{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

Chem 3502/5502 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 × 10 19 J 9.109 × 10

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}