Cracking the SAT Physics Subject Test

The work function for a certain metal is 4.14 eV.

1. What is the threshold frequency required to produce
   photoelectrons from this metal?


2. To what wavelength does the frequency in Question 1
   correspond?


3. 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

1. 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


2. From the equation λf = c, where c is the speed of light, we find that


3. 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