InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A hydrogen-like atom (aatomic number `Z`) is in a hight excited state of quantum number `n` . This excited atom can make a transition to the first excited state by successively emitting two proton of energy `10.20 e V and 17.00 e V`, respectively. Alternatively , the atom from the same excited state can make a transition to the second excited state by successively emitting two proton of energy `4.25 e V and 5.954 e V`, respectively. Determine the value of `n and Z` (ionization energy of hydrogen atom `= 13.6 eV)` |
|
Answer» For the transition from `a` higher state `n` to the first excited state `n_(1) = 2`, the total energy released is `10.2 + 17.0 e V` or `27.2 e V` Thus , for `Delta E = 27.2 e V, n_(1) = 2 and n_(2) = n`, we have `27.2 = 13.6 Z^(2) [(1)/(4) - (1)/(n^(2))]` ...(i) For the eventual transition to the second excited state `n_(1) = 3`, the total energy released is `(4.25 + 5.95) e V or 10.2 e V` Thus, `10.2 = 13.6 Z^(2) [(1)/(9) - (1)/(n^(2))]` ...(ii) Driving Eqs. (i) and (ii), we get `(27.2)/(10.2) = (9 n^(2) - 36)/(4 n^(2) - 36)` Solving we get `n^(2) = 36` or `n = 6` Substituting `n = 6` in any of the above equations, we obtain `Z^(2) = 9` or `Z = 3` Thus, `n = 6` and `Z = 3` |
|