Single Choice

The maximum number of electrons that can have principal quantum number, $$n =3$$ , and spin quantum number, $$m_{s}=-1/2$$, is:

A$$7$$
B$$8$$
C$$9$$
Correct Answer
D$$10$$

Solution

For principal quantum number $$(n=3)$$
Number of orbitals $$=n^2=9$$
So, number of electrons with $$m_s=\dfrac {1}{2}$$ will be 9

SIMILAR QUESTIONS

Atomic Structure

What is the maximum number of orbitals that can be identified with the following quantum numbers? $$n=3,l=+1, m_l=0$$

Atomic Structure

The number of orbitals associated with quantum numbers $$n=5,m_s=+\dfrac{1}{2}$$ is :

Atomic Structure

The quantum number of four electrons are given below- $$I. n=4,l=2, m_l=-2, m_s=-1/2$$ $$II. n=3,l=2, m_l=1, m_s=+1/2$$ $$III. n=4,l=1, m_l=0, m_s=+1/2$$ $$IV. n=3,l=1, m_l=1, m_s=-1/2$$ The correct order of their increasing energy will be:

Atomic Structure

The total number of orbitals associated with the principal quantum number 5 is :

Atomic Structure

What is the maximum number of orbitals that can be identified with the following quantum number? $$n = 3, l = 1, m = 0$$

Atomic Structure

The electrons identified by quantum numbers $$n$$ and $$l$$ can be placed in the order of increasing energy as: 1. $$n=4, l=1$$ 2. $$n=4, l=0$$ 3. $$n=3, l=2$$ 4. $$n=3, l=1$$

Atomic Structure

The correct set of quantum numbers for valence electron of $$Rb$$ (atomic number 37) is:

Atomic Structure

What is the lowest value of n that allows g orbitals to exist?

Atomic Structure

An electron is in one of the 3d orbitals. Give the possible values of n, l and $$\displaystyle { m }_{ l }$$ for this electron.

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