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 2.00 Mu m. The electrons then head toward an array of detectors or a distance 0.9940 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 period of time) with minima of electron intensity on either side, spaced 0.500 cm from the centre of the pattern. What is the wavelength Lambda of one of the electrons in the beam? Recall that the location of the first intensity minima in a single slit diffraction pattern for light is y= L Lambda/a, where L is the distance to the screen (detector) and a is 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

y = L/a

thus, the wavelength is,

= ay/L

= (2.00 * 10-6) * (0.500 * 10-2) / 0.994

=10.0 * 10-9 m

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