The photoelectric effect is very convenient to be the first topic of quantum mechanics because it is perhaps the first solid proof of the validity of QM .It is an effect discovered by Einstein which begs a radically new explanation: energy, and in this case light energy, travels in packages called quanta (quantum singular). In the case of light this packets are called photons.

Plank had already proposed the quantum theory and his formula gives the relationship between frequency of light and the energy of the energy “packet”: the higher the frequency, the more energy , so the larger the quantum.

E= hf

Where he introduced a new constant of nature: planck's constant  h. 

Photo=light , electric- electrons: When light shines on a metal, electrons are ejected. The interesting point is that the photoelectric will depend on the frequency of the light , and not only on its intensity. That means that blue light may pull lots of electrons from a metal surface, while red light pulls none, or very few. The energy needed to remove the electron is called work function (Φ). If the photon carries more energy than needed, the excess is transferred into kinetic energy:

E = h f − Φ

Photons with energy 1.1 × 10^-18 J are incident on a third metal surface. The maximum energy of electrons emitted from the surface of the
metal is 5.1 × 10^-19 J.
Calculate, in eV, the work function of the metal.

According to the formula above we justneed to subtract one from the other.

How intensity depends on frequency

The intensity (I) of the light beam is :

I = Nhf

where N is the number of photons per second per square meter. Remember that the intensity is the power divided by the area.

So how quantum theory explains that the number of ejected electrons depends on frequency of light?

Metal is a special kind of chemical bond where there is a “sea” of electrons (this is studied in chemistry). Atoms are bonded not by means of covalent or ionic bonds, but by loose electrons that move in between the nuclei and shield the electric repulsion between them and allows for bonding. 

When light hits one of these loose electrons they can get enough energy to jump out of the metal: this is the photoelectric effect! The energy needed for this jump depends on the metal and it is called a work function.

So, how is it possible that blue light removes electrons where red light doesn't? The quantum of red light (smaller than the blue) doesn´t have enough energy to p]provide a large enough  jump, and the electrons remain attached to the metal. If the quantum energy is higher than the work function, the excess energy will be transferred into kinetic energy. 

If the photons that are hitting the metal have energies higher than the work function the more photons hitting the surface ,the more photoelectrons are emitted. So in this case, the higher the intensity, the more photoelectrons.

In the case where the photons don´t have enough energy to remove electrons, increasing the intensity of the light beam will not help. You will have more photons, but still none of them capable of removing an electron. If you increase the frequency you may be able to remove electrons, even with low intensity.

IB physics catch question

When they ask if keeping the intensity of light the same but changing the wavelength will change anything. There is a formula, which is not in the IB formula booklet, that I find very useful to analyse this type of question. It was explained above and it is :

I = Nhf

Question: If you increase  the frequency of the light beam  and  keep the intensity the same, what changes?

The detail here is that the rate of emission of photons has to decrease. If your total output of energy does not change but the size of the packages increases, you need less packages.

Photoelectrons are emitted at a certain rate when monochromatic light is incident on a metal surface. Light of the same intensity but of higher frequency is now used. After this change, the rate of emission of electrons from the surface is
A. zero.
B. lower.
C. the same.
D. higher.

This is a catch question because when you use light of a higher frequency (as as a result higher energy photons) you may think that it will cause a higher emission. But the answer is B, lower, because of the argument exposed above.