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Derive the expression for the radius of nth Bohr's orbit in Hydrogen atom. |
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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""(mv_(n)^(2))/(r_(n))=(1)/(4pi in_(0)).(e^(2))/(r_(n)^(2)) rArr v_(n)^(2)=(e^(2))/(4pi in_(0)mr_(n))"...(i)"` and from Bohr.s quantum condition, we have `mv_(n).r_(n)=(nh)/(2pi) or v_(n)=(nh)/(2pi mr_(n))"...(ii)"` Squaring (ii) and then EQUATING it with (i), we get `(n^(2)h^(2))/(4pi^(2)m^(2)r_(n)^(2))=(e^(2))/(4pi in_(0)m.r_(n))` `rArr""r_(n)=(n^(2)h^(2))/(4pi^(2)m^(2))xx(4piin_(0).m)/(e^(2))=(in_(0)h^(2))/(pime^(2)).n^(2)` |
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