Exercises — Hydrogen atom — solving in spherical coordinates

What we check, in order: first $0\le \ell \le n-1$, then $-\ell \le m_{\ell }\le \ell$.

• (i) $n=2\Rightarrow \ell \le 1$. Here $\ell =1$ ✓. Then $|m_{\ell }|\le 1$, and $m_{\ell }=-1$ ✓. Allowed (this is a $2p$ state).
• (ii) $n=2\Rightarrow \ell \le 1$. Here $\ell =2>1$. Illegal — $\ell$ exceeds $n-1$.
• (iii) $n=3,\ell =1\Rightarrow |m_{\ell }|\le 1$. Here $m_{\ell }=2>1$. Illegal — $m_{\ell }$ exceeds $\ell$.
• (iv) $n=1\Rightarrow \ell =0$, $m_{\ell }=0$ ✓. Allowed (the ground state $1s$).

$\ell =0\to s$, $\ell =1\to p$, $\ell =2\to d$, $\ell =3\to f$ (mnemonic Silly People Dance Funny). So $n=3,\ell =2$ is written 3d.

Where this formula comes from (WHAT is $R$?): the full wavefunction factorizes as $\psi =R(r)Y(\theta ,\varphi )$, where $R(r)$ is the radial function — it tells you how the electron's probability amplitude varies with distance $r$ from the nucleus, ignoring direction. A standard trick sets $u(r)=rR(r)$; multiplying by $r$ turns the messy radial equation into a clean 1-D Schrödinger equation. Requiring $u(r)$ to decay to zero as $r\to \infty$ (the electron can't be infinitely far out) is what forces the energy to take only discrete values: $E_1=-\frac{13.6}{1}=-13.6\text{ eV},E_2=-\frac{13.6}{4}=-3.40\text{ eV},E_3=-\frac{13.6}{9}=-1.51\text{ eV}.$ The gaps shrink as $n$ grows — the ladder crowds toward $0$ (the ionization threshold).

Step 1 (WHAT): energy released $=|E_4-E_2|$. $E_4-E_2=-13.6\left(\frac{1}{16}-\frac{1}{4}\right)=-13.6\times (-0.1875)=2.55\text{ eV.}$ Why the sign: the atom falls to a lower (more negative) level, so it loses $2.55$ eV, carried off by the photon. Step 2: $\lambda =1240/2.55=486\text{ nm}$ — the blue-green H-$\beta$ line of the Balmer series.

$m_{\ell }$ runs $-\ell ,\dots ,+\ell$, i.e. $-2,-1,0,+1,+2$: that is $2\ell +1=2(2)+1=$ 5 states.

WHY sum this: every energy level $E_n$ is shared by all its $\ell$ and $m_{\ell }$ (the Coulomb accident), so degeneracy $=$ total count of $(\ell ,m_{\ell })$ pairs. $g(n)={\sum }_{\ell =0}^{n-1}(2\ell +1).$ The terms are the odd numbers $1,3,5,\dots ,(2n-1)$ (put $\ell =0,1,\dots ,n-1$). Algebraic proof that these sum to $n^2$ (no picture needed). Write the sum twice, once forwards and once backwards, and add term by term: $S=1+3+5+\cdots +(2n-1),$ $S=(2n-1)+\cdots +5+3+1.$ Adding the two lines, each of the $n$ column-pairs gives the same total $1+(2n-1)=2n$: $2S=\underset{n\text{ terms}}{\underbrace{2n+2n+\cdots +2n}}=n\cdot 2n=2n^2\Rightarrow S=n^2.$ So $g(n)=n^2$, and $g(4)=4^2=16$. The same fact, pictured: each new odd number fills the next L-shaped border that turns an $(n-1)\times (n-1)$ square into an $n\times n$ square (red border below is the last one added).

First, what is $\mu$ here? $\mu$ is the reduced mass $\mu =\frac{m_e{m}_p}{m_e+m_p}\approx m_e$ — the single effective mass that turns the electron-plus-proton two-body problem into a one-body one. (If you prefer, read it as "the electron mass," since $\mu$ differs from $m_e$ by only $0.05\%$; nothing in this problem changes.) What the extra term is: the centrifugal barrier, a repulsive wall that grows with $\ell$ and diverges as $r\to 0$. The figure shows $V_{\text{eff}}$ for $\ell =0$ vs $\ell =2$: the barrier for $\ell =2$ (red) lifts the curve near the origin, so the well's minimum shifts outward.

Conclusion: larger $\ell$ ⇒ stronger barrier ⇒ electron is held farther from the nucleus (its probability cloud peaks at larger $r$). Physically: more angular motion means more "orbiting," which resists being pulled in — like a spinning skater refusing to be dragged to the centre.

No. For a pure $1/r$ (Coulomb) potential the energy depends only on $n$: $E_3=-1.51$ eV whatever $\ell$ and $m_{\ell }$ are.

• $n$ sets the energy (radial boundary condition).
• $\ell$ sets the size of angular momentum $|\vec{L}|=\sqrt{\ell (\ell +1)}\hslash$.
• $m_{\ell }$ sets the $z$-projection $L_z=m_{\ell }\hslash$. The last two describe orientation/shape of the cloud, not its energy. (Break the symmetry — apply a magnetic field, the Zeeman effect — and $m_{\ell }$ does shift the energy.)

WHAT ionization means: climb from $E_2$ all the way to $E_{\infty }=0$. Maximum wavelength ↔ minimum photon energy ↔ the smallest energy that just reaches the top. $E_{\text{needed}}=0-E_2=0-(-3.40)=3.40\text{ eV.}$ Why this is the minimum: any less and the electron can't escape; any more just gives it leftover kinetic energy (still ionizes, but that's not the threshold). The longest wavelength corresponds to the threshold energy: ${\lambda }_{\max }=\frac{1240}{3.40}=365\text{ nm.}$

Systematic sweep — never skip a case:

• $\ell =0$: $m_{\ell }=0$ → one state, $2s$.
• $\ell =1$: $m_{\ell }=-1,0,+1$ → three states, $2p$.

Total $=1+3=4=2^2=n^2$ ✓. The four states are $\{2s;2p_{-1},2p_0,2p_{+1}\}$.

Why $\mu$ at all: the proton isn't nailed down; both orbit the shared centre of mass. Replacing $m_e$ by $\mu$ turns the Reduced mass and two-body problem into a clean one-body problem. $\frac{\mu }{m_e}=\frac{m_p}{m_e+m_p}=\frac{1836}{1837}=\mathrm{0.999456.}$ So the true energy is $0.999456$ of the naive value — smaller by a fraction $1-0.999456=5.44\times 10^{-4}\approx 0.0544\%.$ Tiny, but measurable — it's why deuterium's spectrum sits slightly shifted from ordinary hydrogen.

Derivation (WHAT + WHY each step):

1. Each stationary state has energy $E_n=-13.6/n^2$ eV — from the radial quantization.
2. A transition conserves energy: photon energy $=|E_{n_i}-E_{n_f}|$, because the atom's lost/gained energy must go somewhere.
3. Substitute: $|\Delta E|=\left|-\frac{13.6}{n_i^2}+\frac{13.6}{n_f^2}\right|=13.6\left|\frac{1}{n_f^2}-\frac{1}{n_i^2}\right|\text{ eV.}$ Check H-$\alpha$ ($n_i=3\to n_f=2$): $|\Delta E|=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=13.6\times 0.13889=1.889\text{ eV},\lambda =\frac{1240}{1.889}=656\text{ nm.}$ That's the deep-red line you see in every hydrogen discharge tube. ✓

Reason from where each number was born:

• $m_{\ell }\in \{0,\pm 1,\pm 2,\dots \}$ came only from single-valuedness of $\Phi (\varphi )=e^{i{m}_{\ell }\varphi }$ after a full turn. That uses no feature of $V(r)$. → Survives, identical rule.
• $\ell$ came from requiring the Spherical harmonics $\Theta (\theta )$ to be finite at the poles — again pure geometry of the sphere. → Survives, identical rule $|m_{\ell }|\le \ell$.
• The radial part is where the force matters. Its solutions are labelled by a radial quantum number $n_r=0,1,2,\dots$, which simply counts how many times the radial function crosses zero (its number of "wiggles"). For the Coulomb atom the combination $n=n_r+\ell +1$ is the usual principal number, which is exactly why $\ell \le n-1$ (since $n_r\ge 0$). Change $V(r)$ and both the energy formula and this combination change: the 3D harmonic well gives evenly spaced levels $E\propto \left(2n_r+\ell +\frac{3}{2}\right)$ instead of $-1/n^2$. Here $n_r$ is the very same "number of radial wiggles" count — only how it bundles with $\ell$ into the energy differs.

Big lesson: angular quantization is a fact about spheres; only the energy pattern fingerprints the force. This is the point of the L1 note in the parent's [!mistake] box.

Step 1: orbital states per shell $=n^2$; with 2 spins each, capacity $=2n^2$.

• $n=1$: $2(1)=2$
• $n=2$: $2(4)=8$
• $n=3$: $2(9)=18$

Step 2 (compare): these match the widths $2,8,18$ that structure Quantum numbers and the periodic table. The real periodic table reorders subshells ($4s$ before $3d$) because in multi-electron atoms energy no longer depends on $n$ alone — the Coulomb degeneracy accident is broken. But the raw capacities $2n^2$ are exactly this hydrogen counting. Beautiful: the whole periodic table's rhythm is $\sum (2\ell +1)$ in disguise.

Why is $\ell$ capped at $n-1$? ::: Because the radial power series terminates into a normalizable polynomial only when $\ell \le n-1$; equivalently $n=n_r+\ell +1$ with $n_r\ge 0$. Where does $m_{\ell }$ quantization come from? ::: From $\Phi (\varphi +2\pi )=\Phi (\varphi )$ — single-valuedness after one full loop. What does energy of hydrogen depend on? ::: Only on $n$ (a Coulomb-specific accident); $\ell ,m_{\ell }$ set shape and orientation.