Saved Bookmarks
| 1. |
In a hydrogen atom, the electron and proton are bound at a distanceof about0.53 Å. (a) Esimate the potential energy of the system in eV, takingthe zero of potential energy of infinite separation of electron from proton. (b) What is the minimum work required to freethe electron, giventhat its KE in the orbit is half the magnitudeof potentialenergyobtained in (a) ? (c ) What are the anwersto (a) and (b) above, if zero of potentitalenergy is takenat 1.06 Å separation ? |
|
Answer» Solution :Here `q_(1) = 1.6xx10^(-19) C , q_(2) = +1.6xx10^(-19) C, r = 0.53 Å = 0.53xx10^(-10) m` Potentialenergy= P.E at `oo -` P.E at r `= 0 - (q_(1) q_(2))/(4pi in_(0) r) = (-9xx10^(9) (1.6xx10^(-19))^(2))/(0.53xx10^(-10)) = -43.47xx10^(-19) joul e` `= (-43.47xx10^(-19))/(1.6xx10^(-19)) EV = -27.16eV` (b) K.E. in the orbit `= (1)/(2) (27.16)eV = 13.58eV` Total energy ` = K.E. + P.E. = 13.58 - 27.16 = -13.58 eV` WORK requiredto freethe ELECTRON`= 13.58 eV` (c ) Potental energyata separation of `r_(1) (= 1.06 Å)` is `= (q_(1) q_(2))/(4pi in_(0) r_(1)) = (9xx10^(9) (1.6xx10^(-19))^(2))/(1.06xx10^(-10)) = 21.73xx10^(-19) J = 13.58 eV` Potentialenergy of the system, when ZERO of P.E. is taken at `r_(1) = 1.06 Å` is `= P.E." at "r_(1) - P.E at r = 13.58-27.16 = -13.58 eV`. By shifting the zero of potentialenergy, workrequiredto freethe electron is not affected. It CONTINUES to bethe same, beingequal to `+13.58 eV`. |
|