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In this case I will assume that $f$ and $\phi$ will remain constant since battery voltage is changing.

The two equations I have are

1) $KE_{max} = hf - \phi = h(f-f_0)$

and

2) $V_se = KE_{max}$

Initially my thought process was that $KE_{max}$ is determined from equation 1 assuming $f$ and $\phi$ remain constant. And because of that, any change in voltage would have no effect on the maximum kinetic energy.

However, another way I was thinking about it was that as the voltage increases the electrons will undergo more repulsion which lowers their kinetic energy.

Is one of these arguments more correct than the other? Or are they both factors in determining the effect on $KE_{max}$?

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The first equation gives you the maximum kinetic energy that the emitted emitted photoelectrons can have.

The second equation relates the the potential difference (stopping potential) between the surface emitting the photoelectrons and an electrode which is collecting the photoelectrons to the maximum kinetic energy of the photoelectrons.
That stopping potential enables one to measure the maximum kinetic energy of the photoelectrons.

As the voltage of the collecting electrode becomes more negative fewer of the photoelectrons have enough kinetic energy to reach the electrode.

When the stopping voltage is reached none of the photoelectrons have enough kinetic energy to reach the collecting electrode.

Making the voltage more negative than the stopping voltage continues to result in no photoelectrons hitting the collecting electrode.

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