*Wow* sir loved your lecture,PLEASE COVER OTHER CHAPTERS AS WELL ,YOUR TEACHING IS #GOD LEVEL, PLEASE SPARE SOME MORE TIME FOR US PHYSICS LOVERS ,IF POSSIBLE.
Sir , When phosporous is added as impurity ,we know that the electron that doesnt get binded moves to conduction band,why shoudnt we think that when electron jumps off it is leaving a hole at phosporous.
A hole is created only when a covalent bond is broken. The electron that moves to the conduction band is not actually bonded with any other atom. So when it moves to the conduction band, No hole is created.
How each quantum state or orbital can have only one electron means according to Pauli exclusion principle not more than two electrons can exit in a orbital.
A quantum state includes spin as well. So no 2 electron can have same quantum state. The other electron that comes will have a different quantum state i.e. spin will be different while n,l,ml quantum nos. will be same.
They are nothing but shells in every orbit of different principal quantum number n = 1,2,3..., each electron occupies one quantum state, out of 2 possible states in s orbital, and so on..
that formation of e-h pair is a function of temperature, say f(T). while the recombination depends on how much e- are available at any instant to combine with hole so recombination is directly proportional to num(electron) and num(holes) recombination is proportional to num(e) * num(h) where num(e) = num(h) = x so for an equilibrium rate of formation of e-h pair = rate of recombination f(T) = k*num(e)*num(h).= k*x*x.......................(i) now consider extrinsic semiconductor, let n-type still the rate of formation of e-h pair is directly proportional to f(T) but now there is already so many electron - conduction band(from P atom)+ due to thermal energy so now in equilibrium f(T) = k* [num(e)+ num(conduction band electron)]*[num(h)] where num(e) = num(h) = y f(T) = k * [y + num(P atom electrons)]*y now y
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