## Kinetic Energy of a Photon With a Given Wavelength

The faster an electron moves, the higher its kinetic anergy and the shorter its wavelength.
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 momentum for low speed electrons, which is usually the case, is given by
$p=m_e v$

where
$m_e$
is the mass of the electron and
$v$
is its speed.
We can write then
$\lambda = \frac{h}{m_e v}$

The kinetic energy is
$KE=\frac{1}{2}m_ev^2$

We can rearrange
$\lambda = \frac{h}{m_e v}$
to give
$v=\frac{h}{m_e \lambda}$

Then
$KE=\frac{1}{2}m_e v^2 =\frac{1}{2}m_e (\frac{h}{m_e \lambda})^2 =\frac{h^2}{2m_e \lambda^2}$

For an electron,
$m_e =9.11 \times 10^{-31} kg$
so if
$\lambda =10^{-8} m$

$KE=\frac{h^2}{2m_e \lambda^2} =\frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (10^{-8})^2} =2.41 \times 10^{-21} J$
.