Saved Bookmarks
| 1. |
In the ground state of hydrogenatom , itsBohr radius is given as 5.3 xx 10^(-11) m. Theatomis excitedsuch athat th radiusbecomes 21.2 xx 10^(-11) m. Find(i) thevalueof principal quantum number, and (ii) thetotal energyof theatom in thisexcited state. |
|
Answer» SOLUTION :Here `a_(0)` (or `r_(1)`) `=5.3 XX 10^(-11) m`and`r_(n)= 21.2 xx10^(-11) m` (i) `becauser_(n) = n^(2) a_(0) rArr n = sqrt((r_(n))/(a_(0)) = sqrt((21.2xx10^(-11))/(5.3 xx 10^(-11))) = 2` (ii)Total energyofatom in the excitedstate correspodingto n=2 . `E_(n) = - (13.6)/(n^(2)) eV =- (13.6)/((2)^(2)) eV = -3.4 eV` |
|