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 . The electrons then head toward an array of detectors a distance 1.001 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 0.528 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 , where is the distance to the screen (detector) and 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.

Explanation / Answer

given that Condition for the minimum intensity in the single slit   diffraction pattern is                   dsin   = m sin   = m /d   ....................(1) Here    d = width of the slit m = 1   first   order = wave length of light when    is small        = tan    = sin   Tan    =   y / D .............(2) From (1) and (2)   m /d   =  y / D                 = yd / D                     here   y   = 0.528*10^-2  m    ,   d   = 2.0*10^-6 m     , D   = 1.001 m here you did not give slit width but I took d as 2C plug all values we get                     = 1.05 * 10-8 m                     or 10.5 nm Condition for the minimum intensity in the single slit   diffraction pattern is                   dsin   = m sin   = m /d   ....................(1) Here    d = width of the slit m = 1   first   order = wave length of light when    is small        = tan    = sin   Tan    =   y / D .............(2) From (1) and (2)   m /d   =  y / D                 = yd / D                     here   y   = 0.528*10^-2  m    ,   d   = 2.0*10^-6 m     , D   = 1.001 m here you did not give slit width but I took d as 2C plug all values we get                     = 1.05 * 10-8 m                     or 10.5 nm Condition for the minimum intensity in the single slit   diffraction pattern is                   dsin   = m sin   = m /d   ....................(1) Here    d = width of the slit m = 1   first   order = wave length of light when    is small        = tan    = sin   Tan    =   y / D .............(2) From (1) and (2)   m /d   =  y / D                 = yd / D                     here   y   = 0.528*10^-2  m    ,   d   = 2.0*10^-6 m     , D   = 1.001 m here you did not give slit width but I took d as 2C plug all values we get                     = 1.05 * 10-8 m                     or 10.5 nm

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