Question

Learning Goal: To understand de Broglie waves and the calculation of wave properties. In 1924, Louis de Broglie postulated that particles such as electrons and protons might exhibit wavelike properties. His thinking was guided by the notion that light has both wave and particle characteristics, so he postulated that particles such as electrons and protons would obey the same wavelength-momentum relation as that obeyed by light: λ=h/p, where λ is the wavelength, p the momentum, and h Planck's constant. Part APart complete Find the de Broglie wavelength λ for an electron moving at a speed of 1.00×106m/s. (Note that this speed is low enough that the classical momentum formula p=mv is still valid.) Recall that the mass of an electron is me=9.11×10−31kg, and Planck's constant is h=6.626×10−34J⋅s. Express your answer in meters to three significant figures. λλ lambda = 7.270×10−10 m Previous Answers Correct Part BPart complete Find the de Broglie wavelength λ of a baseball pitched at a speed of 41.1 m/s . Assume that the mass of the baseball is 0.143kg. Express your answer in meters to three significant figures λλ lambda = 1.13×10−34 m Previous Answers Correct As a comparison, an atomic nucleus has a diameter of around 10−14m. Clearly, the wavelength of a moving baseball is too small for you to hope to see diffraction or interference effects during a baseball game. Part C Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00μm. The electrons then head toward an array of detectors a distance 1.075 m away. These detectors indicate a diffraction pattern, with a broad maximum of electron intensity (i.e., the number of electrons received in a certain area over a certain period of time) with minima of electron intensity on either side, spaced 0.498 cm from the center of the pattern. What is the wavelength λ of one of the electrons in this beam? Recall that the location of the first intensity minima in a single slit diffraction pattern for light is y=Lλ/a, where L is the distance to the screen (detector) and a is the width of the slit. The derivation of this formula was based entirely upon the wave nature of light, so by de Broglie's hypothesis it will also apply to the case of electron waves. Express your answer in meters to three significant figures.

Answer 1

Part A. my x = h = 6.626410-34 J-5 911X10-30Kg x 100x10°mis Ti = 727 X10-10m 6+626 X10-34 J-S 9.143kg X 41.imis [x = 1.13 010-34 m) Part B MV Parte a = 2000 um = 2:00x106m L = 1.075m y = 0.4980o = 4.988103 m 2000X10-6x4.98X10-3 1.075m T= 9.26X10-9m