Phys 274 – c ± Eric B. Szarmes 4 February 2011 Daily Homework 8 SOLUTIONS Q4S.5 a) Helium atoms with a deBroglie wavelength of 0.103 nm have a momentum of p ( = mv ) = h λ = 6 . 626(10-34 ) J · s0 . 103(10-9 ) m = 6 . 43(10-24 ) kg · m/s . (1) Since the mass of a helium atom is m = 4 . 003 atomic mass units = 6.646(10-27 ) kg, the speed of an individual helium atom is approximately v = p/m = 967 m/s . b) In the two-slit experiment described in the text, the separation between the slits was d = 9 μ m = 9000 nm (1 μ m width plus 8 μ m separation). For a distance to the screen of D = 0 . 64 m, the theoretical separation between adjacent interference maxima (using the small angle approximation, eq. Q2.6 on p. 31 of the text) is then s ≈ λ d D = 7 . 3 μ m , (2) in good agreement with the measured separation of 7.7 ² 1.0 μ m quoted in the text. Q4S.6 a) If the water droplets of mass m = 10-18 kg have a velocity of v = 1 . 0 mm/s, then their momentum is p = mv = 10-21 kg · m/s, and their deBroglie wavelength is
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This note was uploaded on 03/07/2011 for the course PHYS 274 taught by Professor Ericszarmes during the Spring '11 term at University of Hawaii, Manoa.