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