(a) Light of wavelength 300 nm is incident upon a metal that has a work function of 1.4 eV. Wha

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

Key Concepts

Effect of Light Intensity

Light intensity refers to the number of photons arriving per unit time on the metal surface. While increasing intensity raises the number of emitted electrons (i.e., the current), it does not affect the energy of individual photons or the maximum kinetic energy of the emitted electrons. The kinetic energy depends solely on the photon energy and the work function, not on the number of photons.

Kinetic Energy of Emitted Electrons

The maximum kinetic energy of the emitted electrons is given by the difference between the energy of the incident photon and the work function of the metal (KE_max = photon energy - work function). This concept uses the conservation of energy principle, where any excess energy from the photon is converted into the kinetic energy of the escaping electron.

Photoelectric Effect

The photoelectric effect is the phenomenon where electrons are emitted from a metal surface when light of sufficient frequency shines on it. The effect demonstrates the quantized nature of light, revealing that light consists of photons, each carrying a discrete amount of energy that can knock electrons out of the material if the energy exceeds the metal's work function.

Photon Energy Calculation

Photon energy is determined by the equation E = hf (or E = hc/?), where h is Planck's constant, c is the speed of light, and ? is the wavelength of the incident light. This formula is fundamental in understanding how changes in the wavelength or frequency of light affect the energy available to liberate electrons in the photoelectric effect.

Work Function

The work function is the minimum amount of energy required to liberate an electron from the surface of a metal. It acts as an energy barrier, so only photons with energies exceeding this threshold can emit electrons. The work function is critical in calculating the kinetic energy of the emitted electrons after accounting for this energy loss.

Transcript

00:01 So we're shining a light of a wavelength of 300 nanometers onto a metal that has a work function of 1 .4 ev.

00:09 And we want to know what's the maximum velocity, maximum speed that the electron that's ejected can have.

00:16 This is related to finding maximum kinetic energy.

00:19 There's just an extra step here.

00:20 So we know that the maximum kinetic energy is going to be the energy of the photon that's shining on there minus the work function of the metal.

00:32 Because that leftover energy can be the kinetic energy that the electron has.

00:37 So we just need to get the frequency associated with 300 nanometers.

00:40 We get that by speed of light divided by wavelength.

00:45 I use 3 times 10 to the 8 meters per second for speed of light.

00:49 It's just easier.

00:54 And we will get 1 times 10 to the 15 hertz for this frequency.

01:02 Multiple that by our plank constant.

01:04 We're going to use ev per hertz because we're given ev for the work function.

01:10 So we're going to get hf is 4 .136 ev.

01:19 We're subtracting the work function 1 .4 from that because that's what's left over.

01:30 So we have a maximum kinetic energy of 2 .736 ev.

01:42 Now how do we get the speed from this? well, let's use the electron mass in the units of mev per c squared.

01:51 You may be familiar with this.

01:54 This is common in high energy and particle physics.

01:59 It's basically related to the rest mass and the rest energy of the electron.

02:03 It's convenient that it's in mev per c squared because we can use our kinetic energy equation and we can still keep everything in ev.

02:14 What we'll end up getting is a ratio v squared to c squared, or we can just multiply what we get by c squared.

02:19 You'll see what i'm talking about.

02:21 So we can say 2 .736 equals 1⁄2 times 0 .511 times 10 to the 6th ev v squared over c squared.

02:39 Because that c squared is implied in the units of the electron mass there.

02:45 All right, so we'll just solve for this algebraically, right? so at two points, you can look at it two ways...

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