Dual Nature and Photoelectric effect

• Photoelectric effect could be explained satisfactorily after it is assumed  that the light consists of particles which travel with the velocity of light and these particles have energy and momentum , where energy is proportional to the frequency of the light and momentum is inversely proportional to the wavelength.
• The behaviour of light  as particle is not a universal  i.e. in every situation the light does not behave as particle but in most situations it behaves as wave.
• So the  light behaving as wave and also as particle, is of dual nature.
• Just like the light has duality in nature, de Broglei proposed that all matter must possesses wave nature also and therefore each matter should be associated with  a wave known as  de Broglei wave whose wavelength  is related to the momentum of the matter in the same manner  for the light where  l = h / p . 
• 
  
• For photons energy    E = h c/ l    and effective mass  = E/ c²                                                       
  

   therefore momentum  p = effective mass  x  c 
  = (E/ c² ) x c

  \ p = E/c                                                                                                                        
  But    E = hc/ l   for photons , so      p  = (hc/ l )/c                                                                    
                                    or    p  = h /  l
   
• When we calculate  wavelength for the matter, we must  not    use   the relation                               E = hc/ l  as this relation is strictly valid only for light.  For matters, first calucate momentum and then using the relation    l = h / p  we can find the deBroglei wavelength.
• For example if we have to find deBroglei wavelength. of an electron  of kinetic energy 4 ev , first we calculate  its momentum  using the relation  P= ( 2 m E )^1/2   and then find its wavelength. 
   


• Specially for the electron, there is a relation  12.27 / ( E )^1/2  angstrom  where E is the kinetic energy of the electron in the electron volt . So  using this relation, we get wavelength of the electron as  12.27/ 2= 6.13  A^{0 .} 
• Suppose  there is a light consisting of the photons of energy  4 ev and we have to find its wavelength. here we will use the relation  
  E = hc/ l  which on simplification comes out to be a simple fomula as   l = 12400 / E  A⁰  and so  wavelength will be 3100 A⁰  which  is clearly different than wavelength of the electron of same kinetic energy.
• Note that the wavelength associated with the electron is much  shorter than  that of the photon of same energy. 
• Lets compare the de Broglei wavelength of a proton and that of a alpha particle having been accelerated to same electric potential :