Atomic StructureHard
Question
The ionization energy of a hydrogen-like atom is 14.4 eV. The amount of energy released when electron jumps from the fourth orbit to the first orbit in this atom is
Options
A.13.5 eV
B.10.8 eV
C.0.9 eV
D.12.75 eV
Solution
$\Delta E = \left( \text{I.E.} \right)\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right) = 14.4 \times \left( \frac{1}{1^{2}} - \frac{1}{4^{2}} \right) = 13.5\text{ eV}$
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