A LEVEL 04年 物理 mc q32

Assume all light energy converts to electrical energy wihtout any loss, the answer is option C.

Intensity of light = nhf
where n is the number of photons per unit time incident onto the photocell
h is Planck's constant
f is the frequency of light used

For blue light, I = (n1)h.(f1)
For yellow light, I = (n2).h.(f2)
Because the two colour lights are of the same intensity,
(n1).h.(f1) = (n2).h.(f2)
or n2/n1 = f1/f2
but f2 < f1
thus, n2 > n1
That is, the no. of photons in the yellow light is larger than that in the blue light
Assume one photon gives off one photo-electron, the no. of electrons produced by yellow light is more than that produced by blue light. In other words, the photo-current produced by yellow light is larger than that produced by blue light.

According to the equation E=hv (where E is the energy of a photon, h is the Planck constant and v is the frequency of the corresponding light), the energy of a photon of yellow light is smaller than that of blue light.

When yellow light is incident on the surface of the cathode, electrons are emitted. The maximum kinetic energy of electrons emitted is given by the equation KE=hv-W (where W is the work function of the metal that makes up the cathode). Hence, the smaller the energy possessed by a photon, the smaller the maximum kinetic energy possessed by a photoelectron emitted. Therefore, option a is wrong.

On the other hand, as the intensity of the light used is kept constant, the number of photons in the yellow light will then be larger than that of the blue light. This results in more photoelectrons emitted from the cathode and thus a larger current flowing in the circuit. Hence, option c is the correct answer.