Atomic Structure Question 402

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}} $



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