The Photoelectric Effect: A Problematic Problem

We use the equation:

$E_x=hv-\phi$ , where $E_x$ is the maximum electron energy, $v$ is the light frequency, $\phi$ is the work function and $h$ is Planck's constant.

We also need to know that $1 eV$ is equivalent to $1.6 \times 10^{-19}J$ , and we must remember that $v=c/\lambda$ , where $c=3 \times 10^8 ms^{-1}$ is the speed of light and $\lambda$ is the wavelength, we get:

$2\pi \times 10^{15} \frac {h}{2}-\phi= 3.36 \times 10^{-19}$

and

$1.2\pi \times 10^{15} \frac{h}{2}-\phi=0.8 \times 10^{-19}$ .

Solving these simultaneous equations gives $\frac{h}{2}=1.02 \times 10^{-34}J$ and $\phi=1.9 eV$ .

We have thus found both the value of $h$ bar and the work function of potassium.