Atomic Structure Question 228
Question: What is the angular velocity $ ( \omega ) $ of an electron occupying second orbit of $ L{i^{2+}} $ ion-
Options:
A) $ \frac{8\pi^{3} m e^{4}}{h^{3}} K^{2}$
B) $ \frac{8{{\pi }^{3}}me^{4}}{9h^{3}}K^{2} $
C) $ \frac{64}{9}\times \frac{{{\pi }^{3}}me^{4}}{h^{3}}K^{2} $
D) $ \frac{9{{\pi }^{3}}me^{4}}{h^{3}}K^{2} $
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Answer:
Correct Answer: D
Solution:
- $ v_{n}=r_{n}\omega $ where $ r_{n}=\frac{n^{2}h^{2}}{4{{\pi }^{2}}me^{2}ZK} $ and $ v_{n}=\frac{2\pi \cdot Z\cdot e^{2}\cdot K}{n\cdot h} $ $ \therefore =\frac{2\pi Ze^{2}\cdot K}{n\cdot h}=\frac{n^{2}h^{2}}{4{{\pi }^{2}}me^{2}Z\cdot K}\times \omega ; $
$ \omega =\frac{8{{\pi }^{3}}m{e}^{4}\cdot {Z}^{2}\cdot {K}^{2}}{{n}^{3}\cdot {h}^{3}} $
$ =\frac{9{{\pi }^{3}}me^{4}\cdot K^{2}}{h^{3}}(\therefore n=2and,Z=3) $
BETA
