Exercises — Derivation of Bohr's radii and energies from electrostatics + quantization

WHAT: plug into $r_n=a_0{n}^2/Z$. WHY: hydrogen means $Z=1$, and $n=3$. $r_3=0.529\times \frac{3^2}{1}=0.529\times 9=4.76\text{A˚}$

WHAT: plug into $E_n=-13.6Z^2/n^2$. $E_2=-13.6\times \frac{1^2}{2^2}=-\frac{13.6}{4}=-3.40\text{eV}$ WHAT IT LOOKS LIKE: on the staircase (left panel of the figure), the $n=2$ rung sits one-quarter of the way down from the top — much shallower than the deep $n=1$ rung.

$r_1=a_0\frac{n^2}{Z}=0.529\times \frac{1}{2}=0.265\text{A˚}$ WHY smaller than hydrogen? The nucleus of He${}^{+}$ carries twice the charge, pulling the electron into an orbit half the size.

WHAT: invert the radius formula. With $Z=1$, $r_n=a_0{n}^2$, so $n^2=\frac{r_n}{a_0}=\frac{4.76}{0.529}=9\Rightarrow n=3.$ WHY square-root: radius grows as $n^2$, so to go backwards from radius to $n$ you take a square root. Only integer answers are physical — here it lands exactly on $n=3$.

$v_2=2.18\times 10^6\frac{Z}{n}=2.18\times 10^6\times \frac{1}{2}=1.09\times 10^6\text{m/s}$ WHY slower in outer orbits: $v\propto 1/n$. Far from the nucleus the pull is weaker, so less speed is needed to stay in circular balance.

WHAT: energy released $=E_{\text{initial}}-E_{\text{final}}=E_3-E_2$. $E_3=-\frac{13.6}{9}=-1.51\text{eV},E_2=-\frac{13.6}{4}=-3.40\text{eV}$ $\Delta E_{\text{released}}=E_3-E_2=-1.51-(-3.40)=1.89\text{eV}$ WHY positive: the electron falls to a deeper (more negative) rung, so the atom loses $1.89\text{eV}$ — carried off as a photon. On the figure's left panel, this is the vertical grey arrow from the $n=3$ rung to the $n=2$ rung.

Total energy: $E\propto 1/n^2$, so $\frac{E_2}{E_1}=\frac{1/2^2}{1/1^2}=\frac{1}{4}.$ The total energy is one-quarter as deep (less bound). Kinetic energy: by the virial result $KE=-\frac{1}{2}PE=-E$ (derived from scratch in L4.2), so $KE\propto 1/n^2$ too: $\frac{K{E}_2}{K{E}_1}=\frac{1}{4}.$ WHY the same factor: for a $1/r$ force, $KE$ and $E$ are locked together ($KE=-E$). They scale identically. The kinetic energy drops to a quarter — the outer electron is slower ($v\propto 1/n$, so $v^2\propto 1/n^2$).

WHAT: $v_n\propto Z/n$ and $r_n\propto n^2/Z$. Divide: $\omega =\frac{v_n}{r_n}\propto \frac{Z/n}{n^2/Z}=\frac{Z}{n}\cdot \frac{Z}{n^2}=\frac{Z^2}{n^3}.$ WHY this matters: the orbital frequency (how many times per second the electron circles) falls off steeply as $n^3$ — the outer orbits are both huge and slow.

WHAT: $E_1=-13.6Z^2$. Set equal and solve. $-13.6Z^2=-122.4\Rightarrow Z^2=\frac{122.4}{13.6}=9\Rightarrow Z=3.$ $Z=3$ is Li${}^{2+}$ (lithium with two electrons stripped). Radius: $r_1=a_0/Z=0.529/3=0.176\text{A˚}$. WHY square-root again: energy carries $Z^2$, so recovering $Z$ needs a square root. The tighter binding ($9\times$ hydrogen) squeezes the orbit to one-third the size.

Step 1 — WHAT: read $m{v}^2$ off the force balance. Multiply the force-balance equation $\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r^2}=\frac{m{v}^2}{r}$ by $r$: mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r}. \tag{$\ast$} WHY: kinetic energy is $\frac{1}{2}m{v}^2$, so if we can express $m{v}^2$ using Coulomb quantities we are almost done.

Step 2 — build $KE$, $PE$ and prove the virial links. Halve $(\ast )$: $KE=\frac{1}{2}m{v}^2=\frac{1}{2}\cdot \frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}.$ The Coulomb potential energy (attractive, hence negative) is $PE=-\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}$ — the very quantity listed in the toolkit above. Comparing the two: $KE=-\frac{1}{2}PE,E=KE+PE=-\frac{1}{2}PE+PE=\frac{1}{2}PE=-KE.$ WHY these hold: they are a direct consequence of the $1/r$ force — the force balance $(\ast )$ forces $m{v}^2$ to equal the magnitude of the Coulomb term, which is what makes $KE$ exactly half of $|PE|$. No external theorem needed; it fell out of the two starting equations.

Step 3 — bring in quantization to fix $r$. The radius that satisfies both starting equations is (parent-note result) $r_n=a_0{n}^2/Z$. Substitute into $KE$: $K{E}_n=\frac{Z{e}^2}{8\pi {\epsilon }_0{r}_n}=\frac{Z{e}^2}{8\pi {\epsilon }_0}\cdot \frac{\pi mZ{e}^2}{{\epsilon }_0{n}^2{h}^2}=\frac{m{Z}^2{e}^4}{8{\epsilon }_0^2{n}^2{h}^2}.$ WHY: without quantization $r$ is free and $KE$ could be anything; the quantized $r_n$ snaps it onto the ladder.

Step 4 — evaluate the constant. The prefactor $\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}=13.6\text{eV}$ (same number that appears in $E_n$), so $K{E}_n=+13.6\frac{Z^2}{n^2}\text{eV}=-E_n.$ WHAT IT LOOKS LIKE: kinetic energy is always positive (it is $\frac{1}{2}m{v}^2$). Its magnitude equals the depth of the total-energy rung. Deep rung ⇒ fast electron.

Step 1 — joules: $K{E}_1=\frac{1}{2}(9.11\times 10^{-31})(2.18\times 10^6{)}^2=2.16\times 10^{-18}\text{J}.$ Step 2 — to eV: $\frac{2.16\times 10^{-18}}{1.602\times 10^{-19}}=13.5\text{eV}\approx 13.6\text{eV}.✓$ WHY it works: two independent routes — energy formula and raw $\frac{1}{2}m{v}^2$ — land on the same $13.6\text{eV}$. That agreement is the internal consistency of the Bohr model.

Step 0 — WHAT is momentum $p$: a particle of mass $m$ moving at speed $v$ carries momentum $p=mv$ (defined in the toolkit above — think of it as "quantity of motion"). We need it because de Broglie's rule is stated in terms of $p$. Step 1 — WHAT is $\lambda$: de Broglie says a particle of momentum $p=mv$ has a wavelength $\lambda =h/p=h/(mv)$. So faster or heavier particles have shorter waves. Step 2 — the standing-wave condition: for a stable orbit, the wave must join up smoothly — an integer number of wavelengths must fit around the circumference: $2\pi r_n=n\lambda =n\frac{h}{mv}.$ Step 3 — rearrange: $mv{r}_n=\frac{nh}{2\pi }=n\hslash .$ WHY this is beautiful: the "arbitrary" quantization rule is exactly the condition that the electron-wave doesn't destructively interfere with itself. Non-integer orbits cancel out and cannot exist. See the figure below.

Step 1 — WHAT: photon carries the level difference. An electron dropping from a high rung $n_{\text{high}}$ to a low rung $n_{\text{low}}$ emits a photon of energy $\Delta E=E_{n_{\text{high}}}-E_{n_{\text{low}}}=-13.6Z^2\left(\frac{1}{n_{\text{high}}^2}-\frac{1}{n_{\text{low}}^2}\right)=13.6Z^2\left(\frac{1}{n_{\text{low}}^2}-\frac{1}{n_{\text{high}}^2}\right)\text{eV}.$ WHY: energy is conserved — the atom's lost binding energy becomes the photon's energy. The sign flip makes $\Delta E$ positive (a real emitted photon).

Step 2 — WHY use $E=hc/\lambda$: a photon's energy and wavelength are tied by $E=hc/\lambda$, so $\frac{1}{\lambda }=\frac{\Delta E}{hc}$: $\frac{1}{\lambda }=\frac{13.6\text{eV}}{hc}{Z}^2\left(\frac{1}{n_{\text{low}}^2}-\frac{1}{n_{\text{high}}^2}\right)=R{Z}^2\left(\frac{1}{n_{\text{low}}^2}-\frac{1}{n_{\text{high}}^2}\right),$ with the Rydberg constant $R=\frac{13.6\text{eV}}{hc}=\frac{13.6}{1240\text{nm}}=1.097\times 10^{-2}{\text{nm}}^{-1}=1.097\times 10^7{\text{m}}^{-1}$. This IS the empirical Rydberg formula — Bohr derived the experimentally-measured $R$.

Step 3 — H$\alpha$ number. For H ($Z=1$), $n_{\text{low}}=2$, $n_{\text{high}}=3$: $\Delta E=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=1.89\text{eV},\lambda =\frac{hc}{\Delta E}=\frac{1240}{1.89}=656\text{nm}.$ WHAT IT LOOKS LIKE: $656\text{nm}$ is deep red — precisely the H$\alpha$ line seen in hydrogen lamps and red nebulae. Bohr's toy model nails the real colour.

Step 1 — WHY mass enters at all. Look at the constants hiding in $a_0$ and $13.6\text{eV}$: $a_0=\frac{{\epsilon }_0{h}^2}{\pi m{e}^2}(\text{has orbiting mass }m\text{ on the bottom}),13.6\text{eV}=\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}(\text{has }m\text{ on top}).$ So radius $\propto 1/m$ and energy $\propto m$. Change the orbiting mass and both master numbers must be rescaled.

Step 2 — WHAT reduced mass is, and WHY we need it. In the Bohr picture we pretended the nucleus was nailed down. Really both particles orbit their common centre of mass — the nucleus wobbles too. The correct single number to put in place of $m$ is the reduced mass $\mu =\frac{m_1{m}_2}{m_1+m_2},$ the "effective" mass of the two-body dance. Here $m_1=m_{\mu }=207m_e$ (the muon) and $m_2=m_p=1836m_e$ (the proton): $\mu =\frac{m_{\mu }{m}_p}{m_{\mu }+m_p}=\frac{207m_e\cdot 1836m_e}{207m_e+1836m_e}=\frac{207\times 1836}{2043}{m}_e=186.0m_e.$ WHY it matters HERE and not for an ordinary electron: for a normal electron $m_e\ll m_p$, so $\mu =\frac{m_e{m}_p}{m_e+m_p}\approx m_e$ (correction $\sim 0.05\%$, safe to ignore). But the muon is only ~9× lighter than the proton, so the proton visibly recoils and $\mu$ drops from the naive $207m_e$ to $186.0{\mathbf{m}}_{\mathbf{e}}$ — a $10\%$ correction we must keep.

Step 3 — (a) the radius. Radius $\propto 1/m$, so replace $m_e$ by $\mu =186.0m_e$ means divide $a_0$ by $186$: $r_1=\frac{a_0}{(\mu /m_e)}\cdot \frac{n^2}{Z}=\frac{0.529\text{A˚}}{186.0}\times \frac{1^2}{1}=2.84\times 10^{-3}\text{A˚}=2.84\times 10^{-13}\text{m}.$

Step 4 — (b) the energy. Energy $\propto m$, so replace $m_e$ by $\mu$ means multiply $13.6\text{eV}$ by $186$: $E_1=-13.6\text{eV}\times (\mu /m_e)\times \frac{Z^2}{n^2}=-13.6\times 186.0\times \frac{1}{1}=-2.53\times 10^3\text{eV}=-2.53\text{keV}.$

WHAT IT MEANS: the heavy muon orbits ~186× closer to the proton and is bound ~186× more tightly. This is real, measured physics — muonic atoms are used to probe nuclear size precisely because the tiny orbit dives right down toward the nucleus.