Question

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

Answer 1

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

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

Answer: 1) 60.