Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6 de Broglie hypothesis Dual nature for de Broglie - de Broglie reasoned that if light can display this waveparticle duality, then matter might also display wave like properties under certain conditions. - de Broglie was able to put his idea into a quantitative scheme - Einstein had shown from relativity theory that the momentum of a photon is p h - de Broglie assumed that both light and matter obey the equation h p (1-32) This equation predicts of mass m moving with a velocity v will have de Broglie wave length given by h mv [1] Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6 Example 10: Calculate the de Broglie wavelength for a baseball 0.14 kg traveling at 40 m/s. Solution: The momentum of the baseball is P= m v=0.14x40=5.6 kg. m. s-1 The de Broglie wavelength is h 6.626 1034 1.2 1034 m p 5.6 - The de Broglie wavelength of the baseball is so small as to be completely undetectable. The reason for this is the large value of m. Example 11: Calculate the de Broglie wavelength of an electron traveling at 1% of the speed of light. The mass of an electron is 9.11 x 10-31kg. Solution: One percent of the speed of light is v= 0.01x3x108= 3x106m/s The momentum of the electron is given by P=m v= (9.11x10-31)(3x106)=2.73 x10-24 kg.m.s-1 [2] Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6 The de Broglie wavelength of this electron is h 6.626 1034 10 2.43 10 m 24 p 2.73 10 = 2.43Å Example12: Through what potential must an electron initially at rest fall in order that λ=10-10 m? Solution: The momentum associated with this wavelength is 6.626 1034 24 1 p 6.626 10 kg . m . s 1010 h The energy of an electron with this momentum is 24 2 6.626 10 2.411017 J 1 p E mv2 2 2m 2 9.111031 2.411017 150eV = 1.602 1019 ,and hence the voltage through which the electron must fall is 150 V. [3] Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6 Heisenberg uncertainty principle - The Heisenberg uncertainty principle states that it is not possible to specify both the position and the momentum of a particle simultaneously with infinite precision - This is means that if we located an electron within a distance Δ x, then this causes uncertainty in the momentum of the electron Δ p x p h (1-33) This equation called uncertainty principle for Heisenberg. Example 13: Calculate the uncertainty in the position of a baseball, where uncertainty in the momentum is 5.6x10-8 kg.m.s-1. Solution: 8 p 5.6 10 kg .m .s 1 The uncertainty in the position of the baseball is h 6.625 1034 26 x 1.2 10 m 8 p 5.6 10 [4] Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6 Example14: What is the uncertainty in momentum and in speed if we located an electron within an atom, Δx=50 x10-12m? Solution: h 6.626 1034 23 1 p 1.3 10 kg . m . s x 50 1012 Because p=mv and the mass of electron is 9.11x10-31kg, p 1.3 1023 7 1 v= 1.4 10 m . s m 9.111031 This is very large uncertainty in speed. [5]