IB — physicsMedium3 min read

Atomic Structure and Energy Levels

Bohr model; energy levels; photon emission and absorption; line spectra; E = hf

Tutor AI Team
Tutor AI Team

# Atomic Structure and Energy Levels — IB Physics

1. Atomic Models

Rutherford: Dense positive nucleus, electrons orbit around it. Based on alpha scattering.

Bohr: Electrons in discrete energy levels (orbits). They can only transition between levels by absorbing or emitting specific amounts of energy.

2. Energy Levels

Electrons can only exist at discrete energy values. The ground state is the lowest level (n=1n = 1).

To move to a higher level: absorb a photon of exact energy ΔE\Delta E. To drop to a lower level: emit a photon of energy ΔE\Delta E.

ΔE=hf=hcλ\boxed{\Delta E = hf = \frac{hc}{\lambda}}

Where h=6.63×10−34h = 6.63 \times 10^{-34} J·s.

3. Line Spectra

Emission spectrum: Bright lines on dark background. Atoms are excited and emit specific wavelengths.

Absorption spectrum: Dark lines on continuous spectrum. Atoms absorb specific wavelengths.

Each element has a unique spectrum → used for identification.

4. The Photoelectric Effect

hf=ϕ+Ek,max⁡hf = \phi + E_{k,\max}

  • ϕ=hf0\phi = hf_0 = work function
  • Below threshold frequency: no electrons emitted
  • Above: max KE increases linearly with frequency
  • Intensity affects number, not energy of electrons

5. Wave-Particle Duality

Light: waves (interference, diffraction) and particles (photoelectric effect) Electrons: particles (deflection) and waves (electron diffraction)

de Broglie wavelength: λ=h/p=h/(mv)\lambda = h/p = h/(mv)

Worked Example: Example 1

Problem

An electron drops from −1.5 eV to −3.4 eV. Find the photon wavelength. ΔE=3.4−1.5=1.9\Delta E = 3.4 - 1.5 = 1.9 eV =1.9×1.6×10−19=3.04×10−19= 1.9 \times 1.6 \times 10^{-19} = 3.04 \times 10^{-19} J λ=hc/ΔE=6.63×10−34×3×108/3.04×10−19=6.54×10−7\lambda = hc/\Delta E = 6.63 \times 10^{-34} \times 3 \times 10^8/3.04 \times 10^{-19} = 6.54 \times 10^{-7} m = 654 nm (red)

Solution

Worked Example: de Broglie

Problem

Electron accelerated through 150 V. Find its de Broglie wavelength. KE=eV=1.6×10−19×150=2.4×10−17KE = eV = 1.6 \times 10^{-19} \times 150 = 2.4 \times 10^{-17} J p=2mKE=2×9.11×10−31×2.4×10−17=6.61×10−24p = \sqrt{2mKE} = \sqrt{2 \times 9.11 \times 10^{-31} \times 2.4 \times 10^{-17}} = 6.61 \times 10^{-24} kg m/s λ=h/p=6.63×10−34/6.61×10−24=1.0×10−10\lambda = h/p = 6.63 \times 10^{-34}/6.61 \times 10^{-24} = 1.0 \times 10^{-10} m = 0.1 nm

Solution

7. Practice Questions

    1. Explain why atomic spectra consist of discrete lines. (3 marks)
    1. A photon of wavelength 434 nm is emitted. Calculate its energy in eV. (2 marks)
    1. State two pieces of evidence for wave-particle duality. (2 marks)

    Answers

    1. Electrons exist at discrete energy levels. Transitions between levels emit/absorb photons of specific energy (E=hfE = hf). Specific energies → specific frequencies → discrete spectral lines.
    1. E=hc/λ=6.63×10−34×3×108/434×10−9=4.58×10−19E = hc/\lambda = 6.63 \times 10^{-34} \times 3 \times 10^8/434 \times 10^{-9} = 4.58 \times 10^{-19} J = 2.86 eV.

Want to check your answers and get step-by-step solutions?

Get it on Google PlayDownload on the App Store

Summary

  • Discrete energy levels → discrete spectra
  • ΔE=hf=hc/λ\Delta E = hf = hc/\lambda
  • Photoelectric: hf=ϕ+KEmax⁡hf = \phi + KE_{\max}
  • de Broglie: λ=h/(mv)\lambda = h/(mv)
Tags:
#ib#physics#nuclear#atomic-structure#energy-levels#photon#spectra

Related Topics

Ready to Ace Your IB physics?

Get instant step-by-step solutions to any problem. Snap a photo and learn with Tutor AI — your personal exam prep companion.

Get it on Google PlayDownload on the App Store