This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: b. 650 km/s c. 420 km/s d. 1480 km/s Solution b K max = hf = hc/  = 1240/2005=1.2 eV = 1.92 x 1019 J=> Remember electron mass = 9.11 x 1031 kg v max = 2 K max m = 2 " 1.92 " 10 # 19 9.11 " 10 # 31 = 650 km / s 3. The light intensity on a metallic surface with a work function of 3 eV produces photoelectrons with a maximum kinetic energy of 2 eV. The frequency of light is doubled. Determine the maximum kinetic energy of photoelectrons in eV. a. 3 b. 7 c. 2 d. 4 Solution: b K max = hf => hf = 5 eV If f is doubled K max = 2hf = 10eV 3 eV = 7 eV. 4. A neutron has de Broglie wavelength is 1.4 x 1010 m. What is its kinetic energy in eV? The neutron mass is 939.573 MeV/c 2 = 1.67 x 1027 kg. a. 4 b. 0.4 c. 0.04 d. 40 Solution c. " = h p # K = p 2 2 m = h 2 2 m 2 = ( hc ) 2 2 mc 2 2 = 1240 2 2 $ 939.573 $ 10 6 $ 0.14 2 = 0.04 eV...
View
Full
Document
This note was uploaded on 04/29/2008 for the course PHYS 208 taught by Professor Rzchowski during the Spring '08 term at Wisconsin.
 Spring '08
 Rzchowski
 Physics, Power, Photon, Light

Click to edit the document details