How is wavelength of light related to observability? One of the things I learned from Heisenberg's Uncertainty Principle is that when you try to observe the position of a microscopic particle, you have to use a light of smaller wavelength (which implies high energy), and that causes a large change in the particle's momentum. I do not however understand how does light of smaller wavelength help us observe such particles clearly? What exactly happens when such energetic light is incident on a particle, and how is it different from when the light is not so energetic? We essentially see scattered light from objects, so does it mean that light of high wavelength does not scatter off small enough particles? How can we understand this if we consider light as photons?
 A: 
One of the things I learned from Heisenberg's Uncertainty Principle is that when you try to observe the position of a microscopic particle, you have to use a light of smaller wavelength (which implies high energy), and that causes a large change in the particle's momentum.

Be careful here. Don't mix up the HUP with the observer effect. The HUP relates position and momentum uncertainties of the same state, not uncertainties before and after measurements. Additionaly, the HUP is not dependent on the mechanism of observation.

I do not however understand how does light of smaller wavelength help us observe such particles clearly? What exactly happens when such energetic light is incident on a particle, and how is it different from when the light is not so energetic? We essentially see scattered light from objects, so does it mean that light of high wavelength does not scatter off small enough particles?

It essentially comes down to diffraction. The smaller the wavelength of the light you are using to see something, the less diffraction will matter. If diffraction is too prevalent, then it becomes impossible to resolve nearby points that are emitting, scattering, etc. light that you want to see.
This is evident in everyday vision. For example your phone is made up of small atoms, and the light that bounces off of your phone that allows you to see it has wavelengths that are much larger than those atoms. This doesn't mean that you can't see your phone though; it just means that your eyes can't distinguish between the individual atoms that make up your phone.
A: For light (photons) $E = pc$ and $p = \frac{h }{\lambda}$we can see that photons of higher momentum have higher energy, but momentum is inversely proportional to wavelength. In decreasing the frequency of the light, we are using light of higher wavelength since $c = f \lambda$. This diminishes the light's ability to probe a material, meaning smaller detail and  resolution.
Think of trying to image say a human cell. If we bombard it with light that has a wavelength comparable to that of the actual cell size, the image we get will be terribly blurred with basically no detail or resolution. If we did the exact same thing but using photons of much smaller wavelength (higher frequency) than the cell, we will pick up a lot more detail.
Light (photons) that have higher frequency (energy), are more able to penetrate objects and regions of fine detail. So high frequency light can carry more information (per unit time) than lower frequency light.
