Question

Matter Waves - Double Slit Interference Previously we learned the wave behaviors of light - the single-slit diffraction, and the double-slit interference. In this problem, you will explore the wave behaviors of matter (particles). Double Slit Screen Interference y=Ltane 0 L bright fringes: sine Dark fringes above central sin 8 = (m m-0, 11, 12(constructive) (destructive). Matter Waves - Double Slit electrons Max Men Max photons Electrons pass through a double slit separated by d = 0.0100 um, 1um = 10-6, m. The First dark fringe above the central bright fringe is formed on the screen at an angle of 14.18° relative to the incident light. (a) Find the wavelength of the electrons in nm, 1nm = 109 m. Keep 3 decimal place. Enter a number nm Submit (5 attempts remaining) The Planck constant is 6.626x10^-34 Js. (b) Calculate the electrons' momentum. Write the result in terms of 10-24 kg • m/s . Keep 3 decimal places. Enter a number x10-24kg m/s Submit (5 attempts remaining) (c) The mass of an electron is 9.11x10^-31 kg. Use the classical physics formula for kinetic energy to calculate the kinetic energy of the electrons. Write the result in terms of 10-19 J. Keep 4 decimal places. Enter a number *10-19 Submit (5 attempts remaining)

Answer 1

Date : Page No 0.01 -Дm 0 14.18 = for dark fring = "8915 elitu Louise . na a = THI 2 a a 10.0pt sin (14.1897 und ele la) -g 10633|810 mu (b) 6.626 X 10 kemst p = 1:633X le -24 50.406 x 10 b kg mil KE = I mv²= p² am (0.406x70247 2(9:118,103) L KE to 0.904 X16 J