Atomic Structure Question 266
Question: The $ L{i^{2+}} $ ion is moving in the third stationary state, and its linear momentum is $ 7.3\times {10^{-34}}gm{s^{-1}} $ . Angular momentum is.
Options:
A) $ 1.158\times {10^{-45}}kgm^{2}{s^{-1}} $
B) $ 1.158\times {10^{-48}}kgm^{2}{s^{-1}} $
C) $ 1.158\times {10^{-47}}kgm^{2}{s^{-1}} $
D) $ 12\times {10^{-45}}kgm^{2}{s^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Z= 3 for $ L{i^{2+}} $ ions So $ r_{n}=\frac{52.9\times n^{2}}{Z}pm $
$ n=3,Z=3 $
$ r_{n}=\frac{52.9\times {{(3)}^{2}}}{3}pm $
$ =158.7pm $ Also, linear momentum (mv) $ =7.3\times {10^{-34}}kgm{s^{-1}} $ Then angular momentum will be $ \omega =( mv )\times r $
$ =(7.3\times {10^{-34}}kgm{s^{-1}})(158.7pm) $
$ =7.3\times {10^{-34}}kgm{s^{-1}}\times (158.7\times {10^{-12}}m) $
$ =11.58\times {10^{-48}}kgm^{2}{s^{-1}} $
BETA
