A single slit has a width of 5x10 -5 m. Electrons with kinetic energy E pass thr

Question

A single slit has a width of 5x10-5m. Electrons with kinetic energy E pass through the slit and produce an interference pattern on a distant screen. If the first minimum on one side of the central diffraction maximum is at an angle of 0.080 rad (measured from the line from the slit to the center of the central maximum), what is the energy E of the electrons?

***Please provide ALL concepts/formulas used***I have no idea how to setup this problem to where I could get the energy of electrons as an answer....the correct answer is 1.51x10-26 J (= 9.42x10-8 eV)***

Explanation / Answer

    Diffraction due to single slit               mimimum condition   is    dsin = m              For first mimima    m =1                = 0.08 rad (   1800 / )    = 4.590       d   = width of the slit   = 5*10-5 m                 solving for    wave length      =     dsin    / 1                             =   ( 5*10-5 m ) sin( 4.590  )   = 0.4 *10-5 m                debroglie wave length    of the electron      =  12.3 *10-8   / sqrt (E)                        energy of the electron beam    E   =   (   12.3 *10-8 m / 0.4 *10-5 m  )2 eV                              E   = 9.45*10-8 eV                      E   =   (  9.45*10-8 eV  )   ( 1.6*10-19 J   /eV)   = 1.51*10-26 J

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