(4) Electrons are no longer ejected from a sample of metal whenever it is illuminated by light with wavelength greater than 344nm. (a) Based on the table below, what is the most likely metal being illuminated? (b) If the sample is illuminated by light with wavelength 48.3nm, what is the stopping potential? (c) In the event this sample of metal were replaced by a metal with a smaller work function, but still illuminated with light with 48.3nm, would the kinetic energy of the ejected electrons increase or decrease?

A light wave of a particular frequency hits a metal.  The light wave is made of discrete photons, each with energy hf = hc/λ.  An electron will either completely absorb a photon, or not at all. An electron in a metal needs an energy to leave the metal -- the work function (W).  For the W given, calculate the maximum light wavelength (in nm) that will eject an electron.  I will use electron-volts (eV) for energy, because the computer gives me trouble with very low numbers. hc = 1240 eV nm W = 3.9 eV

In designing an experiment, you want a beam of photons and a beam of electrons with thesame wavelength of 0.281 nm, equal to the separation of the Na and Cl ions in a crystal ofNaCl. Find the energy of the photons and the kinetic energy of the electrons in electrovolts(eV).

In an x-ray tube, the maximum photon energy is given by hf = qV. Would it be technically more correct to say hf = qV + BE, where BE is the binding energy of electrons in the target anode? Why isn’t the energy stated the latter way?