Photo-Electricity The Liberation of Electrons by Light With Chapters on F

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Norah Ali Al moneef 7. Norah Ali Al moneef 7 8. When light that is above the threshold frequency strikes the photoemissive material, electrons are ejected and current flows. Light of low frequency does not cause current flow … at all. As with line spectra, the photoelectric effect cannot be explained by classical physics.

Norah Ali Al moneef 8 9. Norah Ali Al moneef 9 The Photoelectric Effect Explained The electrons in a photoemissive material need a certain minimum energy to be ejected. Short wavelength high frequency, high energy photons have enough energy per photon to eject an electron.

Norah Ali Al moneef 10 Norah Ali Al moneef 11 Norah Ali Al moneef 12 Norah Ali Al moneef 13 Photons easily explain the results of both photoelectric effect experiments. This minimum energy is called the work function w of the material. Norah Ali Al moneef 14 Photon: A packet or bundle of energy is called a photon. Properties of photons: i A photon travels at a speed of light c in vacuum.

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Norah Ali Al moneef 15 Norah Ali Al moneef 16 Norah Ali Al moneef 17 Norah Ali Al moneef 18 Norah Ali Al moneef 19 Norah Ali Al moneef 20 Norah Ali Al moneef 21 Example The work function of silver is 4. EM radiation with a frequency of 1. What is the speed of the electrons that are emitted? Norah Ali Al moneef 22 II When nm light falls on a metal, the current through a photoelectric circuit is brought to zero at a stopping voltage of 1.

What is the work function of the metal T. Norah Ali Al moneef 23 What is threshold frequency of a material with a work function of 10eV? Norah Ali Al moneef 24 What is the maximum wavelength of electromagnetic radiation which can eject electrons from a metal having a work function of 3 eV?

We just need to convert that to wavelength. Norah Ali Al moneef 25 Norah Ali Al moneef 26 In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than nm. At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so we have The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy T. Norah Ali Al moneef 27 What is the energy associated with a light quantum of wavelength 5. Norah Ali Al moneef 28 If a monochromatic light beam with quantum energy value of 3.

Origins of Quantum Theory

Norah Ali Al moneef 29 Norah Ali Al moneef 30 The wavelength of the radiation is 1. What is the energy of one photon of this microwave radiation? Norah Ali Al moneef 31 A photon at nm will kick out an electron with an amount of kinetic energy, KE If the wavelength is halved to nm and the photon hits an electron in the metal with same energy as the previous electron, the energy of the electron coming out is a. A voltage diff of 1. The number of electrons ejected therefore depends upon the number of photons, i.

Some of the energy in the packet is used to overcome the binding energy of the electron in the metal. What is the velocity of this photoelectron? From which region of the electromagnetic spectrum is this photon? The fact that the same quantization constant could be derived from two very different experimental observations was very impressive and made the concept of energy quantization for both matter and light credible. Notice that in the above process we have obtained something very interesting.

We do not know this just yet, because it is different for every different material, and we cannot get a general theory of that now. Formulation of a general theory of the properties of different substances—their natural frequencies, and so on—is possible only with quantum atomic mechanics. Also, different materials have different properties and different indexes, so we cannot expect, anyway, to get a general formula for the index which will apply to all substances. However, we shall discuss the formula we have obtained, in various possible circumstances.

First of all, for most ordinary gases for instance, for air, most colorless gases, hydrogen, helium, and so on the natural frequencies of the electron oscillators correspond to ultraviolet light. Then we find that the index is nearly constant.

Work function formula class 11

So for a gas, the index is nearly constant. This is also true for most other transparent substances, like glass. The index is higher for blue light than for red light. That is the reason why a prism bends the light more in the blue than in the red. The equation for the index of refraction as a function of frequency is called a dispersion equation. So we have obtained a dispersion equation. Our dispersion equation suggests other interesting effects. This would occur, for example, if we take a material like glass, say, and shine x-ray radiation on it.

In fact, since many materials which are opaque to visible light, like graphite for instance, are transparent to x-rays, we can also talk about the index of refraction of carbon for x-rays. All the natural frequencies of the carbon atoms would be much lower than the frequency we are using in the x-rays, since x-ray radiation has a very high frequency.

A similar situation would occur if we beam radiowaves or light on a gas of free electrons.

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In the upper atmosphere electrons are liberated from their atoms by ultraviolet light from the sun and they sit up there as free electrons. Can that be correct? It is correct. This just means that the phase shift which is produced by the scattered light can be either positive or negative. It can be shown, however, that the speed at which you can send a signal is not determined by the index at one frequency, but depends on what the index is at many frequencies.

What the index tells us is the speed at which the nodes or crests of the wave travel.

The node of a wave is not a signal by itself. In a perfect wave, which has no modulations of any kind, i. In order to send a signal you have to change the wave somehow, make a notch in it, make it a little bit fatter or thinner. That means that you have to have more than one frequency in the wave, and it can be shown that the speed at which signals travel is not dependent upon the index alone, but upon the way that the index changes with the frequency.

Then we will calculate for you the actual speed of signals through such a piece of glass, and you will see that it will not be faster than the speed of light, although the nodes, which are mathematical points, do travel faster than the speed of light. Just to give a slight hint as to how that happens, you will note that the real difficulty has to do with the fact that the responses of the charges are opposite to the field, i. The formula says that when the electric field is pulling in one direction, the charge is moving in the opposite direction. How does the charge happen to be going in the opposite direction? It certainly does not start off in the opposite direction when the field is first turned on. When the motion first starts there is a transient, which settles down after awhile, and only then is the phase of the oscillation of the charge opposite to the driving field.

And it is then that the phase of the transmitted field can appear to be advanced with respect to the source wave.

Emission Mechanism

In Fig. You will see from the diagram that the signal i. Let us now look again at our dispersion equation. We should remark that our analysis of the refractive index gives a result that is somewhat simpler than you would actually find in nature. To be completely accurate we must add some refinements. First, we should expect that our model of the atomic oscillator should have some damping force otherwise once started it would oscillate forever, and we do not expect that to happen.

We have worked out before Eq. We need a second modification to take into account the fact that there are several resonant frequencies for a particular kind of atom. It is easy to fix up our dispersion equation by imagining that there are several different kinds of oscillators, but that each oscillator acts separately, and so we simply add the contributions of all the oscillators. Perhaps you have noticed something a little strange about the last form Eq. What does that mean? We can see what such a complex index means when there is only one resonant frequency by going back to Eq.

Also, the exponent is negative, so the factor is a real number less than one. As the wave goes through the material, it is weakened.