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A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV) |
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Answer» Correct Answer - `n=6, Z=3` From the given conditions : `E_(n)-E_(2)=(10.2+17)eV=27.2 eV` and `E_(n)-E_(3)=(4.25+5.95)eV=10.2 eV` Equations (i)-(ii) gives. `E_(3)-E_(2)=17.0 eV` or `Z^(2)(13.6)((1)/(4)-(1)/(9))=17.0` `rArrZ^(2)(13.6)(5//36)=17.0` `rArr Z^(2)=9` or `Z=3` From equation (i), `Z^(2)(13.6)((1)/(4)-(1)/(9))=27.2` or `(3)^(2)(13.6)((1)/(4)-(1)/(n^(2)))=27.2` or `((1)/(4)-(1)/(n^(2)))=0.222` or `(1)/(n^(2))=0.0278` or `n^(2)=36` `:. n=6` |
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