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Using Bohr's postulates of the atomic model, derive the expression for radius of th electron orbit. Thus, obtain the expression of Bohr's radius. |
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Answer» Solution :We know that when an electron revolves in a stable orbit, the centripetal force is provided by ELECTROSTATIC force of attraction acting on it due to a proton present in the nucleus. `therefore""(m v_(n)^(2))/(r_(n)) = (1)/(4pi in_(0)) .(e^(2))/(r_(n)^(2))or v_(n) = (NH)/( 2pi m r_(n))"".....(i)` andfromBohr.s quantum CONDITION , we have . `m v_(n) r_(n) =(n h)/(2 pi) or v_(n) =(n h)/( n h)/(4 pi in_(0) m r_(n))""........(ii) ` Squaring(ii) and then equatingit with (i) , we get . `(n^(2)h^(2))/(4 pi^(2) m^(2) r_(n)^(2)) = (e^(2))/(4 pi in_(0) m.r_(n))rArrr_(n) = (n^(2) h^(2))/(4pi^(2) m^(2)) xx (4 pi in_(0) m)/(e^(2)) = ( in_(0)h^(2))/( pi m e^(2)) = n^(2)` In stable orbitof hydrogen atom n = 1and then radiusof 1st ORBITIS called Bohr.sradius`a_(0)` . Obviously ` a_(0)= (in_(0) h^(2))/(pi m e^(2))` |
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