Saved Bookmarks
| 1. |
What is the minimum energy that must be given to a H atom in ground state so that it can emit an H_(gamma) line in Balmer series?If the angular momentum of the system is conserved, what would be the angular momentum of such H_(gamma) photon ? |
|
Answer» Solution :For Balmer series of atomic spectrum of hydrogen, `(1)/(lambda)=R((1)/(2^(2))-(1)/(n^(2)))` where `n=3,4,5,6,....oo` respectively for `H_(ALPHA) , H_(beta), H_(gamma), H_(delta),....H_(oo)` lines ). Here for `H_(gamma)` lines n=5 and so minimum energy to be supplied to an electron LYING in n = 1 energy LEVEL will be equal to, `E_(5)-E_(1)=-(13.6)/(n^(2))-(13.6)` `=13.6-(13.6)/(n^(2))` `=13.6-(13.6)/(n^(2))` `=13.6-(13.6)/((5)^(2))( :.n=5)` `=13.6(1-(1)/(25))` `=(13.6xx24)/(25)` `=13.056eV` Angular momentum of emitted `H_(gamma)` photon, = decrease in angular momentum of electron `=l_(5)-l_(2)` `=(5h)/(2pi)-(2h)/(2pi)` `=(3h)/(2pi)` `=(3xx6.625xx10^(-34))/(2xx3.14)` `=3.165xx10^(-34)Js` |
|