# Question Video: Finding the Maximum Kinetic Energy of Emitted Photoelectrons Physics • 9th Grade

A polished metal surface in a vacuum is illuminated with light from a laser, causing electrons to be emitted from the surface of the metal. The light has a frequency of 2.00 × 10¹⁵ Hz. The work function of the metal is 1.40 eV. What is the maximum kinetic energy that the electrons can have? Use a value of 4.14 × 10⁻¹⁵ eV⋅s for the Planck constant. Give your answer in electron volts.

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### Video Transcript

A polished metal surface in a vacuum is illuminated with light from a laser, causing electrons to be emitted from the surface of the metal. The light has a frequency of 2.00 times 10 to the 15 hertz. The work function of the metal is 1.40 electron volts. What is the maximum kinetic energy that the electrons can have? Use a value of 4.14 times 10 to the negative 15 electron volt seconds for the Planck constant. Give your answer in electron volts.

Let’s suppose that this here is the polished metal surface from the question. And we’re told that this surface is illuminated with light from a laser. One of the important features of light that comes from a laser is that that light is monochromatic, which means that it has just a single frequency. In this case, we’re told that that frequency is 2.00 times 10 to the 15 hertz. Let’s label this frequency of the laser light as 𝑓. The other thing that we’re told is that the work function of this metal is 1.40 electron volts. Let’s call this work function capital 𝑊.

We can recall that the work function of a metal is the energy that binds an electron to the surface of that metal. We’re told in the question that the light from the laser causes electrons to be emitted from the surface of the metal. So the situation we’ve got here is that this light from the laser is incident on this polished metal surface. And that’s causing these electrons to be emitted from that surface. Now, we know that because of the work function of the metal, the electrons are bound to its surface. Because the work function is the energy which binds an electron to the surface, then this is equal to the energy that’s required to remove an electron from that surface.

So what’s happening here is that the light from the laser is transferring energy to the electrons on the surface of the metal. Since we know that this causes the electrons to be emitted from the surface, then we know that the energy transferred must be at least equal to the work function of the metal. It turns out that we can actually calculate the energy of this light because we’ve got all of the information that we need to do this.

We can recall that, as well as thinking of light as being like a wave, we can also think of it as being made up of particles known as photons. In this particle-based model, if we have light with a frequency of 𝑓, then the energy of a single photon, 𝐸 subscript 𝑝, is equal to 𝑓 multiplied by a constant called Planck constant, which is typically denoted as ℎ. We’re told in the question to use a value of 4.14 times 10 to the negative 15 electron volt seconds for the Planck constant. So we’ve got a value for the quantity ℎ in this equation.

As a quick aside, we may perhaps be more used to seeing the Planck constant expressed in units of joule-seconds. In these SI units, the Planck constant is approximately equal to 6.63 times 10 to the negative 34 joule-seconds. However, we were given the work function of the metal in units of electron volts. And we’re also asked to give our answer to this question in units of electron volts. So, as it turns out, having the Planck constant ℎ given to us in units of electron volt seconds is going to be helpful.

As well as having a value for the Planck constant ℎ, we also know the frequency 𝑓 of the light. This means that we can take our values for ℎ and 𝑓 and substitute them into this equation to calculate the energy of a photon of this light. When we do this, we find that the energy 𝐸 subscript 𝑝 is equal to 4.14 times 10 to the negative 15 electron volt seconds, that’s the value for the Planck constant, multiplied by 2.00 times 10 to the 15 hertz, that’s the frequency 𝑓.

Now, in terms of the units, we have electron volt seconds and we have hertz. We’ll find it useful to recall that units of hertz are equal to the inverse of units of seconds. So, in place of the units of hertz in this expression, we can instead write units of one over seconds. Or we could say units of per second. The units of seconds from the first term and the units of per second from the second term then cancel each other out. That means that we’re left with energy units of electron volts, which are a suitable choice of unit to express the energy 𝐸 subscript 𝑝 of a single photon.

Before evaluating this expression, we can see that we can make our lives a little bit easier. We have a 10 to the power of negative 15 from the first term, which multiplies by a 10 to the power of 15 from the second term. 10 to the power of negative 15 multiplied by 10 to the power of 15 is simply equal to one. And so multiplying by both of these two terms is just the same as multiplying by one, which we know will do nothing to our result. So we can say that the 10 to the negative 15 and the 10 to the 15 cancel each other out. Then, tidying up what’s left, we have that the energy of a photon is equal to 4.14 times 2.00 electron volts. This works out as an energy of 8.28 electron volts.

We can notice that this result for the energy of a photon is greater than the work function of the metal. This is a reassuring result because it means that if a photon transfers its energy to an electron on the surface of the metal, then that electron has enough energy to overcome this work function. In other words, a photon from the laser light can transfer enough energy to an electron for that electron to be emitted from the metal. Such an electron is called a photoelectron, since it was ejected from the metal surface due to energy transferred by a photon.

So far then, we’ve confirmed that this incident light from the laser will indeed cause electrons to be emitted from the surface of the metal. What we’re asked to work out is the maximum kinetic energy that these emitted electrons can have. Let’s clear ourselves some space on the board so that we can see how to go about working this out.

We know that each individual photon of the light has an energy of 8.28 electron volts. And we know that the energy it takes to remove one electron from the surface of the metal is 1.40 electron volts. An incident photon is able to transfer all of its energy to an electron on the surface of the metal. In this case, the electron then gains an amount of energy equal to the photon energy.

We can recall that energy is a quantity that is always conserved. This means that if an electron gains an amount of energy equal to the energy 𝐸 subscript 𝑝 of a photon and it has to use a minimum amount of energy equal to the work function 𝑊 in order to leave the metal’s surface, then so long as 𝑊 is smaller than 𝐸 subscript 𝑝, there’s going to be some leftover energy, which we know can’t just disappear because energy is always conserved. This leftover energy becomes kinetic energy of the emitted electron. And we have that 𝐸 subscript 𝑝 minus 𝑊 is equal to the maximum value of this kinetic energy.

Alternatively, we could write this equation in words to say that the energy transferred to an electron by an incident photon minus the minimum amount of energy required for that electron to leave the surface of the metal is equal to the maximum kinetic energy of the electron once it’s been emitted.

This maximum kinetic energy, or the value of 𝐾𝐸 subscript max, is exactly what we’re asked to find in this question. Since we know the values of both of the quantities on the left-hand side of this equation, that’s 𝐸 subscript 𝑝 and 𝑊, we can go ahead and substitute those values into the equation to calculate this maximum kinetic energy. When we do this, we find that the maximum kinetic energy of an emitted electron is equal to 8.28 electron volts, that’s our value of the photon energy 𝐸 subscript 𝑝, minus 1.40 electron volts, the work function 𝑊 of the metal.

Evaluating this expression gives a result of 6.88 electron volts. This value is then our answer for the maximum amount of kinetic energy that an emitted electron can have.