A) \[\sum\limits_{l=0}^{l=n}{2(2l+1)}\]
B) \[s\sum\limits_{l=n+1}^{l=n-1}{2(2l+1)}\]
C) \[\sum\limits_{l=0}^{l=n+1}{2(2l+1)}\]
D) \[\sum\limits_{l=0}^{l=n-1}{2(2l+1)}\]
Correct Answer: D
Solution :
Number of sub-shells (l) in nth shell = 0 to n -1. Number of orbitals in sub-shell \[=2l+1\]. Number of electrons in each orbital = 2. Hence, number of electrons in \[{{\text{n}}^{\text{th}}}\] shell. \[\sum\limits_{l=0}^{l=n-1}{2(2l+1)}\]
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