max} = (1.11)(1.60 \times 10^{-19}) = 1.78 \times 10^{-19} \text{ J}

Part (b): Stopping potential

eVs=KEmaxeV_s = KE_{max} Vs=KEmaxe=1.11 VV_s = \frac{KE_{max}}{e} = 1.11 \text{ V}

Answer:

  • (a) KE_max = 1.11 eV = 1.78 × 10⁻¹⁹ J
  • (b) V_s = 1.11 V

5Problem 5medium

❓ Question:

Light of wavelength 400 nm strikes a metal surface. (a) What is the energy of each photon? (b) If the work function is 2.0 eV, what is the maximum kinetic energy of ejected electrons? Use h = 6.63 × 10⁻³⁴ J·s, c = 3.0 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J.

💡 Show Solution

Solution:

Given: λ = 400 nm = 4.0 × 10⁻⁷ m, φ = 2.0 eV = 3.2 × 10⁻¹⁹ J

(a) Photon energy: E = hf = hc/λ E = (6.63 × 10⁻³⁴)(3.0 × 10⁸)/(4.0 × 10⁻⁷) E = (19.9 × 10⁻²⁶)/(4.0 × 10⁻⁷) E = 4.98 × 10⁻¹⁹ J = 3.1 eV

(b) Maximum kinetic energy (Photoelectric effect): KE_max = E - φ KE_max = 3.1 - 2.0 = 1.1 eV

Or in Joules: KE_max = 4.98 × 10⁻¹⁹ - 3.2 × 10⁻¹⁹ = 1.78 × 10⁻¹⁹ J

6Problem 6hard

❓ Question:

A hydrogen atom electron transitions from n=3 to n=2. Find (a) energy of emitted photon, (b) wavelength of light.

💡 Show Solution

Given:

  • Initial state: ni=3n_i = 3
  • Final state: nf=2n_f = 2
  • Hydrogen energy levels: En=−13.6/n2E_n = -13.6/n^2 eV

Part (a): Photon energy

Energy levels: E3=−13.632=−13.69=−1.51 eVE_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV} E2=−13.622=−13.64=−3.40 eVE_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.40 \text{ eV}

Photon energy (emission, so high → low): Ephoton=E3−E2=−1.51−(−3.40)=1.89 eVE_{photon} = E_3 - E_2 = -1.51 - (-3.40) = 1.89 \text{ eV}

In Joules: E=(1.89)(1.60×10−19)=3.02×10−19 JE = (1.89)(1.60 \times 10^{-19}) = 3.02 \times 10^{-19} \text{ J}

Part (b): Wavelength

λ=hcE=(6.626×10−34)(3.00×108)3.02×10−19\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{3.02 \times 10^{-19}} λ=6.56×10−7 m=656 nm\lambda = 6.56 \times 10^{-7} \text{ m} = 656 \text{ nm}

Answer:

  • (a) E_photon = 1.89 eV
  • (b) λ = 656 nm (red light, Balmer series!)

This is the famous H-alpha line in hydrogen spectrum.

7Problem 7hard

❓ Question:

In the hydrogen atom, an electron transitions from n = 3 to n = 2. (a) Calculate the energy of the emitted photon using E_n = -13.6 eV/n². (b) What is the wavelength of the emitted light? (c) What region of the spectrum is this?

💡 Show Solution

Solution:

(a) Photon energy: E₃ = -13.6/3² = -13.6/9 = -1.51 eV E₂ = -13.6/2² = -13.6/4 = -3.40 eV

ΔE = E₃ - E₂ = -1.51 - (-3.40) = 1.89 eV

Or: 1.89 eV × 1.6 × 10⁻¹⁹ J/eV = 3.02 × 10⁻¹⁹ J

(b) Wavelength: E = hc/λ λ = hc/E = (6.63 × 10⁻³⁴)(3.0 × 10⁸)/(3.02 × 10⁻¹⁹) λ = (19.9 × 10⁻²⁶)/(3.02 × 10⁻¹⁹) λ = 6.59 × 10⁻⁷ m = 659 nm

(c) Spectrum region: λ = 659 nm is in the visible red region (visible: 400-700 nm)

This is the first line of the Balmer series (transitions to n = 2).

8Problem 8hard

❓ Question:

A hydrogen atom electron transitions from n=3 to n=2. Find (a) energy of emitted photon, (b) wavelength of light.

💡 Show Solution

Given:

  • Initial state: ni=3n_i = 3
  • Final state: nf=2n_f = 2
  • Hydrogen energy levels: En=−13.6/n2E_n = -13.6/n^2 eV

Part (a): Photon energy

Energy levels: E3=−13.632=−13.69=−1.51 eVE_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV} E2=−13.622=−13.64=−3.40 eVE_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.40 \text{ eV}

Photon energy (emission, so high → low): Ephoton=E3−E2=−1.51−(−3.40)=1.89 eVE_{photon} = E_3 - E_2 = -1.51 - (-3.40) = 1.89 \text{ eV}

In Joules: E=(1.89)(1.60×10−19)=3.02×10−19 JE = (1.89)(1.60 \times 10^{-19}) = 3.02 \times 10^{-19} \text{ J}

Part (b): Wavelength

λ=hcE=(6.626×10−34)(3.00×108)3.02×10−19\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{3.02 \times 10^{-19}} λ=6.56×10−7 m=656 nm\lambda = 6.56 \times 10^{-7} \text{ m} = 656 \text{ nm}

Answer:

  • (a) E_photon = 1.89 eV
  • (b) λ = 656 nm (red light, Balmer series!)

This is the famous H-alpha line in hydrogen spectrum.

Explain using:
🎴

Practice with Flashcards

Rate this topic's cards with spaced repetition. Cards join your deck when you finish a topic's lesson and take its exit quiz.

📖

Browse All Topics

Explore more AP Physics 2 topics

❓ Frequently Asked Questions

What is Photons and Atomic Physics?▾
Photoelectric effect, photons, atomic models, energy levels, nuclear physics
How can I study Photons and Atomic Physics effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Photons and Atomic Physics study guide free?▾
Yes — all study notes, flashcards, and practice problems for Photons and Atomic Physics on Study Mondo are free to access. No account is needed.
What course covers Photons and Atomic Physics?▾
Photons and Atomic Physics is part of the AP Physics 2 course on Study Mondo, specifically in the Modern Physics section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Photons and Atomic Physics?▾
Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.