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Using Bohr's atomic model, derive the expression for the radius of nth orbit of the revolviting electron in a hydrogen atom. |
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Answer» Solution :We know that when an electron revolves in a STABLE ORBIT, the centripetal force is provided electrostatic force of attraction acting on it due to a portion present in the nucleus. `therefore (m v_(N)^(2))/(r_(n))=(1)/(4pi epsi_(0)). (e^(2))/(r_(n)^(2)) rArr v_(n)^(2) =(e^(2))/(4pi epsi_(0) m r_(n))` and form Bohr.s quantum condition, we have `m v_(n). r_(n) =(nh)/(2pi) or v_(n)=(nh)/(2pi m r_(n))` Squaring (ii) and then equating it with (i), we get `(n^(2) h^(2))/(4pi^(2) m^(2) r_(n)^(2))=(e^(2))/(4pi epsi_(0), m. r_(n)) rArr r_n =(n^(2) h^(2))/(4pi^(2) m^(2)) xx (4pi epsi_(0) m)/(e^(2)) =(epsi_(0) h^(2))/(PI me^(2)) .n^(2)` |
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