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: Lambda = h/p, where Lambda is the wavelength, p the momentum, and h Planck\'s constant. Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00 mu m. The electrons then head toward an array of detectors a distance 1.013 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 0.519 cm from the center of the pattern. What is the wavelength Lambda 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 = Llambda/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. Solution lambda = h / p xmin = L lambda / a lambda = a * xmin / L = (2 * 10-6 * 0.519 * 10-2) / 1.013 = 1.02 * 10-8 m.