Kinetics energy expression of photon

I need to calculate the kinetics energy of a photon and I'm trying to deduce a correct expression for calculating it.

So we have that

$K= \frac{1}{2} m_fv_f^2$

So $v_f= c$

For De Broglie equation we have that $m_f = \frac{h}{\lambda c}$

And then

$K = \frac{hc}{2 \lambda}$

But apparently the correct expression is:

$K = \frac{hc}{\lambda}$

Can someone help me figure out what I'm doing wrong? Thanks!

• Well, photons have no mass, so using the de Broglie equation is inappropriate. Jun 8 '16 at 15:06
• The whole $mv^2\over2$ thing stops working as we approach the speed of light. Jun 8 '16 at 15:26

The energy of a photon is given by (as you already stated) $$E = \frac{hc}{\lambda}$$ where $h$ is Planck's constant, $c$ the speed of light and $\lambda$ the wavelength of the photon.
You can derive it from the relativistic energy $$E^2 = p^2c^2 + m^2c^4$$ with $m =0$ and using the fact that for a photon $$p = \hbar k = \frac{h\nu}{c} = \frac{h}{\lambda}.$$