In the photoelectric effect, the work function represents the amount of work that the incoming photon must do in order to extract one electron from the metal.
The energy of the incoming photon is equal to:
[tex]E=hf[/tex]
where [tex]h=6.63\cdot 10^{-34} m^2 kg/s[/tex] is the Planck's constant, and f is the photon frequency.
Part of the energy of the photon is used to extract the electron from the metal (work function [tex]\phi[/tex]), and the rest is converted into kinetic energy K of the electron:
[tex]hf=\phi+K[/tex]
We can find the threshold frequency [tex]\phi _0[/tex], i.e. the minimum frequency needed to extract the electron, by requiring K=0:
[tex]hf_0 = \phi[/tex]
Using [tex]\phi= 7.4 \cdot 10^{-19} J[/tex], we find
[tex]f_0 = \frac{\phi}{h}= 1.12 \cdot 10^{15} Hz [/tex]
The problem asks to find the corresponding wavelength (i.e. the maximum wavelength able to extract the electron from the metal), which is
[tex]\lambda _0 = \frac{c}{f_0} = \frac{3\cdot 10^8 m/s}{1.12 \cdot 10^{15} Hz}=2.68\cdot 10^{-7}m=268 nm [/tex]
where c is the speed of light.
