Consider a beam of electrons in a vacuum, passing through a very narrow slit of

Question

Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 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. L is the distance to the screen (detector) and y=L lambda /a, where 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 cm from the center of the pattern. What is the wavelength 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 0.496 2.00 ;mu{rm m}. The electrons then head toward an array of detectors a distance0.9080

Explanation / Answer

y = .496 cm (convert to meters) L = .9080 m wavelength = ? a = slit of the width (in meters)
wavelength = (y*a)/L

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