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In deriving the maximum kinetic energy of photoelectrons in the photoelectric effect, Young & Freedman says that "As an electron moves from the cathode to the anode, the potential decreases by $V_0$ and negative work $-eV_0$ is done on the (negatively charged) electron."

For context, $V_0$ refers to the stopping potential, and the potential of the anode relative to the cathode is negative, and electrons are repelled from the anode.

I'm confused because as the electrons move from the cathode to the anode i.e. from (+) to (-), shouldn't its potential increase by $V_0$, not decrease? Also, why is there negative work being done on the electron?

I attach the diagram from the book for extra clarification: enter image description here

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Remember that electric potential describes the difference in the electric field between two points. You can think of it sort of like elevation on a topographical map--high elevation is analogous to high potential, and vice versa. So a particle moving from a cathode (positive electric field) to an anode (negative electric field) will experience a net decrease in electric potential--sort of like a ball rolling down a hill experiences a net decrease in elevation, or gravitational potential.

The reason negative work is being done is because the electron is negatively charged. It experiences a force in the opposite direction from the potential drop. Going back to the topography analogy, an electron would be like a ball that spontaneously rolls uphill. Electrons accelerate "up" an electric potential (negative to positive), while protons accelerate "down" an electric potential (positive to negative).

In even more basic terms, a negatively charged region (anode) will repel approaching electrons, so they will slow down and eventually reverse direction. This slowing down reduces kinetic energy and therefore negative work is being done on the electron.

Hope that helps.

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The electrons have the greatest potential energy at the negative terminal. As they start to move from the negative terminal to the positive terminal, their potential energy must decrease while their kinetic energy increases. Think of the electrons as marbles at the top of a hill. They have greatest potential energy there. As they role down the hill, their potential energy given by $mgh$ decreases until it is zero at the bottom of the hill. Of course there, they will have maximum kinetic energy.

why is there negative work being done on the electron?

The work done by a force in moving a charge from point 1 to 2 is defined as $$W_{1\rightarrow 2}=\Delta{\text{potential energy}}=U_2-U_1$$

At point 1 (cathode) the electrons have potential energy $+qV$ and when they reach point 2 (anode) their potential energy is zero. The change in potential energy is therefore $0-qV=-qV$ and so the work done $$W_{1\rightarrow 2}=-qV$$

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  • $\begingroup$ Thank you for your answer, I really appreciate it. Unfortunately, this answer is incorrect and caused me more confusion. But I got it now. Electrons have 0 potential energy at the cathode and maximum potential energy at the anode. $\endgroup$
    – Freddie
    Oct 31, 2021 at 7:07
  • $\begingroup$ It’s actually not wrong. It caused you confusion because I never mentioned the fact that the force is in the opposite direction of the change in potential from point one to point two. But no worries and good luck with your studies. Cheers. $\endgroup$
    – joseph h
    Oct 31, 2021 at 8:31
  • $\begingroup$ You said in your answer that "The electrons have the greatest potential energy at the cathode. As they start to move from the cathode to the anode, their potential energy must decrease while their kinetic energy increases." But if you look at the diagram in my post, the cathode is positive and the anode is negative. So, as the electrons move from the cathode to the anode, their potential energy INCREASES while their kinetic energy DECREASES, just the opposite of what you said. But thanks for the answer. $\endgroup$
    – Freddie
    Oct 31, 2021 at 9:21
  • $\begingroup$ Ah I'm so sorry. That was truly me mixing up cathode and anode. I have edited my answer. Thanks and again sorry for mix up. Cheers. $\endgroup$
    – joseph h
    Oct 31, 2021 at 9:29

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