Physics 6C Ch28worksheet Solutions

Physics 6C Ch28worksheet Solutions - PHYSICS 6C Ch.28...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
PHYSICS 6C Ch.28 Worksheet Solutions clas.ucsb.edu/staff/vince/ 1) A flashlight emits 1.5W of power. Assuming a frequency of 5.2x10 14 Hz for the light, determine the number of photons given off per second. The energy of each photon is given by E=hf, or E=(6.63x10 -34 J-s)(5.2x10 14 Hz)=3.45x10 -19 J Dividing 1.5W by this energy gives 4.4x10 18 photons per second. 2) The work function of gold is 4.58 eV. What frequency of light must be used to eject electrons from a gold surface with a maximum kinetic energy of 6.48x10 -19 J? Is this light visible to the human eye? Converting to Joules, (4.58eV)(1.6x10 -19 J/eV)=7.33x10 -19 J This is the energy necessary to just barely eject the electrons from gold. We need to add the extra kinetic energy to get the energy of the photon. E photon =1.38x10 -18 J. The frequency is given by E=hf. f photon =2.1x10 15 Hz This light is in the ultraviolet range of the spectrum and cannot be seen by humans. 3)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/09/2011 for the course PHYSICS 6c taught by Professor Staff during the Spring '11 term at UCSB.

Page1 / 2

Physics 6C Ch28worksheet Solutions - PHYSICS 6C Ch.28...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online