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When light of a wavelength of 125 nm strikes a certain metal, electrons having a velocity of 8.2 x 10^5 m/s are
emitted from the surface of the metal. What is the threshold frequency of the metal?
The answer is 1.9 x 10^15 s^-1
I'm not sure how to do this. The equations that I tried to use were E=1/2mv^2, E=hf-hf0, E=hc/lamda but it didn't work

  • Chemistry -

    KE = (hc/w)-hf
    KE = 1/2(m)(v^2) so
    1/2(m)(V)^2 = (hc/wavelength) - hf.
    You have m in kg, v in m/s, h, c, and wavelength (w in meters). Solve for f.
    I get 1.9 x 10^15 Hz.

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