JEE Main 2024 — Structure Of Atom Question with Solution

To find the total number of electrons with the given quantum numbers, let's first understand what each quantum number represents:

• n (Principal Quantum Number): This specifies the energy level or shell of the electron in the atom. Here, $$n=4$$ means we are dealing with the fourth energy level.
• l (Azimuthal Quantum Number or Orbital Angular Momentum Quantum Number): This defines the shape of the orbital and is dependent on the value of n. It can take values from 0 to $$n-1$$. For $$n=4$$, $$l$$ can be 0 (s orbital), 1 (p orbital), 2 (d orbital), or 3 (f orbital).
• $m_l$ (Magnetic Quantum Number): This describes the orientation of the orbital in space and can take values from $$-l$$ to $$+l$$, including zero. The condition $$\left|{\mathrm{m}}_l\right|=1$$ implies $$m_l=-1$$ or $$+1$$.
• $m_s$ (Spin Quantum Number): This indicates the spin direction of the electron, with possible values of $$+\frac{1}{2}$$ or $$-\frac{1}{2}$$. Here, $${\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$$ specifies electrons with a downward spin.

Given $$\left|{\mathrm{m}}_l\right|=1$$, this directly implies that we are not considering s orbitals (which have $$l=0$$ and hence $$m_l=0$$). We are considering p, d, and f orbitals which can have $$m_l=1$$ or $$m_l=-1$$.

For $l=1$ (p orbitals):

• This corresponds to p orbitals, where $$\left|{\mathrm{m}}_l\right|=1$$ can be achieved with $$m_l=-1$$ or $$m_l=+1$$.
• With $${\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$$, we only take half of the available m_l values since each orbital can host two electrons with different spin quantum numbers.
• So, we get 2 electrons for p orbitals with the specified conditions.

For $l=2$ (d orbitals):

• This corresponds to d orbitals, where again $$\left|{\mathrm{m}}_l\right|=1$$ indicates $$m_l=-1$$ or $$m_l=+1$$.
• Similarly, we consider only electrons with a spin quantum number of $$-\frac{1}{2}$$, leading us to again have 2 electrons.

For $l=3$ (f orbitals):

• f orbitals also can have $$m_l=-1$$ or $$m_l=+1$$ satisfying the $$\left|{\mathrm{m}}_l\right|=1$$.
• With $${\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$$, like before, this limits us to 2 electrons.

Adding up all the electrons from the p, d, and f orbitals that meet the given criteria, we get:

• $$2(\mathrm{p}\mathrm{orbitals})+2(\mathrm{d}\mathrm{orbitals})+2(\mathrm{f}\mathrm{orbitals})=6\mathrm{electrons}$$

Therefore, the total number of electrons having quantum numbers with $$\mathrm{n}=4,\left|{\mathrm{m}}_l\right|=1$$ and $${\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$$ is 6.