The work function for a certain metal is 4.14 eV.
- What is the threshold frequency required to produce
photoelectrons from this metal? - To what wavelength does the frequency in Question 1
correspond? - Light with frequency 2 × 10^15 Hz is directed onto the metal
surface. Describe what would happen to the number of
photoelectrons and their maximum kinetic energy if the intensity of
this light were increased by a factor of 2.
Here’s How to Crack It
- We know from the statement of the question that for a photon to be
successful in liberating an electron from the surface of the metal, its energy
cannot be less than 4.14 eV. Therefore, the minimum frequency of the incident
light—the threshold frequency—must be - From the equation λf = c, where c is the speed of light, we find that
- Since the frequency of this light is higher than the threshold frequency,
photoelectrons will be produced. If the intensity (brightness) of this light is
then increased by a factor of 2, that means the metal surface will be hit with
twice as many photons per second, so twice as many photoelectrons will be
ejected per second. Thus, making the incident light brighter releases more
photoelectrons. However, their maximum kinetic energy will not change. The