Question

Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00?m. The electrons then head toward an array of detectors a distance 1.080m 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 0.508cm 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.

Express your answer in meters to three significant figures.

Answer 1

according to debroglie's equation,

= h / p                                { eq. 1 }

where, h = planck's constant

p = momentum

the first intensity minima in a single slit diffraction pattern for light is given as :

y_min = L / a                     { eq. 2 }

where, L = distance to the screen (detector) = 1.08 m

a = width of the slit = 0.508 cm = 0.508 x 10^-2 m

y_min = first intensity minima = 2 m = 2 x 10^-6 m

wavelength of one of the electrons in this beam is given as :

= y_min a / L                                { eq. 3 }

inserting the values in eq.3

= (2 x 10^-6 m) (0.508 x 10^-2 m) / (1.08 m )

= 1.016 x 10^-8 / 1.08 m

= 9.407 x 10^-9 m

or = 9.41 x 10^-9 m                     { Answer in three significant figure }