## 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 \[\lambda =10^{-8} m\]

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

.