When to use E = h(nu) and not to

[JTieu_1L]

Hi,
I don't if I am understanding this correctly, can someone verify if the following statement is true?

To my understanding E = h(nu) can only be applied to find the energy in joules of a photon, however, this formula can be applied to find the energy required to remove an electron only if the kinetic Energy is 0. If energy of photon is not equal to energy required to remove an electron, then to find the energy to remove an electron, or work function, is setting it equal to (energy of photon) - (excess energy or kinetic energy).

[Audrey Han 3L]

Yes this is correct, however you'd only use E(photon)-Ek if you know that the electron was in fact ejected from the metal surface.

[Sabina House 2A]

Yes, you are correct. Because E(photon) - work function = Ek, you could rearrange this so work function = E(photon) - Ek. This means that the energy required to remove an electron is the (energy of a photon) - (kinetic energy of the ejected electron).

[Samuel Flores 1E]

Yes, this is correct. If the kinetic energy of the emitted electron (during the Photoelectric Experiment) is KE=0, then we can conclude that E=hv is equal to the Work Function of the Metal.
Therefore, while E=hv represents the the energy of a photon when we know its frequency (v), this equation can also be used to find the Work function, in the case that the KE of the electron is zero.

Hope this helps!