Answer to Question #83189 in Quantum Mechanics for Philip Friday

Question

Calculate the debroglie wavelengths for the particles
A. Alpha particle accelerated 10mev
B. Electron is silicon crystal with average velocity corresponding to T=300

Answer 1

A. The de Broglie wavelength of a particle is given by the formula λ = h/p, where h is the Planck constant, and p is the relativistic momentum of a particle. The latter is expressed through the particle energy E and mass m as p=√(E^2-m^2 c^4 )⁄c, where c is the speed of light. Hence, the de Broglie wavelength is given by λ=h c⁄√(E^2-m^2 c^4 ). The alpha-particle mass is approximately m = 3.7 GeV/c2 = 3.7 × 109 eV/c2. The value of the Planck constant in the relevant units is h = 4.1 × 10–15 eV s, and the speed of light is c = 299792458 m/s. Substituting these numerical values, and also the value of the alpha-particle energy E = 10 MeV = 107 eV, into the last formula for the de Broglie wavelength, we obtain λ=h c⁄√(E^2-m^2 c^4 )≈3.3×〖10〗^(-16) m. Answer: 3.3 × 10–16 m.

B. At temperatures T = 300 K, electrons with high precision can be regarded as non-relativistic particles because the characteristic energy kBT = 2.3 × 10–2 eV at this temperature is much smaller than the electron’s rest energy mc2 = 0.5 × 106 eV. Here, kB = 8.6 × 10–5 eV/K is the Boltzmann constant, and m = 0.5 × 106 eV/c2 is the electron mass. The thermal average velocity is then given by v=√(k_B T⁄m), and the de Broglie wavelength is λ=h⁄p=h⁄mv=h⁄√(k_B Tm), where h = 4.1 × 10–15 eV s is the Planck constant. Substituting here all numerical values, we obtain λ = 1.1 × 10–8 m. Answer: 1.1 × 10–8 m.