In a double-slit experiment with two slits of width 0.1µm spaced 4µm apart, supp

Question

In a double-slit experiment with two slits of width 0.1µm spaced 4µm apart, suppose a beam of electrons is incident on the slits after being accelerated from rest through a potential difference of 100V.
a) Since wavelength= h/p, What is the wavelength of the electrons in this problem?
b) If the pattern of detected electrons is observed on a fluorescent screen 20m from the slits, what is the width of the central diffraction maximum?
c) How many interference fringes will be observed within the central diffraction maximum?

Explanation / Answer

Given width of two slits is   d = 0.1 m distance between two slits is 4 m    potential difference     V = 100 V 1)     energy of incident electron                     E = q V        here q is charge of electron    ( 1.6 *10 -19 C )                 E = ( 1.6 *10 - 19 C ) ( 100 V )                      =100 eV                       = 1.6*10-17 J             wave length of electron is                E = h c /                      = h c /E                         = ( 6.635* 10 ^ -34 Js ) ( 3 * 10 ^8 m /s ) / ( 1.6 *10 ^ -17 J )                          =  12.440625 n m 2)    distance between slit and screen is   L = 20 m                 d sin = m                        sin    = m / d                               = 12.44 *10 ^ -9 m / 4 *10 ^ -6 m                         =       0.17819 ^ o          width of the central diffraction maximum                     y = L tan                        = 20 m tan ( 0.17819 )                         =   0.0622 m 3)       d sin = m               sin 1                m = d /                     =  4 *10 ^ -6 m   / 12.44 *10 ^ -9 m                       = 321

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