Question Video: Finding the Current Produced by Photoelectrons | Nagwa Question Video: Finding the Current Produced by Photoelectrons | Nagwa

# Question Video: Finding the Current Produced by Photoelectrons Physics • Third Year of Secondary School

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A 20.0 mW laser that produces 250 nm light is used to illuminate a block of silver. This causes electrons to be ejected from the surface of the silver. If all of the photons produced by the laser each eject one electron from the silver, what is the total current of the photoelectrons? Silver has a work function of 4.26 eV. Use a value of 4.14 × 10⁻¹⁵ eV⋅s for the Planck constant and a value of 1.6 × 10⁻¹⁹ C for the charge of one electron. Give your answer to two decimal places.

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### Video Transcript

A 20.0-milliwatt laser that produces 250-nanometer light is used to illuminate a block of silver. This causes electrons to be ejected from the surface of the silver. If all of the photons produced by the laser each eject one electron from the silver, what is the total current of the photoelectrons? Silver has a work function of 4.26 electron volts. Use a value of 4.14 times 10 to the negative 15 electron volt seconds for the Planck constant and a value of 1.6 times 10 to the negative 19 columbs for the charge of one electron. Give your answer to two decimal places.

So in this question, we’ve got light from a laser that’s illuminating a block of silver. We’re being asked to find the total current of the photoelectrons. So that’s the electrons that are rejected from the silver as a result of this incident laser light. To do this, we’ll need to clear ourselves some space on the board. But as we clear this space, let’s make a note of the important information that we’re given.

We’re told that the laser has a power, which we’ve labeled as capital 𝑃, of 20.0 milliwatts. We’re also told that it produces light with a wavelength of 250 nanometers, which we’ve labeled as 𝜆. Now one of the important features of light produced by a laser is that that light is monochromatic. That means that it consists of just a single wavelength. So in the case of the laser from this question, all of the light has this wavelength of 250 nanometers. This laser light is being used to illuminate a block of silver.

And we’re told that the work function of silver, which we’ve labeled as capital 𝑊, is equal to 4.26 electron volts. We can recall that the work function of a metal is the amount of energy which binds electrons to the surface of that metal. The final bits of information that we’re given that we should use a value of 4.14 times 10 to the negative 15 electron volt seconds for the Planck constant ℎ and a value of 1.6 times 10 to the negative 19 columbs for the charge of one electron, which we’ve labeled as 𝑞 subscript e.

So what’s happening here is that we’ve got a block of silver, and then there’s this light from a laser shining at the surface of this block. Now on the surface of the silver, there’s all of these electrons. And they’re bound to the surface by the work function 𝑊. We’re told in the question though that this light from the laser causes these electrons to be ejected from the surface.

To understand how this happens, we need to recall that, as well as thinking of light as being like a wave, we can also think of it as acting like a particle. In this particle model of light, the light is made up of these particles known as photons. So we can think of this laser light as being like a stream of photons heading toward the surface of the silver. Each one of these photons has an energy 𝐸 subscript p equal to the Planck constant ℎ multiplied by the frequency 𝑓 of the light. The frequency and wavelength of light are related by this equation, which says that the speed of light 𝑐 is equal to the frequency multiplied by the wavelength.

If we divide both sides by 𝜆, then on the right-hand side, the 𝜆’s cancel out. And so we see that frequency 𝑓 is equal to the speed of light 𝑐 divided by the wavelength 𝜆. We can use this equation to replace the 𝑓 in the expression for the photon energy by 𝑐 divided by 𝜆. So then we have that the photon energy 𝐸 subscript p is equal to ℎ multiplied by 𝑐 divided by the wavelength 𝜆.

Since laser light is monochromatic, containing just a single wavelength 𝜆, this means that all of the photons will have the same value of energy 𝐸 subscript p. When a photon from the light collides with an electron on the surface of the metal, the photon can transfer all of its energy 𝐸 subscript p to the electron. If this energy transferred is greater than or equal to the energy which binds the electron to the surface, then the electron has enough energy to leave the surface.

Let’s recall that the energy binding the electron to the surface is the work function of the metal. So that’s this value 𝑊. That means that the incident photons of laser light will cause electrons to be ejected from the surface of the metal, provided that the photon energy 𝐸 subscript p is greater than or equal to the work function 𝑊.

Let’s now use this equation to calculate the value of 𝐸 subscript p for the photons from this question so that we can check whether it does indeed exceed the work function 𝑊. On the right-hand side of the equation, we’re told the value to use for the Planck constant ℎ. And we know the wavelength 𝜆 of the light. We can also recall that 𝑐, the speed of light in vacuum, is equal to 3.0 times 10 to the eight meters per second. We need the units on the right-hand side to agree with each other, and we’ve got a speed of light in meters per second but a wavelength of light in nanometers. So before putting our values into the equation, let’s convert our wavelength 𝜆 from nanometers into meters.

We can recall that one nanometer is equal to 10 to the negative nine meters. So to convert from nanometers to meters, we need to multiply the value by 10 to the negative nine. So then in units of meters, we have that 𝜆 is equal to 250 times 10 to the negative nine meters. We can also write that as 2.5 times 10 to the negative seven meters.

If we now take our values for 𝜆, ℎ, and 𝑐 and substitute them into this equation, we get that the photon energy 𝐸 subscript p is equal to 4.14 times 10 to the negative 15 electron volt seconds multiplied by 3.0 times 10 to the eight meters per second divided by 2.5 times 10 to the negative seven meters. In terms of the units, we’ve got meters in the numerator, which cancels with the meters in the denominator. And then we’ve also got seconds and per second in the numerator. So these to cancel out as well. That leaves us with units of electron volts.

Evaluating the expression, we get a result for the photon energy of exactly 4.968 electron volts. Comparing this photon energy against the work function 𝑊 of the metal, which is equal to 4.26 electron volts, we find that 𝐸 subscript p is indeed greater than or equal to 𝑊, which means that electrons will be ejected from the surface.

These ejected electrons are known as photoelectrons. And we’re being asked in this question to find the total current of them. Let’s clear some space on the board and think about how we can work this out.

We can recall that the electron is a charged particle. And specifically, we’re told to use a value for its charge of 1.6 times 10 to the negative 19 columbs. We can also recall that current is defined as the rate of flow of charge over time. If a total amount of charge 𝑄 passes a given point during a time of 𝑡, then the current 𝐼 through that point is equal to 𝑄 divided by 𝑡. The total charge of the ejected photoelectrons must be equal to the number of ejected electrons multiplied by the charge of each individual electron.

We can also think about what’s happening per second of time, and then the total charge per second must equal the number of electrons ejected per second multiplied by the charge of each electron. Since charge per unit of time is simply equal to the current, then this total charge per second is just the current of the photoelectrons. And this current is exactly what we’re asked to find.

We already know the charge of each electron. So in order to find the current, we just need to work out the number of electrons that are rejected per second. We were told in the question that every photon causes one electron to be ejected from the surface of the metal. That means that the number of electrons ejected per second is equal to the number of photons incident on the surface per second. And if we assume that the laser is directed at the silver blocks so that all of the photons hit the surface, then that’s equal to the number of photons produced by the laser per second.

Now we were told the power of the laser, and we’ve already worked out the energy of each photon. We can recall that power is defined as the rate of energy transfer. We can express this as power 𝑃 is equal to energy 𝐸 divided by time 𝑡. In the case of the laser light, 𝑃 is just the power of the laser. So that’s our value of 20 milliwatts. On the right-hand side, 𝐸 over 𝑡 is the total energy of the photons produced per unit of time. This total energy must be equal to the number of photons produced multiplied by the energy 𝐸 subscript p of each photon.

So if we label the total number of photons as 𝑁, then we have that 𝑃 is equal to 𝑁 multiplied by 𝐸 subscript p divided by 𝑡. We can also write this right-hand side as 𝑁 over 𝑡 multiplied by 𝐸 subscript p. Then this equation is saying that the power 𝑃 of the laser is equal to 𝑁 over 𝑡, which is the number of photons produced per second of time multiplied by 𝐸 subscript p, the energy of each photon. Notice that we just said this quantity here, 𝑁 divided by 𝑡, is equal to the number of photons produced by the laser per second. And we’ve already seen that this is equal to the number of electrons ejected from the metal per second, which is what we need in order to work out this current.

In this equation here then, we can replace this number of electrons ejected per second by this quantity 𝑁 divided by 𝑡. And while we’re at it, let’s rewrite this charge of one electron using the symbol 𝑞 subscript e. So now we’ve got this expression for the current of the photoelectrons, where we know the value of 𝑞 subscript e. And 𝑁 over 𝑡 comes from this equation here. If we take this equation and divide both sides by the photon energy 𝐸 subscript p, then on the right-hand side, the two 𝐸 subscript p’s cancel each other out. Then we can write this equation the other way around to say that 𝑁 over 𝑡 is equal to 𝑃 divided by 𝐸 subscript p.

Before we put our values of 𝑃 and 𝐸 subscript p into this equation, we need to make sure that the units make sense. In order to get a quantity with units of per second, we need a power with units of watts and we need a photon energy with units of joules. At the moment though, the power is in milliwatts, and the energy is an electron volts. We can recall that one milliwatt is equal to 10 to the negative three watts. So to convert from milliwatts to watts, we multiply by 10 to the negative three. So the power 𝑃 of the laser becomes 20.0 times 10 to the negative three watts. And we can also write that as 2.00 times 10 to the negative two watts.

Then to convert our photon energy into joules, we need to recall that one electron volt is equal to 1.6 times 10 to the negative 19 joules. So we need to take this value and multiply it by 1.6 times 10 to the negative 19. This gives a result for the photon energy in joules of 7.9488 times 10 to the negative 19 joules. Substituting this value along with our value for the power 𝑃 into this equation, we get this expression for 𝑁 over 𝑡, the number of photons produced per second of time. This works out as 2.516 etcetera times 10 to the 16 per second.

So now in this equation here, we know both the values of 𝑞 subscript e and 𝑁 over 𝑡. So let’s clear ourselves some space so we can substitute these values in. Putting our values of 𝑁 over 𝑡 and 𝑞 subscript e into this equation, we get this expression for the total current of the photoelectrons. The overall units here are columbs per second, which is equivalent to units of amperes. Then evaluating the expression gives a result of 4.026 etcetera times 10 to the negative three amperes.

Since one milliampere is equal to 10 to the negative three amperes, then we can rewrite this current by replacing the 10 to the negative three amperes by units of milliamperes. The last thing to do is to recall that we were asked to give our answer to two decimal places. Rounding our result then gives us our final answer that the total current of the photoelectrons is equal to 4.03 milliamperes.

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