HWkey1 - Chem 3502/5502 Physical Chemistry II (Quantum...

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Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Answers to Homework Set 1 To be collected on September 17 2010 From lecture 1: What is the maximum number of electrons that can be emitted if a potassium surface of work function 2.45 eV absorbs 3.20 x 10 –3 J of radiation at a wavelength of 250 nm? What is the kinetic energy and velocity of the emitted electrons? Frequency ν is related to wavelength λ by the equation c = λν where c is the speed of light (2.998 x 10 8 m sec –1 ). Thus, the frequency of the incident light is The energy of photons having this frequency is computed from the equation E = h ν where h is Planck’s constant (6.626 x 10 –34 J s). Carrying out the relevant multiplication gives E = 7.946 x 10 –19 J. Using the conversion factor 1 J = 6.24 x 10 18 eV gives an energy of 4.95 eV. If the work function is 2.45 eV, this leaves 2.50 eV (4.021 x 10 –19 J) of energy that is converted into electron kinetic energy. To determine the velocity of the electrons, we use v = 2 T m e = 2 × 4.021
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This note was uploaded on 12/18/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

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HWkey1 - Chem 3502/5502 Physical Chemistry II (Quantum...

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