Homework 1B. 15

[April Meza 3K]

The velocity of an electron that is emitted from a
metallic surface by a photon is 3.6 3 103 km.s2^-1. (a) What is
the wavelength of the ejected electron? (b) No electrons are
emitted from the surface of the metal until the frequency of
the radiation reaches 2.50 3 1016 Hz. How much energy is
required to remove the electron from the metal surface?

(c) What is the wavelength of the radiation that caused photoejection of the electron? (d) What kind of electromagnetic radiation was used?

Can someone explain how I should approach this problem? Please and thanks!

[Chem_Mod]

a) You can use de Broglie's equation, where lambda = h / p = h / mv (v is velocity here).
b) This is essentially asking for the work function, but think back to week 1, where E = hv (v is frequency here).
c) This is the photoelectric effect, where the photon energy is used to eject the electron and give the electron some kinetic energy. Look at the photoelectric effect equations to determine the energy of the photon needed to "overcome" the work function (part b) and kinetic energy.
d) You'll have to use a reference chart for this.

If you are confused, please attend a UA session or office hours!

[Alexa Mugol 3I]

For a), use the DeBroglie wavelength equation (lambda/mv) using the velocity given. For b), you use E=h*(frequency), using the frequency given. Then you can use the threshold energy equation to find the work function. For c), use E_k=mv₂/2 to find the energy then use E=hc/lambda to find the wavelength. For d), just match the type of radiation to the wavelength you calculated.