Question

UN 6 of 17 Constants Part Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00 pm. The electrons then head toward an array of detectors a distance 1.076 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 area over a certain period of time) with minima of electron intensity on either side, spaced 0.524 cm 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=LX/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. 10 AED ? & VE V X X XI X.10 1 = 1.00.108 m Submit Previous Answers Request Answer X Incorrect: Try Again: 4 attempts remaining 8:51 PM 7/30/2020

Answer 1

Given a = 2 об ит - 2 00 x 66m L 1.07 от ac - о. 524 см - Б. 5 x 10” ил is given by и first intensity minima Я - 15 6 х2 00 x 10 m 0.5 4х б?, ya 2 - J. 076 m = 9. 74 x 169 Hence wavelength a of the one of the electrons is 4.74 x 109 т.