Kinetic energy of an electron with de Broglie wavelength $\lambda$ If an electron has de Broglie wavelength $\lambda$, can I write it's kinetic energy $E = \frac{hc}{\lambda}$? If not then what energy does the equation represent and when can the equation be used ?
 A: The problem is that for an electron $\nu \ne c/\lambda$. You need to replace $c$ by the phase velocity, which is somewhat involved to calculate. However you can get an approximate equation for the energy and wavelength by noting that the de Broglie wavelength is given by:
$$ \lambda = \frac{h}{p} $$
For a non relativistic electron we have:
$$ E = \frac{p^2}{2m} $$
Giving us:
$$ E = \frac{h^2}{2m\lambda^2} $$
For a relativistic electron you need to use the relativistic equation for the total energy:
$$ E^2 = p^2c^2 + m^2c^4 $$
And substituting for $p$ gives:
$$ E^2 = \frac{h^2}{\lambda^2}c^2 + m^2c^4 $$
Not that it's particularly relevant, but we can recover the non-relativistic equation by writing the above equation as:
$$ E^2 = m^2c^4\left(\frac{h^2}{m^2c^2\lambda^2} + 1\right) $$
so:
$$ E = mc^2\left(\frac{h^2}{m^2c^2\lambda^2} + 1\right)^{1/2} $$
If we assume the energy is much less than the rest mass we can use the binomial expansion on the square root to get:
$$ E \approx mc^2\left(1 + \frac{h^2}{2m^2c^2\lambda^2}\right) $$
or:
$$ E \approx mc^2 + \frac{h^2}{2m\lambda^2} $$
If we identify this as the energy equal to the rest energy plus the kinetic energy that gives us:
$$ KE \approx \frac{h^2}{2m\lambda^2} $$
Just as we concluded above.
A: the energy of a debroglie wave can be gotten from the equation below 
                  E=hv/2l    where l=wavelength 
it is obvious that a change in velocity will lead to a change in wavelength and energy. Therefore E=f(v,l)
Using partial derivatives, the change in energy, dE is expressed through the following equation
                        dE=h/2l(dv.(v/l).dl)

