1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: What is the work function for silver?

  1. Mar 30, 2008 #1
    The wavelength associated with the cut-off frequency, for silver is 325 nm.

    a) What is the work function for silver?
    b) Find the maximum kinetic energy and speed of the electrons ejected from a silver surface by ultraviolet light of wavelength 254 nm.

    So far for part a) i have:

    f = c / 325 = 9.25 * 10 ^ 14

    After that im not to sure if im suppose to use this formula

    9.25 * 10^14 = hc/325e - work function
  2. jcsd
  3. Mar 31, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

    The equation describing the photoelectric effect is:

    [tex]hf=E_{max}+\phi [/tex] where [tex]\phi[/tex] is the work function.

    Now, what is E_max for the case of the cutoff frequency?
  4. May 14, 2008 #3
    im having the same problem where im supposed to find its max. kinetic energy of the photoelectron in electron volts(eV) and joules(J), but im completely stuck.

    if i use hf = Emax + work, i have planks constant (h) and frequency, but there are two variables left to find...
  5. May 16, 2008 #4
    Go through the link that G01 has given. Jot down the data given to you and make the appropriate substitution in the photoelectric equation and obtain the KE. [itex]h, m_e[/itex] are known constants.
  6. May 17, 2008 #5
    i dont see anything in that link... for my problem i have the wavelength, cutoff potential and the frequency, im only confused on the equations to reach to the maximun Ek of the ejected electrons in electron volts(eV) and joules(J).
  7. May 18, 2008 #6
    Well then how is the cut-off potential related to the kinetic energy? can you determine it from the given quantities?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook