A hydrogen atom is in the \(4 F_{5 / 2}\) state. Find (a) its energy in units of the ground-state energy, (b) its orbital angular momentum in units of \(\hbar\), and (c) the magnitude of its total angular momentum in units of \(\hbar.\)

The principal quantum number, symbolized as 'n', is of paramount importance when understanding quantum states of atoms, specifically seen in the hydrogen atom. An atom's principal quantum number defines its energy level or 'shell'.

For a hydrogen atom, 'n' can take on any positive integer value starting from 1, moving upwards (2, 3, 4 and so on). Higher values of 'n' imply the electron resides in an outer shell, further away from the nucleus, and has greater potential energy. As 'n' increases, the energy levels become less negative, suggesting that the electron is less tightly bound to the atom's nucleus and requires less energy to be removed, or ionized. In the example provided, the hydrogen atom in the state of '4 F_{5 / 2}' has a principal quantum number of 4, meaning the electron is found in the fourth energy level.

Next comes the azimuthal quantum number, denoted by 'l', which is associated with the orbital angular momentum of an electron within an atom. Often referred to as the angular momentum quantum number, it dictates the subshell or orbital type in which the electron can be found.

Each energy level 'n' has 'n' subshells ranging from 0 to 'n-1'. These subshells are labeled as 's', 'p', 'd', 'f', and so on, corresponding to 'l' values 0, 1, 2, 3, etc. For our hydrogen atom in the '4 F_{5 / 2}' state, the term symbol indicates that 'F' corresponds to an 'l' value of 3, situating the electron in an 'f'-type orbital. The azimuthal quantum number has implications on the shape of the electron's probability cloud around the nucleus.

The orbital angular momentum of an electron within an atom is derived from the azimuthal quantum number. This quantity, rooted in quantum mechanics, deals with the momentum due to the electron's 'orbit' around the nucleus—though in quantum terms, it is more about probability distributions than actual orbits.

The magnitude of the orbital angular momentum is quantized and given by the expression \(\sqrt{l(l+1)}\hbar\), where 'l' is the azimuthal quantum number, and \(\hbar\) is the reduced Planck's constant. In the '4 F_{5 / 2}' state example, substituting \(l=3\) gives us a magnitude of the orbital angular momentum as \(\sqrt{3(3+1)}\hbar\), which underscores the quantized nature of these angular momenta in atom.

When discussing the total angular momentum quantum number, symbolized as 'J', we're referring to the combined effect of both the electron's orbital angular momentum and its intrinsic spin angular momentum. This value is a crucial aspect of an electron's quantum state.

The total angular momentum for any state is quantified by the expression \(\sqrt{j(j+1)}\hbar\) where 'j' represents the total angular momentum quantum number. It can take any half-integer or whole number value from |l-s| up to l+s, where 's' is the spin quantum number, typically \(\pm1/2\) for electrons. For our '4 F_{5 / 2}' example, the total angular momentum quantum number is \(j=\frac{5}{2}\), leading to a computed magnitude that incorporates both the electron's orbital and spin characteristics.

Regarding the energy levels of the hydrogen atom, the electrons are found at specific levels of energy, quantized and determined by the principal quantum number (n). The energy of an electron in a hydrogen atom is calculated using the formula \(-\frac{13.6}{n^2}\) eV.

This equation shows that as the principal quantum number increases, the energy becomes less negative (higher in value), but overall less in absolute magnitude. For instance, using the provided exercise states '4 F_{5 / 2}', with a principal quantum number of 4, plugging 'n=4' into the energy equation grants us an energy level of \(-\frac{13.6}{4^2}\) eV. This result, a fraction of the ground-state energy (when n=1), reflects a higher, less bound state of the electron within the atom.