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Nuclear and Quantum Physics - Quantum Physics

Grade 12IBPhysics

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

🔑Concepts

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The Photoelectric Effect: Light behaves as discrete packets of energy called photons. Each photon has energy E=hfE = hf. When a photon hits a metal surface, it can eject an electron if its energy is greater than the work function Φ\Phi.

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Threshold Frequency (f0f_0): The minimum frequency of incident radiation required to liberate an electron from a metal surface. At this frequency, hf0=Φhf_0 = \Phi.

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Stopping Potential (VsV_s): The potential difference required to stop the most energetic photoelectrons from reaching the anode, defined by Emax=eVsE_{max} = eV_s.

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Wave-Particle Duality: Particles like electrons exhibit wave-like properties (diffraction/interference), and waves like light exhibit particle-like properties (photoelectric effect).

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De Broglie Wavelength: Every moving particle has an associated wavelength λ\lambda inversely proportional to its momentum pp.

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Atomic Energy Levels: Electrons in an atom exist in discrete, quantized energy states. Photons are emitted or absorbed when electrons transition between levels: ΔE=hf\Delta E = hf.

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Heisenberg's Uncertainty Principle: It is fundamentally impossible to know both the exact position Δx\Delta x and momentum Δp\Delta p of a particle simultaneously: ΔxΔp≥h4π\Delta x \Delta p \ge \frac{h}{4\pi}.

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The Wave Function (ψ\psi): The state of a quantum particle is described by ψ\psi. The value of ∣ψ∣2|\psi|^2 at a point represents the probability density of finding the particle at that location.

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Quantum Tunneling: A phenomenon where a particle has a non-zero probability of passing through a potential barrier even if its total energy is less than the barrier height.

📐Formulae

E=hf=hcλE = hf = \frac{hc}{\lambda}

hf=Φ+Emaxhf = \Phi + E_{max}

Emax=eVsE_{max} = eV_s

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

ΔE=∣Efinal−Einitial∣\Delta E = |E_{final} - E_{initial}|

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

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

Ek=p22mE_k = \frac{p^2}{2m}

💡Examples

Problem 1:

Calculate the maximum kinetic energy of photoelectrons ejected from a calcium surface (work function Φ=2.90 eV\Phi = 2.90 \text{ eV}) when illuminated by light of frequency 1.20×1015 Hz1.20 \times 10^{15} \text{ Hz}.

Solution:

  1. Convert energy of the photon to eV: E=hf=(6.63×10−34 J s)×(1.20×1015 Hz)=7.956×10−19 JE = hf = (6.63 \times 10^{-34} \text{ J s}) \times (1.20 \times 10^{15} \text{ Hz}) = 7.956 \times 10^{-19} \text{ J}.
  2. E(eV)=7.956×10−191.60×10−19=4.97 eVE(\text{eV}) = \frac{7.956 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.97 \text{ eV}.
  3. Use Einstein's equation: Emax=hf−Φ=4.97 eV−2.90 eV=2.07 eVE_{max} = hf - \Phi = 4.97 \text{ eV} - 2.90 \text{ eV} = 2.07 \text{ eV}.

Explanation:

Einstein's photoelectric equation relates the incident photon energy to the work function and the maximum kinetic energy of the emitted electrons.

Problem 2:

An electron is accelerated from rest through a potential difference of 150 V150 \text{ V}. Determine its de Broglie wavelength.

Solution:

  1. Kinetic energy Ek=qV=1.60×10−19×150=2.40×10−17 JE_k = qV = 1.60 \times 10^{-19} \times 150 = 2.40 \times 10^{-17} \text{ J}.
  2. Momentum p=2mEk=2×(9.11×10−31)×(2.40×10−17)≈6.61×10−24 kg m s−1p = \sqrt{2mE_k} = \sqrt{2 \times (9.11 \times 10^{-31}) \times (2.40 \times 10^{-17})} \approx 6.61 \times 10^{-24} \text{ kg m s}^{-1}.
  3. λ=hp=6.63×10−346.61×10−24≈1.00×10−10 m\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} \approx 1.00 \times 10^{-10} \text{ m}.

Explanation:

The wavelength is calculated by first finding the kinetic energy from the accelerating voltage, then the momentum, and finally applying the de Broglie relation.

Problem 3:

The position of an electron is measured with an uncertainty of 1.0×10−10 m1.0 \times 10^{-10} \text{ m}. Calculate the minimum uncertainty in its momentum.

Solution:

  1. Use Heisenberg's Uncertainty Principle: Δp≥h4πΔx\Delta p \ge \frac{h}{4\pi \Delta x}.
  2. Δp≥6.63×10−344π×1.0×10−10\Delta p \ge \frac{6.63 \times 10^{-34}}{4 \pi \times 1.0 \times 10^{-10}}.
  3. Δp≥5.28×10−25 kg m s−1\Delta p \ge 5.28 \times 10^{-25} \text{ kg m s}^{-1}.

Explanation:

This demonstrates the fundamental limit on measurement precision at the quantum scale.

Quantum Physics - Revision Notes & Key Formulas | IB Grade 12 Physics