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InterviewSolution
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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)A. 6 and 6B. 3 and 3C. 6 and 3D. 3 and 6 |
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Answer» Correct Answer - C In first case , transition is from `n^(th)` state to `2^(nd) (1 ^(st excite)` state. `:. (10.2 + 17.0) eV = 13.6 xx Z^(2) [(1)/(2^(2))/(1)/n^(2)]` In `2^(nd)` case , transition is from `n^(th)` state to `3^(rd)`state. `:. (4.25 + 5.95) eV = 13.6 xx Z^(2) [(1)/(3^(2))/(1)/(n^(2))]` Solving above equation , we get `n = 6 and Z = 3`. |
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