A) \[1.158\times {{10}^{-45}}kg{{m}^{2}}{{s}^{-1}}\]
B) \[1.158\times {{10}^{-48}}kg{{m}^{2}}{{s}^{-1}}\]
C) \[1.158\times {{10}^{-47}}kg{{m}^{2}}{{s}^{-1}}\]
D) \[12\times {{10}^{-45}}kg{{m}^{2}}{{s}^{-1}}\]
Correct Answer: B
Solution :
[b] 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) \[=7.3\times {{10}^{-34}}kg\text{ }m{{s}^{-1}}\] Then angular momentum will be \[\omega =\left( mv \right)\times r\] \[=(7.3\times {{10}^{-34}}kgm{{s}^{-1}})(158.7pm)\] \[=7.3\times {{10}^{-34}}kgm{{s}^{-1}}\times (158.7\times {{10}^{-12}}m)\] \[=11.58\times {{10}^{-48}}kg{{m}^{2}}{{s}^{-1}}\]You need to login to perform this action.
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