Atomic Structure Question 371
Question: What is the angular velocity $ ( \omega ) $ of an electron occupying second orbit of $ L{i^{2+}} $ ion-
Options:
A) $ \frac{8{{\pi }^{3}}me^{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} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ v_{n}=r_{n}\omega $ where $ r_{n}=\frac{n^{2}h^{2}}{4{{\pi }^{2}}me^{2}Z.K} $ and $ v_{n}=\frac{2\pi \cdot Z\cdot e^{2}\cdot K}{n\cdot h\cdot } $
$ \therefore =\frac{2\pi Ze^{2}\cdot K}{n\cdot h\cdot }=\frac{n^{2}h^{2}}{4{{\pi }^{2}}me^{2}Z\cdot K}\times \omega ; $
$ \omega =\frac{8{{\pi }^{3}}me^{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) $
