Question

Electrons are ejected from a metallic surface with speeds ranging up to 4.9 105 m/s when light with a wavelength of λ = 625 nm is used. (a) what is the work function of the surface?

Answer 1

In the photoelectric effect, part of the energy of the photon (E=hf) of the incoming light is used to extract the photoelectron from the metal (work function: $\phi$) and part is given to the electron as kinetic energy: $K= \frac{1}{2}mv^2$:
$hf = \phi + \frac{1}{2}mv^2$
The frequency of the photon is related to the wavelenght: $f= \frac{c}{\lambda}$, and so we can re-write the formula as
$\phi = \frac{hc}{\lambda} - \frac{1}{2} mv^2$
where:
$h=6.6 \cdot 10^{-34}Js$ is the Planck constant
$c=3 \cdot 10^8 m/s$ is the speed of light
$\lambda = 625 nm = 625 \cdot 10^{-9}m$ is the wavelength of the light
$m=9.1 \cdot 10^{-31}kg$ is the electron mass
$v= 4.9 \cdot 10^5 m/s$ is the electron speed.

Substituting these numbers into the equation, we can find the work function of the surface:
$\phi = \frac{hc}{\lambda}- \frac{1}{2}mv^2=3.2 \cdot 10^{-19}J-1.1\cdot 10^{-19}J=2.1 \cdot 10^{-19}J$