# Why does Einstein's photoelectric effect equation assume $\text{KE}_{max}=\frac{1}{2}mv_{max}^2$

The derivation of Einstein's photoelectric equation is shown as such in the textbook I am currently using:

The work function $$W$$ of a metal is related to its threshold frequency $$f_o$$ by: $$W=hf_o$$ If frequency $$f$$ of incident radiation is greater than $$f_o$$ then the energy of the photon off radiation: $$hf>hf_o$$ And the emitted photoelectrons would have a: $$\text{maximum kinetic energy}=hf-hf_o$$ $$E_{max}=\frac{1}{2}m(v_{max})^2=hf-W$$ $$=h(f-f_o)$$

However, I do not understand how the maximum kinetic energy could be $$\text{KE}_{max}=\frac{1}{2}mv_{max}^2$$?

As energy is obviously required by the electron to reach the surface of the metal and any residual energy left greater than that of the work function would then be used into kinetic energy and the electron would not reach maximum velocity.

• Maximal velocity is reached when all the excess of $hf - W$ is transferred to kinetic energy. Hence the definition of $v_{max}$ – nwolijin Dec 13 '20 at 20:48
• That is because the work function is the minimum potential energy the electron has to overcome to be emitted as a photoelectron-it is the electron on the surface of the metal which has the least force of attraction to the metal. – SK Dash Dec 14 '20 at 0:53

The work function depends on the configurations of atoms at the surface of the material. For example, on polycrystalline silver, the work function is $$4.26$$ eV, but on silver crystals, it varies for different crystal faces as $$(100)$$ face: $$4.64$$ eV, $$(110)$$ face: $$4.52$$ eV, $$(111)$$ face: $$4.74$$ eV. Ranges for typical surfaces are shown in the table below.
$$T_{max}=h\nu-\phi_{min}$$