# What equation shows the relationship between gap energy and wavelength

### Absorption and emission (video) | Khan Academy

electromagnetic radiation with shorter wavelengths is more energetic. The relation- ship between energy and frequency is given by the equation, E = hν, where h. This equation can also be solved with wavelength instead of frequency, since wavelength Qualitatively, this shows that the energy of the band gap is inversely. To relate this to the radiated energy averaged over some period of time, you will have to By the simple relationship between wavelength and frequency ν, The Parseval theorem, to which I have referred, is relevant in that it shows that the How can I calculate the band gap from UV-Vis absorption spectra of thin films.

When it does that, it's going to emit a photon. It's going to emit light. When the electron drops from a higher energy level to a lower energy level, it emits light.

This is the process of emission. I could represent that photon here. This is how you usually see it in textbooks. We emit a photon, which is going to have a certain wavelength. Lambda is the symbol for wavelength. We need to figure out how to relate lambda to those different energy levels. The energy of the photon is, the energy of the emitted photon is equal to the difference in energy between those two energy levels. We have energy with the third energy level and the first energy level.

The difference between those So, the energy of the third energy level minus the energy of the first energy level. That's equal to the energy of the photon. This is equal to the energy of that photon here.

We know the energy of a photon is equal to h nu. Let me go ahead and write that over here. Energy of a photon is equal to h nu. H is Planck's constant, this is Planck's constant.

Nu is the frequency. We want to think about wavelength. We need to relate the frequency to the wavelength. The equation that does that is of course, C is equal to lambda nu. So, C is the speed of light, lambda is the wavelength, and nu is the frequency. So, if we solve for the frequency, the frequency would be equal to the speed of light divided by lambda.

Then, we're going to take all of that and plug this in to here. We get the energy of a photon is equal to Planck's constant, h, I'll write that in here, times the frequency, and the frequency is equal to c over lambda.

## Absorption and emission

Now we have the energy of the photon is equal to hc over lambda. Instead of using E3 and E1, let's think about a generic high energy level. Let's call this Ej now, so this is just a higher energy level, Ej. The electron falls back down to a lower energy level, which we'll call Ei.

Instead of using E3 and E1, let's make it more generic, let's do Ej and Ei. Let's go ahead and plug that in now. The energy of the photon would be equal to the higher energy level, Ej minus the lower energy which is Ei. So now we have this equation, let me go and highlight it here. We have hc over lambda is equal to Ej minus Ei. Let's get some more room, and let's see if we can solve that a little bit better here. So let me write this down here. We have hc over lambda is equal to the energy of the higher energy level, minus the energy of the lower energy level, like that.

Alright, so in an earlier video, I showed you how you can calculate the energy at any energy level. We derived this equation. The energy at any energy level, n, is equal to E1 divided by n squared.

## Energy of Photon

So, if we wanted to know the energy when n is equal to j, that would be just E1 over j squared. We could take that and we could plug it in to here. Alright, if I wanted to know the energy for the lower energy level, that was Ei, and that's equal to E1 divided by i squared. I could take all of this, I could take this and I could plug it into here. Let's, once again, get some more room.

Let's write what we have so far. We have hc over lambda is equal to Ej was E1 over j squared and Ei was E1 over i squared.

### HubbleSite - Reference Desk - FAQs

Okay, we could pull out an E1 on the right. Electrons and holes in semiconductors Pure undoped semiconductors can conduct electricity when electrons are promoted, either by heat or light, from the valence band to the conduction band.

The hole, which is the absence of an electron in a bonding orbital, is also a mobile charge carrier, but with a positive charge. The motion of holes in the lattice can be pictured as analogous to the movement of an empty seat in a crowded theater. An empty seat in the middle of a row can move to the end of the row to accommodate a person arriving late to the movie if everyone moves over by one seat.

Because the movement of the hole is in the opposite direction of electron movement, it acts as a positive charge carrier in an electric field. The opposite process of excitation, which creates an electron-hole pair, is their recombination. When a conduction band electron drops down to recombine with a valence band hole, both are annihilated and energy is released. This release of energy is responsible for the emission of light in LEDs.

At equilibrium, the creation and annihilation of electron-hole pairs proceed at equal rates. We can write a mass action expression: Note the similarity to the equation for water autodissociation: This is about 12 orders of magnitude lower than the valence electron density of Al, the element just to the left of Si in the periodic table. Thus we expect the conductivity of pure semiconductors to be many orders of magnitude lower than those of metals. This kind of plot, which resembles an Arrhenius plot, is shown at the right for three different undoped semiconductors.

Very small amounts of dopants in the parts-per-million range dramatically affect the conductivity of semiconductors. For this reason, very pure semiconductor materials that are carefully doped - both in terms of the concentration and spatial distribution of impurity atoms - are needed. This is exactly the right number of electrons to completely fill the valence band of the semiconductor. Introducing a phosphorus atom into the lattice the yellow atom in the figure at the right adds an extra electron, because P has five valence electrons and only needs four to make bonds to its neighbors.

The extra electron, at low temperature, is bound to the phosphorus atom in a hydrogen-like molecular orbital that is much larger than the 3s orbital of an isolated P atom because of the high dielectric constant of the semiconductor. The energy needed to ionize this electron - to allow it to move freely in the lattice - is only about meV, which is not much larger the thermal energy 26 meV at room temperature.

Therefore the Fermi level lies just below the conduction band edge, and a large fraction of these extra electrons are promoted to the conduction band at room temperature, leaving behind fixed positive charges on the P atom sites.

### Semiconductors: band gaps, colors, conductivity and doping - Chemistry LibreTexts

Boron has only three valence electrons, and "borrows" one from the Si lattice, creating a positively charged hole that exists in a large hydrogen-like orbital around the B atom. This hole can become delocalized by promoting an electron from the valence band to fill the localized hole state. Again, this process requires only meV, and so at room temperature a large fraction of the holes introduced by boron doping exist in delocalized valence band states.

These substitutions introduce extra electrons or holes, respectively, which are easily ionized by thermal energy to become free carriers. The Fermi level of a doped semiconductor is a few tens of mV below the conduction band n-type or above the valence band p-type.