In an atom, total number of electrons having quantum numbers $\mathrm{n}=4,\left|\mathrm{~m}_l\right|=1$ and $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$ is _________.

<p>To find the total number of electrons with the given quantum numbers, let's first understand what each quantum number represents:</p> <ul> <li><strong>n (Principal Quantum Number):</strong> 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.</li> <li><strong>l (Azimuthal Quantum Number or Orbital Angular Momentum Quantum Number):</strong> 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).</li> <li><strong>$m_l$ (Magnetic Quantum Number):</strong> 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 $+1$.</li> <li><strong>$m_s$ (Spin Quantum Number):</strong> 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.</li> </ul> <p>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 $m_l = -1$.</p> <p><strong>For $l = 1$ (p orbitals):</strong></p> <ul> <li>This corresponds to p orbitals, where $\left|\mathrm{~m}_l\right|=1$ can be achieved with $m_l = -1$ or $m_l = +1$.</li> <li>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.</li> <li>So, we get 2 electrons for p orbitals with the specified conditions.</li> </ul> <p><strong>For $l = 2$ (d orbitals):</strong></p> <ul> <li>This corresponds to d orbitals, where again $\left|\mathrm{~m}_l\right|=1$ indicates $m_l = -1$ or $m_l = +1$.</li> <li>Similarly, we consider only electrons with a spin quantum number of $-\frac{1}{2}$, leading us to again have 2 electrons.</li> </ul> <p><strong>For $l = 3$ (f orbitals):</strong></p> <ul> <li>f orbitals also can have $m_l = -1$ or $m_l = +1$ satisfying the $\left|\mathrm{~m}_l\right|=1$.</li> <li>With $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$, like before, this limits us to 2 electrons.</li> </ul> <p>Adding up all the electrons from the p, d, and f orbitals that meet the given criteria, we get:</p> <ul> <li>$$2\ (\mathrm{p\ orbitals}) + 2\ (\mathrm{d\ orbitals}) + 2\ (\mathrm{f\ orbitals}) = 6\ \mathrm{electrons}$$</li> </ul> <p>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 <strong>6</strong>.</p>