Video Transcript
A tunable laser is used to illuminate the surface of a metal with different wavelengths of light. When the wavelength of the light is shorter than a certain value, electrons are emitted from the surface of the metal. The graph shows the maximum kinetic energy of the electrons emitted against the wavelength of the photons. What is the maximum wavelength of light for which electrons will be emitted from the surface of the metal?
Looking at our graph, we see that this curve shows us the maximum kinetic energy of electrons ejected from the surface of a metal caused by incident photons of a given wavelength. This overall process, where photons are incident on the surface of a metal and through their absorption by the metal lead to the ejection of electrons from the surface, is called the photoelectric effect. In order for this effect to happen, the incident photons must have a minimum amount of energy. This is the energy required to be transferred to an electron in the metal so that it’s able to overcome the energy barrier tending to bind it to the metal and can be released.
On the horizontal axis of our graph though, we don’t have the energy of incident photons but rather their wavelength. We can recall though that these two quantities, energy and wavelength, are related through the relationship for the energy of a photon. The energy of a photon is equal to the Planck constant multiplied by the speed of the photon, the speed of light in vacuum, divided by the photon’s wavelength. Therefore, photon energy and wavelength are inversely proportional to one another. This is why our question talks about the maximum wavelength of light rather than the minimum wavelength for which electrons will be emitted from the metal surface.
Considering our graph, we see that the graph goes to zero right here at 300 nanometers. And that means that any photons incident on our metal surface with a longer wavelength and therefore a smaller amount of energy are not energetic enough to cause the ejection of electrons from the metal. So, even though our graph indicates that there are many different wavelengths of light which would cause the ejection of electrons from the metal, these values do have a maximum, and that maximum wavelength is 300 nanometers. Knowing this, let’s look now at part two of our question.
What is the work function of the metal? 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 to two decimal places.
In the first part of our question, we found out the maximum photon wavelength that could cause the ejection of an electron from our metal surface. Here, we want to solve for the work function of our metal, which is an amount of energy rather than a wavelength. We can think of the work function as an energy barrier that electrons in the metal must overcome in order to leave the metal. The energy for electrons in the metal to do this is supplied by incident photons.
A photon that transfers energy equal to the work function of the metal will have the longest possible photon wavelength, while still leading to the ejection of an electron from that surface. In other words, the work function of our metal corresponds to the energy of a photon with a wavelength of 300 nanometers. This, our graph shows us, is the wavelength of an incident photon that would eject an electron from our metal but just barely. In fact, according to our graph, that ejected electron would have zero kinetic energy.
To solve for this photon’s energy and therefore the work function of the metal, we can use this equation from earlier. So then, 𝐸 is our metal’s work function. ℎ, the Planck constant, has a value of 4.14 times 10 to the negative 15 electron volt seconds. The speed of light 𝑐 in vacuum has a value of 3.0 times 10 to the eight meters per second. And the wavelength 𝜆 of our photon is the longest wavelength that will still eject an electron from the surface of our metal.
Before we calculate this work function, let’s convert the wavelength of our photon from units of nanometers to units of meters. We recall that 10 to the nine, or one billion nanometers, equals one meter, and therefore 300 nanometers equals 300 times 10 to the negative nine meters. Or, writing this in scientific notation by shifting the decimal place two spots to the left, we have 3.00 times 10 to the negative seven meters.
Notice that in this fraction, we have 3.0 times some power divided by 3.00 times some power. This then is equal to one times 10 to the eight divided by 10 to the negative seven. We can note that 10 to the eight divided by 10 to the negative seven equals 10 to the eight minus negative seven, which equals 10 to the positive 15. Therefore, our fraction simplifies to one times 10 to the 15 with units of meters per second per meter. If we multiply numerator and denominator by one divided by meters, then meters multiplied by one divided by meters both in numerator and denominator cancel out. And our fraction simplifies to one times 10 to the 15 inverse seconds.
But then notice that if we multiply one over seconds by seconds in the units for Planck’s constant, the units of seconds cancel out entirely, and we’re left with units of electron volts. Looking further into this product, we see that we’re multiplying a value times 10 to the negative 15 by one times 10 to the positive 15. The energy 𝐸 then equals 4.14 times one multiplied by 10 to the negative 15 plus 15 electron volts. But then negative 15 plus 15 is zero, and 10 to the zero is one. And so our answer simplifies to 4.14 electron volts. This is the work function of our metal.
