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  
\[\lambda = \frac{h}{p}\]

\[h=6.626 \times 10^{-34} Js\]
  is Planck's constant and  
  is the electrons momentum. The momentum for low speed electrons, which is usually the case, is given by  
\[p=m_e v\]
  is the mass of the electron and  
  is its speed.
We can write then  
\[\lambda = \frac{h}{m_e v}\]

The kinetic energy is  

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

\[KE=\frac{1}{2}m_e v^2 =\frac{1}{2}m_e (\frac{h}{m_e \lambda})^2 =\frac{h^2}{2m_e \lambda^2}\]

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

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

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