A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the 
surface is (c = velocity of light) 

The radiation energy is given by  $E=\frac{hc}{\lambda }$

Initial momentum of the radiation is $p_i=\frac{h}{\lambda }=\frac{E}{c}$

The reflected momentum is  $p_r=-\frac{h}{\lambda }=-\frac{E}{c}$

So, the change in momentum of light is  

$\Delta p_{\text{light }}=p_r-p_i=-\frac{2E}{c}$

Thus, the momentum transferred to the surface is   $\Delta p_{\text{light }}=\frac{2E}{c}$.