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Nuclear and Quantum Physics - Electron Diffraction and Quantum Theory

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Wave-particle duality: Light and matter exhibit both wave-like and particle-like properties depending on the experiment performed.

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The Photoelectric Effect: The emission of electrons from a metal surface when light of a sufficient frequency ff shines on it, demonstrating the particle nature of light (photons).

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Photon Energy: The energy of a single photon is directly proportional to its frequency, given by E=hfE = hf, where hh is Planck's constant 6.63×10−34 J s6.63 \times 10^{-34} \text{ J s}.

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De Broglie Hypothesis: Any moving particle with momentum pp has an associated wavelength λ\lambda, confirming the wave nature of matter.

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Electron Diffraction: When a beam of electrons passes through a thin graphite film, they produce a circular interference pattern. This serves as experimental evidence for the wave nature of electrons, as diffraction is a wave phenomenon.

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Atomic Energy Levels: Electrons in atoms exist in discrete, quantized energy states. Transitions between these states involve the absorption or emission of photons with energy E=hf=∣Efinal−Einitial∣E = hf = |E_{final} - E_{initial}|.

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Heisenberg Uncertainty Principle: It is fundamentally impossible to simultaneously know the exact position Δx\Delta x and momentum Δp\Delta p of a particle with absolute precision, expressed as ΔxΔp≥h4π\Delta x \Delta p \geq \frac{h}{4\pi}.

📐Formulae

E=hf=hcλE = hf = \frac{hc}{\lambda}

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

Ek=eV=12mv2=p22mE_k = eV = \frac{1}{2}mv^2 = \frac{p^2}{2m}

λ=h2meV\lambda = \frac{h}{\sqrt{2meV}}

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

ΔxΔp≥h4π\Delta x \Delta p \geq \frac{h}{4\pi}

ΔEΔt≥h4π\Delta E \Delta t \geq \frac{h}{4\pi}

💡Examples

Problem 1:

An electron is accelerated from rest through a potential difference of V=100 VV = 100\text{ V}. Calculate its de Broglie wavelength λ\lambda.

Solution:

First, calculate the kinetic energy Ek=eV=1.60×10−19 C×100 V=1.60×10−17 JE_k = eV = 1.60 \times 10^{-19} \text{ C} \times 100 \text{ V} = 1.60 \times 10^{-17} \text{ J}. Next, use the relation between momentum and kinetic energy: p=2meEkp = \sqrt{2m_e E_k}. λ=hp=6.63×10−342×9.11×10−31×1.60×10−17\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-17}}}. λ≈1.23×10−10 m\lambda \approx 1.23 \times 10^{-10} \text{ m}.

Explanation:

The electrical potential energy converted into kinetic energy allows us to find the momentum of the electron, which is then used in the de Broglie equation to find the wavelength.

Problem 2:

An electron in a hydrogen atom drops from an energy level of −1.51 eV-1.51\text{ eV} to −3.40 eV-3.40\text{ eV}. Determine the wavelength of the emitted photon.

Solution:

The energy of the emitted photon is ΔE=∣−3.40−(−1.51)∣=1.89 eV\Delta E = |-3.40 - (-1.51)| = 1.89 \text{ eV}. Convert this to Joules: 1.89×1.60×10−19=3.024×10−19 J1.89 \times 1.60 \times 10^{-19} = 3.024 \times 10^{-19} \text{ J}. Using λ=hcΔE\lambda = \frac{hc}{\Delta E}, λ=6.63×10−34×3.00×1083.024×10−19≈6.58×10−7 m\lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.024 \times 10^{-19}} \approx 6.58 \times 10^{-7} \text{ m} (or 658 nm658 \text{ nm}).

Explanation:

The difference in the discrete energy levels of the atom dictates the specific energy (and thus wavelength) of the photon emitted during the transition.

Electron Diffraction and Quantum Theory Revision - Grade 11 Physics IB