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Answer to Question #88103 in Atomic and Nuclear Physics for Sridhar

Question #88103
If the elements with principal quantum number n>4 where not allowed in the nature the number of possible elements would be 1)60 2)32 3)2 4) 64
1
2019-04-22T11:06:42-0400

In order to answer the question, one has to calculate the number of different electronic states with the top value n = 4 for the principal quantum number (because electrons occupy these states one by one for every subsequent atom). Let us recall that every atom can be characterised by a set of electron occupation numbers written in the form

@$(n,l,m_l,m_s)@$

Taking into account that ranges for these numbers vary between

@$m_s = \pm \frac{1}{2}\\ m_l = -l, -l+1, .., 0, .., l-1,l\\ l = 0,..,n-1@$

we obtain

@$N = \sum_{n=1}^{4} \sum_{l=0}^{n-1} \sum_{m_l = -l}^{l} \sum_{m_s = \pm \frac{1}{2}} 1 = \sum_{n=1}^{4} \sum_{l=0}^{n-1} 2(2l+1) = \sum_{n=1}^{4} 2n^2 = 60 @$

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