1 answer

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

Question:

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 1.086 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 0.501 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.

in meters


Answers


alwidth of slith = 2.00 Mm = 2x10m, L = 1.086 mi - y = 0.501 cm 9242 | 2X 6 X 0:1X 246 1.086 mi = 0.92265 X lom 19 a 9.228109

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