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 $
