(a) Light of wavelength 300 nm is incident on a metal that has a work function of 1.4 eV. What i

(a) Light of wavelength 300 nm is incident on a metal that...

Key Concepts

Kinetic Energy and Electron Speed

Once an electron absorbs a photon with energy exceeding the work function, the excess energy is converted into the kinetic energy of the emitted electron. The kinetic energy (K.E.) is given by K.E. = photon energy - work function, and it can be related to the electron's speed through the relation K.E. = (1/2)mv².

Effect of Light Intensity

Light intensity is related to the number of photons incident per unit time, not the energy of each individual photon. Doubling the light intensity increases the number of electrons that are ejected (assuming sufficient photon energy) but does not affect the maximum kinetic energy or speed of each individual electron, as these depend solely on the energy of the individual photons and the work function.

Work Function

The work function is the minimum energy required to eject an electron from the surface of a metal. It is a material-specific parameter that sets the threshold for the photoelectric effect. Only photons with energy exceeding the work function can result in the emission of electrons.

Photon Energy Calculation

Photon energy is calculated using the equation E = hc/?, where h is Planck's constant, c is the speed of light, and ? is the wavelength of the incident light. This calculation is key to determining whether the light has enough energy to overcome the material's work function.

Photoelectric Effect

The photoelectric effect is the phenomenon where electrons are emitted from a material, usually a metal, when light of sufficient energy shines on it. This effect demonstrates the particle nature of light and is explained by the quantized photon energy interacting with electrons in the material.

Transcript

00:01 For the first part of this question, we have a photon with wavelength of 300 nanometers incident on the surface of a metal.

00:09 We want to find out what the velocity of an ejected electron would be from the surface.

00:16 And we are given that the work function of the metal is 1 .4 ev.

00:22 So we need to use the fact that we know that the kinetic energy of the electron is going to be the energy left over from the photon after the energy.

00:31 Has been removed to get the electron out of the surface of the metal, which is what the work function tells us.

00:37 The work function is the minimum energy required to eject the electron from the surface of the metal.

00:42 So what we can do is put this into an expression and say that the kinetic energy of the electron is equal to the energy of the incident photon, take away the work function.

00:53 So the first thing we need to do is find the energy of this 300 nanometer photon.

00:58 We can do this by using the planck formula that says that e is equal to hc over lambda, where h is planck's constant and c is the speed of light.

01:09 So we can simply plug in what we've got and say that the energy is equal to 6 .63 times 10 to the minus 34, which is planck's constant, times 3 times 10 to the a, which is the speed of light, divided by the wavelength, which we must convert to metres, so multiply by 10 to the minus 9, so 300 times 10 to the minus 9.

01:47 And also divide this by 1 .6 times 10 to the minus 19 to make sure we get the answer in electron volts.

01:57 This is the charge on an electron, which gives us a value of 4 .14 electron volts for the energy of the incident photon.

02:11 So the kinetic energy of the ejected electron is going to equal, 4 .14 take away the work function.

02:22 So 4 .14 minus 1 .4 ev is equal to 2 .74 electron volts.

02:35 So we know that this 2 .74 electron volts is the kinetic energy of the electron, as it's the energy that's left over.

02:43 So what we need to do is convert this back into joules and then equate that to 1 .5 mb squared, which is the expression for kinetic energy.

02:51 So what we can do, is we can say that 2 .74 times 1 .6 times 10 to the minus 19 to convert back into joules is equal to 1 half mv squared, where this velocity is what we're trying to find out...

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