Atomic Structure Question 377

Question: The energy of an electron in first Bohr orbit of Pi- atom is $ -13.6eV $ . The energy value of electron in the excited state of $ L{i^{2+}} $ is:

Options:

A) $ -27.2eV $

B) $ 30.6eV $

C) $ -30.6eV $

D) $ 27.2eV $

Show Answer

Answer:

Correct Answer: C

Solution:

  • For $ L{i^{2+}} $ ion $ E=-13.6\times \frac{Z^{2}}{n^{2}}=-13.6\times \frac{{{(3)}^{2}}}{{{(2)}^{2}}} $

$ =\frac{-13.6\times 9}{4}=-30.6eV $



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