You really need to read about a technique before you start using it. 

The peaks you see at around 1keV are from the Zn L lines, at 8.601 keV is Zn Ka, 9.5 keV is the Zn Kb line.

You should also see a peak at around 0.5 keV from O Ka. You do not mention anything below 1 KeV. I have added a link to show you where peaks should be - the software on the instrument should do this for you (but it doesn't always get it right, so you need to know what you are doing). 

There should not be anything from ZnO around 2 or 3 keV - you probably have some contaminant. Please post an image of your your spectrum and we may be able to help further.

I have added a second link to start you off in your reading to learn this technique. 

Please avoid calling it EDAX analysis. EDAX is a brand name, a company that is part of AMETEK Materials Analysis Division. The technique should properly be called EDX or EDS (the two terms are synonymous).

http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_2/4_2_1.html

http://www.ammrf.org.au/myscope/analysis/introduction/

Hi Sukumar, In SEM-EDAX  ZnO presented 2 peaks (major Intensity peak Zn K-alpha in lower keV and minor Intensity peak Zn L-alpha in a higher keV). It means the energy used is appropriated to exclude the electron from inner shell (K-level) as a high intensity (1-3 keV) and a small quantitiy of the electron back to L-level (from higher level) by emitting energy 8 -  10 keV. Hoping this information will be useful.

Rudy, that is not quite right.

In the first part you have mixed up 'lower keV and 'higher keV'. The Zn K is at a higher energy than Zn L.

Also you have mixed up the second part as well. The incident electron will remove a Zn K-shell electron, electrons fall back from L and M shells to give discreet X-ray quanta; from Liii to K gives Zn Ka1 at 8.63 keV and from Lii to K gives Ka2 (this line is not resolvable in EDX), the transition from Miii to K gives a higher energy line Zn Kb at 9.5 keV. There are many transition possiblilities from M to L but the most intense will be Mv to Liii to give Zn La1 at 1.02 keV - the L lines are at lower energy. 

Generally the Ka lines will have a greater intensity than the L lines - BUT this depends on the value of the energy of the incident beam. If the accelerating voltage is reduced then the L lines can be at a higher intensity than the K lines! The intensity ratios between Ka and Kb and between the L series lines will always be the same.