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 2.00m. The electrons then head toward an array of detectors a distance 1.079 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 0.513 cm from the center of the pattern. What is the wavelength 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=L/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.

Explanation / Answer

Given as

   beam of electrons in a vacuum

   slit width a = 2.00 m, spacing between the central bright fringe to the dark next fringe, from the centre of the pattern is y = 0.513 cm

   L = 1.079 m

   from the formula y=L/a ==> = y*a/L
substituting the data
           = (0.513*10^-2*2*10^-6)/1.079 = 9.5088*10^-9 m

wavelength of the electron beam is 9.5088 nm

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