light Intensity and its effects on energy and emitted electron on metal surface
what if i double the intensity of light , what would happen to the
  number of electrons emitted and the average energy of the electrons?


This is known as the "Photoelectric Effect" and is described by the following equation

$$\ce{E_{max}=h\nu- WF_{M}}$$

where $\ce{E_{max}}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\ce{WF_{M}}$ is the workfunction for the particular metal. Your guess is part right, doubling the intensity of the incoming beam doubles the number of electrons ejected from the metal surface, but their kinetic energy distribution remains unchanged.  The maximum kinetic energy of an emitted electron is still determined by the above equation and therefor still capped by $\ce{E_{max}}$ and the kinetic energies of all electrons emitted are still distributed from 0 to $\ce{E_{max}}$.  If you'd like more information, here's a good, concise reference.