Kinetic Energy of an Electron With a Given Frequency

The faster an electron moves, the higher its kinetic anergy and the higherits frequency.
The wavelength is given by  
\[\lambda = \frac{h}{p}\]

where  
\[h=6.626 \times 10^{-34} Js\]
  is Planck's constant and  
\[p\]
  is the electrons momentum. The wave equation is  
\[v = f \lambda\]
  and the mometum is  
\[p=mv\]
  so that  
\[\frac{v}{f} = \frac{h}{mv} \rightarrow mv^2 = hf \]

But  
\[KE=\frac{1}{2}mv^2\]

so  
\[KE=\frac{1}{2} hf\]

for an electron.
For an electron wtih a freuency  
\[10^20 Hz\]

 
\[KE=\frac{1}{2}hf=\frac{1}{2} \times 6.626 \times 10^{-34} \times 10^20 =3.313 \times 10^{-14} J\]
.