Question 6

  In the photoelectric effect, suppose that the intensity of the light is increased, while the frequency is kept constant. The frequency is greater than the minimum frequency f_o. State whether each of the following will increase, decrease, or remain constant, and explain your choice: (a) the current in the phototube, (b) the number of electrons emitted per second from the metal surface, (c) the maximum kinetic energy that an electron could have, (d) the maximum momentum that an electron could have, and (e) the minimum de Broglie wavelength that an electron could have.

SOLUTION:
(a) According to the photoelectric effect, the intensity of light corresponds to the density of photons; if you have a more intense beam of light, you have more photons. Since the frequency is higher than the minimum frequency required to overcome the work function of the metal, more photons implies that more electrons will be ejected from the metal. Therefore, the current in the phototube will be higher.
(b) Current is defined as charge per unit time. Electrons are charged, and we said in part (a) that the current would increase, so the number of electrons emitted from the metal surface will increase as well.
(c) The maximum kinetic energy will not change because the energy of the incident photons is constant at E = hf and the work function (the energy taken away by the metal) is a constant depending on the particular metal.
(d) Momentum and kinetic energy, you will note, are basically the same thing since

$$KE = mv^2/2 = \frac{(mv)^2}{2m} = p^2/2m .$$

So we get the same answer as in part (c); namely, the maximum momentum will remain the same independent of the intensity.
(e) The definition of the de Broglie wavelength is

$$\lambda = \frac{h}{p}$$

so, since the maximum momentum p_max is constant, and h is Planck's constant, the minimum de Broglie wavelength is constant.