Atomic StructureHard
Question
The excitation energy of an electron from second orbit to third orbit of a hydrogen-like atom or ion with +Ze nuclear charge is 47.2 eV. If the energy of H-atom in the lowest energy state is –13.6 eV, then the value of Z is
Options
A.4
B.5
C.6
D.7
Solution
$\Delta E = 13.6z^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)eV$
Or, $47.2 = 13.6 \times z^{2}\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) \Rightarrow z \approx 5$
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