Fusion releases MORE energy per nucleon than fission!
Radioactive Decay
Unstable nuclei decay spontaneously:
Types:
Alpha (α): 4He nucleus, A↓4, Z↓2
Beta (β⁻): Electron, neutron→proton, A same, Z↑1
Gamma (γ): High-energy photon, A and Z same
Half-lifet1/2:
N(t)=N0(21)t/t1/2
After one half-life: 50% remain
After two: 25% remain
After three: 12.5% remain
Conservation Laws
All nuclear reactions must conserve:
Mass-energy: Total E (including mc²) conserved
Charge: Total Z conserved
Mass number: Total A conserved
Momentum: Total p conserved
Problem-Solving Strategy
Photoelectric:
Find photon energy: E=hf or E=hc/λ
Apply: KEmax=hf−ϕ
Check threshold: if f<f0, no electrons!
Atomic transitions:
Find energy levels: En=−13.6/n2 eV
Energy difference: ΔE=∣Ef−Ei∣
Photon: λ=hc/ΔE
Nuclear:
Check conservation (A and Z)
Calculate mass defect: Δm
Energy: E=Δmc2
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 nm =5.00×10−7 m
Planck's constant: h=6.626×10−34 J·s
Speed of light: c=3.00×108 m/s
Solution:
Photon energy:
E=λhc=
Convert to eV:
E=1.60×10−193.98×10
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 nm = m
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 nm m
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 nm m
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=3
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
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.
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