Math Problem Statement

A particle of mass 0.5 MeV/c 2 has kinetic energy 100 eV. Find its de -Broglie wavelength and group velocity, where c is the velocity of light.

Solution

To find the de Broglie wavelength and group velocity of a particle given its mass and kinetic energy, we'll follow these steps:

Given Data:

  • Mass (mm) = 0.5 MeV/c²
  • Kinetic Energy (KEKE) = 100 eV
  • Speed of light (cc) = 3×1083 \times 10^8 m/s

1. Convert mass from MeV/c² to kg:

The conversion factor is 1 MeV/c2=1.783×1030 kg1 \text{ MeV/c}^2 = 1.783 \times 10^{-30} \text{ kg}. m=0.5 MeV/c2×1.783×1030 kg/MeV/c2=0.8915×1030 kgm = 0.5 \text{ MeV/c}^2 \times 1.783 \times 10^{-30} \text{ kg/MeV/c}^2 = 0.8915 \times 10^{-30} \text{ kg}

2. Convert kinetic energy from eV to joules:

The conversion factor is 1 eV=1.602×1019 J1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}. KE=100 eV×1.602×1019 J/eV=1.602×1017 JKE = 100 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 1.602 \times 10^{-17} \text{ J}

3. Find the particle’s momentum pp:

The non-relativistic kinetic energy formula KE=p22mKE = \frac{p^2}{2m} can be rearranged to find pp: p=2mKEp = \sqrt{2m \cdot KE} p=2×0.8915×1030 kg×1.602×1017 Jp = \sqrt{2 \times 0.8915 \times 10^{-30} \text{ kg} \times 1.602 \times 10^{-17} \text{ J}} p=2×0.8915×1.602×1047p = \sqrt{2 \times 0.8915 \times 1.602 \times 10^{-47}} p=2.854×1047p = \sqrt{2.854 \times 10^{-47}} p1.689×1023 kgm/sp \approx 1.689 \times 10^{-23} \text{ kg} \cdot \text{m/s}

4. Calculate the de Broglie wavelength λ\lambda:

The de Broglie wavelength is given by: λ=hp\lambda = \frac{h}{p} where hh is Planck’s constant (6.626×1034 Js6.626 \times 10^{-34} \text{ Js}). λ=6.626×10341.689×1023\lambda = \frac{6.626 \times 10^{-34}}{1.689 \times 10^{-23}} λ3.922×1011 m\lambda \approx 3.922 \times 10^{-11} \text{ m} λ0.0392 nm\lambda \approx 0.0392 \text{ nm}

5. Calculate the group velocity vgv_g:

The group velocity vgv_g for a non-relativistic particle can be approximated by the classical velocity vv: v=2KEmv = \sqrt{\frac{2 \cdot KE}{m}} v=2×1.602×10170.8915×1030v = \sqrt{\frac{2 \times 1.602 \times 10^{-17}}{0.8915 \times 10^{-30}}} v=3.594×1013v = \sqrt{3.594 \times 10^{13}} v1.895×106 m/sv \approx 1.895 \times 10^6 \text{ m/s}

Summary:

  • de Broglie wavelength: λ0.0392 nm\lambda \approx 0.0392 \text{ nm}
  • Group velocity: vg1.895×106 m/sv_g \approx 1.895 \times 10^6 \text{ m/s}

Do you want details or have any questions?

Here are some related questions:

  1. How is the de Broglie wavelength affected by the particle's mass?
  2. What would happen to the de Broglie wavelength if the kinetic energy were doubled?
  3. How does the de Broglie wavelength relate to the particle's wave-particle duality?
  4. What is the significance of the group velocity in wave mechanics?
  5. How would the group velocity change if the mass of the particle were halved?
  6. What is the difference between phase velocity and group velocity?
  7. How is the concept of de Broglie wavelength used in electron microscopy?
  8. Can you explain the concept of wave-particle duality with another example?

Tip: When dealing with particle physics, always ensure the units are correctly converted to maintain consistency in calculations.

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Math Problem Analysis

Mathematical Concepts

Quantum Mechanics
De Broglie Wavelength
Group Velocity

Formulas

de Broglie wavelength formula: λ = h / p
Non-relativistic kinetic energy: KE = p^2 / (2m)
Group velocity approximation: vg ≈ √(2KE/m)

Theorems

de Broglie hypothesis

Suitable Grade Level

Advanced Physics