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UsingBohr's postulates , obtain the expreesion for the total energy in the stationary states of hydrogen atom . Hence , draw the energylevel diagram showinghow the line spectra correspondingto Balmer series occur duetotransition betweenenergy levels. |
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Answer» Solution :According to Bohr.s firstpostulate , an electron is revolvingaroundthe nucleus ofhydrogenin a circularorbitsuch thatcentripetal force is provided by electronstaticforce of attraction experiencedby electron due toproton . Thus, we have ` (m v_(n)^(2))/(r_(n)) =(1)/(4 pi in_(0)) .(e^(2))/(r_(n)^(2)) rArrm v_(n)^(2) r_(n) = (e^(2))/(4 pi in_(0))"".......(i)` Again according to Bohr.s quantum condition , we have `m v_(n) r_(n) =n (h)/(2PI)""........(ii)` Dividing (i) by (ii) , weget`v_(n) = (e^(2))/(2 in_(0) n h) ` andfrom(ii) `r_(n) =(n h)/(2 pi m v_(n)) = (nh)/(2pi m((e^(2))/(2in_(0) nh))) = (in_(0) n^(2) h^(2))/(pi m e)` `THEREFORE ` Kineticenergyof revolvingelectron `k_(n) = ( 1)/(2) mv_(n)^(2) = (1)/(2) m ((e^(2))/(2 in_(0) n h)) = (m e^(4))/( 8 in_(0)^(2) n^(2) h^(2))` Again electrostatic potential energy of electron (charge `q_(1)` = -e) PROTON (charge `q_(2)` = + e) system `U_(n) = (1)/(4 pi in_(0)) .(q_(1) q_(2))/(r_(n)) =- (1)/(4 pi in_(0))(e^(2))/(r_(n)) = - ( me^(4))/( 4 in_(0)^(2) n^(2) h^(2))` `therefore `Total energy of the revolving electron in nth orbit will be `E_(n)= K_(n) + U_(n) = (me ^(4))/(8 in_(0)^(2) n^(2) h^(2)) - (m e^(4))/(4 in_(0)^(2) n^(2) h^(2)) =- (me^(4))/( 8 in_(0)^(2) n^(2) h^(2))` The energy level diagram is given in Fig. 12.08. |
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