## 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
$10^20 Hz$

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