Why can't a bose-einstein condensate ever be produced?

Pretty technical question. So, we'll have to have a pretty technical answer.  

Bose-Einstein condensation occurs when matter is cooled down, and there are just not enough many 'states' for all the particles 'to fit'. So, being bosons, which can squeeze into the same state, any excess bosons end up squeezing into the same, ground state. This is what is meant by 'Bose-Einstein condensation'.  

So why doesn't it happen with photons? Because there is no such law as 'conservation of photons', they can be created and destroyed at will by interacting with matter. So, when you try to cool down a photon gas to achieve a B-E condensate, they just get absorbed by the material they are in contact with. Therefore photons can not go under B-E condensation.

They do, it's called a laser, where the photons all have the same wave-function, except it's not REAL Bose-Einstein condensation. The difference is that no phase transition takes place in a laser. A gas of rubidium-87 atoms say, can be made to condense into a B-E condensate because as they are cooled below the critical temperature, the chemical potential goes through zero. Lasers on the other hand don't 'condense'.

Hope that helps You, Good Luck

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.