Photons and Atomic Physics

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

⚛️ 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: $E = hf = \frac{hc}{\lambda}$

where:

h = Planck's constant = $6.626 \times 10^{-34}$ 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:

Threshold frequency $f_0$ : Below this, NO electrons (even intense light!)
Instantaneous: Electrons ejected immediately
KE depends on f, not intensity
Intensity affects number of electrons, not their energy

Cannot be explained by classical wave theory!

Einstein's Photoelectric Equation

$KE_{max} = hf - \phi$

where:

$KE_{max}$ = maximum kinetic energy of ejected electron
$hf$ = photon energy
$\phi$ = work function (minimum energy to eject electron)

At threshold: $hf_0 = \phi$ (just enough to eject, KE = 0)

Stopping potential $V_s$ : $eV_s = KE_{max}$

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!

$\lambda = \frac{h}{p} = \frac{h}{mv}$

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: $r_n = n^2 a_0$ where $a_0 = 0.529$ Å
Quantized angular momentum: $L = n\hbar$ where $\hbar = h/2\pi$

Energy levels (hydrogen): $E_n = -\frac{13.6 \text{ eV}}{n^2}$

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: $E_{photon} = |E_f - E_i| = hf$

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!

$\Delta x \Delta p \geq \frac{\hbar}{2}$

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

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

Mass-Energy Equivalence

Einstein's most famous equation:

$E = mc^2$

Mass and energy are interchangeable!

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

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

More binding energy → more stable

Nuclear Reactions

Fission:

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

Fusion:

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

Fusion releases MORE energy per nucleon than fission!

Radioactive Decay

Unstable nuclei decay spontaneously:

Types:

Alpha (α): $^4\text{He}$ 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 $t_{1/2}$ : $N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}$

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

Conservation Laws

All nuclear reactions must conserve:

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 $E = hc/\lambda$
Apply: $KE_{max} = hf - \phi$
Check threshold: if $f < f_0$ , no electrons!

Atomic transitions:

Find energy levels: $E_n = -13.6/n^2$ eV
Energy difference: $\Delta E = |E_f - E_i|$
Photon: $\lambda = hc/\Delta E$

Nuclear:

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

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

Solution:

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

Convert to eV: $E = \frac{3.98 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.49 \text{ eV}$

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

Solution:

Convert to eV: $E = \frac{3.98 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.49 \text{ eV}$

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: $\lambda = 400$ nm $= 4.00 \times 10^{-7}$ m
Work function: $\phi = 2.0$ eV $= 3.20 \times 10^{-19}$ J

Part (a): Maximum kinetic energy

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

Einstein's photoelectric equation: $KE_{max} = hf - \phi = 3.11 - 2.0 = 1.11 \text{ eV}$

In Joules: $KE_{max} = (1.11)(1.60 \times 10^{-19}) = 1.78 \times 10^{-19} \text{ J}$

Part (b): Stopping potential

$eV_s = KE_{max}$ $V_s = \frac{KE_{max}}{e} = 1.11 \text{ V}$

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: $\lambda = 400$ nm $= 4.00 \times 10^{-7}$ m
Work function: $\phi = 2.0$ eV $= 3.20 \times 10^{-19}$ J

Part (a): Maximum kinetic energy

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

Einstein's photoelectric equation: $KE_{max} = hf - \phi = 3.11 - 2.0 = 1.11 \text{ eV}$

In Joules: $KE_{max} = (1.11)(1.60 \times 10^{-19}) = 1.78 \times 10^{-19} \text{ J}$

Part (b): Stopping potential

$eV_s = KE_{max}$ $V_s = \frac{KE_{max}}{e} = 1.11 \text{ V}$

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: $n_i = 3$
Final state: $n_f = 2$
Hydrogen energy levels: $E_n = -13.6/n^2$ eV

Part (a): Photon energy

Energy levels: $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV}$ $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.40 \text{ eV}$

Photon energy (emission, so high → low): $E_{photon} = E_3 - E_2 = -1.51 - (-3.40) = 1.89 \text{ eV}$

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

Part (b): Wavelength

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

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

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: $n_i = 3$
Final state: $n_f = 2$
Hydrogen energy levels: $E_n = -13.6/n^2$ eV

Part (a): Photon energy

Energy levels: $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV}$ $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.40 \text{ eV}$

Photon energy (emission, so high → low): $E_{photon} = E_3 - E_2 = -1.51 - (-3.40) = 1.89 \text{ eV}$

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

Part (b): Wavelength

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

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.

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