# Question Video: Finding the Frequency of Light in the Photoelectric Effect Physics • 9th Grade

Lead contained in a vacuum is illuminated with light from a laser, causing electrons to be emitted from the surface of the metal. Lead has a work function of 4.25 eV. The maximum kinetic energy of the electrons is 4.03 eV. What frequency of light does the laser emit? Use a value of 4.14 × 10⁻¹⁵ eV⋅s for the value of the Planck constant. Give your answer in scientific notation to two decimal places.

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### Video Transcript

Lead contained in a vacuum is illuminated with light from a laser, causing electrons to be emitted from the surface of the metal. Lead has a work function of 4.25 electron volts. The maximum kinetic energy of the electrons is 4.03 electron volts. What frequency of light does the laser emit? Use a value of 4.14 times 10 to the negative 15 electron volt seconds for the value of the Planck constant. Give your answer in scientific notation to two decimal places.

We’re told that there’s some lead contained within a vacuum. Let’s imagine that this here is the surface of that lead. And we’re told that it’s illuminated with light from a laser. Now, one of the important features of light produced by a laser is that that light is monochromatic, which means that it consists of just a single frequency. The question is asking us to work out what the value of this frequency is. For now, let’s label this unknown frequency as 𝑓.

We’re told that this light causes electrons to be emitted from the surface of the metal. So, on our sketch, let’s add in a couple of emitted electrons. When light that’s incident on a metal surface causes electrons to be emitted in this way, it’s known as the photoelectric effect. And the emitted electrons are often referred to as photoelectrons.

In order to understand the photoelectric effect, we need to recall that, as well as acting like a wave, light also acts like a particle. A particle of light is known as a photon, and we can picture the light from the laser as being like a stream of photons heading toward the metal surface. On this surface, there’s a whole load of electrons that these incident photons can hit. When one of the photons of the light collides with one of the electrons on the surface, then the photon can transfer its energy to the electron.

Now, the electrons have some amount of energy that binds them to the surface of the metal. And this energy is known as the work function of that metal. If the energy transferred from a photon to an electron exceeds the work function, that is, it exceeds the amount of energy binding the electron to the surface, then the electron has been given enough energy to leave the surface. And this is the basic principle of the photoelectric effect.

Since we’re told in the question that the light from the laser does cause electrons to be emitted from the surface of the metal, then we can assume that this condition is met. That is, the energy of a photon in the laser light, which it transfers to an electron in the metal, is at least equal to the work function of that metal. And in this case, that’s the work function of lead.

It’s important to keep in mind that this transfer of energy is a complete one. That is, the energy transferred from the photon to the electron must be equal to all of the energy that the photon had. Another thing that’s important to keep in mind is that, for laser light, which has a single frequency, all of the photons in that light will have the same amount of energy, because the energy of a photon of light, which we’ve labeled as 𝐸 subscript p, is equal to the frequency 𝑓 of that light multiplied by a constant ℎ, called the Planck constant.

Now, in this question, the frequency 𝑓 of the light is what we’re trying to find. So, if we can work out the energy of a photon, 𝐸 subscript p, then we can use this equation to calculate the frequency. Let’s clear ourselves some space and see how we can go about calculating the value of 𝐸 subscript p.

Now, we said earlier that if a photon transfers its energy to an electron, then it transfers all of its energy. So the energy transferred is equal to the photon energy, 𝐸 subscript p. And so we can write this statement here. We also know that there’s this thing called a work function that we’ll label as 𝑊. This is the energy that binds an electron to the surface of the metal and is equal to the minimum amount of energy that’s required to remove an electron from that surface.

We can also recall that energy is a quantity that is always conserved. That means that if the energy 𝐸 subscript p transferred to an electron is greater than the work function 𝑊 of the metal, then once the electron has used the energy 𝑊 to escape the surface, there’ll be an amount of energy left over equal to 𝐸 subscript p minus 𝑊.

Since energy is always conserved, then this leftover energy can’t just disappear. And actually this leftover energy becomes kinetic energy of the electron. So we have that 𝐸 subscript p minus 𝑊 is equal to the maximum kinetic energy of an emitted electron. If we take this equation and add the work function 𝑊 to both sides of it, then on the left-hand side of the equation, the 𝑊 and the minus 𝑊 cancel each other out. This leaves us with an equation that says 𝐸 subscript p is equal to 𝑘𝐸 max plus 𝑊.

We’re told in the question that the work function of lead is equal to 4.25 electron volts. So that’s our value for the quantity 𝑊. We’re also told that the emitted electrons have a maximum kinetic energy of 4.03 electron volts. So 4.03 electron volts is our value for the quantity 𝑘𝐸 max. We can now take these two values and substitute them into this equation to calculate the value of 𝐸 subscript p. We have that 𝐸 subscript p is equal to 4.03 electron volts plus 4.25 electron volts, which works out as 8.28 electron volts. This result is equal to the amount of energy transferred to an electron by a photon. And that’s equal to the amount of energy of a single photon of the light.

We can now use our result for 𝐸 subscript p in this equation in order to calculate the frequency 𝑓 of the light. To do this, we’re going to want to make 𝑓 the subject of the equation, which means dividing both sides of the equation by the Planck constant ℎ. On the right-hand side, the ℎ in the numerator cancels with the ℎ in the denominator. So we find that the frequency 𝑓 is equal to 𝐸 subscript p divided by ℎ.

We’ve worked out that the value of 𝐸 subscript p is 8.28 electron volts. And we’re told in the question to use a value for the Planck constant of 4.14 times 10 to the negative 15 electron volt seconds. Taking these two values and substituting them into this equation, we get this expression for the frequency 𝑓 of the light. The units of electron volts in the numerator and denominator cancel each other out. This leaves us with units of one over seconds, which we can recall is exactly the same as units of hertz.

Evaluating the expression gives a result of exactly 2.00 times 10 to the 15 hertz. As requested, we’ve given this answer to two decimal places and in scientific notation. So then our answer is that the light from the laser has a frequency of 2.00 times 10 to the 15 hertz.