7th 1B.15 When do you use E = hv or Ek = 1/2m v^2?

[LuisG14A1J]

I am trying to follow the steps of the similar practice problem in the textbook of Example 1.B3 but I have no idea how they somehow get electron volts for their (b) section. I understand how to use the Ek function, but I do not know exactly why or when. I'm just trying to trace the steps of the example problem but I don't know how or why they are using the "work function" and then subtracting it from hv.
The main question is why do they use Ek = 1/2 m_ev^2 and then also hv, if E = hv is meant to tell you the energy of each individual photon only? Also, where do they get the work function from and how should I be using it to solve this problem?

[Elisa Bass 4L]

It looks like the problem gave the electron volts for the (b) section of the Example 1B.3. They went from Joules from electron volts using a conversion factor. The Ek function is used specifically to find the kinetic energy of an ejected electron, otherwise known as the excess energy the photon transferred to the electron that was not used to remove the electron from the metal. The “work function” is the amount of energy used to remove the electron from the metal and it is being subtracted from the energy of the photon, which you correctly listed as E = hv.

If you are given the velocity of the ejected electron, and need to find the kinetic energy of it, it is a good idea to use the Ek = 0.5mv^2 equation. The mass of the electron, m, is constant and can be found online. The equation Ek = Energy(photon) - Energy(remove e-) can be manipulated based on the problems. It is also written Ek = hv - “work function”, or with Ek labeled as “excess energy”.

[gillianozawa4I]

I agree with this. When given velocity, use Ek = 1/2 mv^2. It's also be confusing because "v" in E=hv is frequency, while v in Ek = 1/2mv^2 is velocity. E=hv is the energy brought by the photon, while Ek is E=hv minus the work function (threshold value).

[Tony Ong 3K]

You can think of the Kinetic Energy as the "excess" energy emitted if that helps. Professor Lavelle mentioned that there can only be Kinetic energy released if the E (photon) > Work function. So knowing this, if E(photon) equals threshold/work function, then kinetic energy is 0. So whenever they are asking you to find the energy "emitted" and the question is dealing with the photoelectric effect, you are dealing with the function Ek= 1/(2mv^2).

[Chloe Qiao 4C]

In other words, use E=hv when finding the energy of a photon(the inputing energy), and Ek=1/2mv^2 as the energy of the ejected electron(energy left), which is the energy of the photon minus the threshold energy of the electron.