Chemistry-
General
Easy
Question
The velocity of electron in third excited state of Be3+ will be-
(2.188×108)ms–1
- (2.188×106)ms–1
- (2.188×106)Kms–1
- (2.188×103)Kms–1
(2.188×108)ms–1
(2.188×106)ms–1
The correct answer is: (2.188×103)Kms–1
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