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1.2.8Atomic Structure (Classical)

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

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WHAT are we deriving?

  • The allowed radii rnr_n of the electron's circular orbit.
  • The allowed total energies EnE_n of the atom.

We do it for a hydrogen-like atom: one electron, nuclear charge +Ze+Ze (H is Z=1Z=1, He+^+ is Z=2Z=2, ...).

HOW: the two ingredients

WHY these two? Ingredient 1 is pure classical physics (what keeps the electron circling). Ingredient 2 is the new quantum rule Bohr bolted on to stop the electron from spiralling into the nucleus and to explain sharp spectral lines.

Derivation of the radius rnr_n

Step 1 — Simplify Coulomb = centripetal. Multiply both sides by rr: \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r} = m v^2 \tag{A} Why this step? We want a relation between rr and vv that we can combine with quantization.

Step 2 — Get vv from quantization. From mvr=nh2πmvr=\dfrac{nh}{2\pi}: v = \frac{nh}{2\pi m r} \tag{B} Why this step? The quantization rule is the only extra info that will pin down the value of rr.

Step 3 — Substitute (B) into (A). 14πε0Ze2r=m(nh2πmr)2=n2h24π2mr2\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r} = m\left(\frac{nh}{2\pi m r}\right)^2 = \frac{n^2h^2}{4\pi^2 m r^2} Why this step? Kill vv so only rr remains.

Step 4 — Solve for rr. Cancel one rr and rearrange: Ze24πε0=n2h24π2mr  ⇒  r=ε0n2h2πmZe2\frac{Ze^2}{4\pi\varepsilon_0} = \frac{n^2 h^2}{4\pi^2 m r} \;\Rightarrow\; r = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}

Read it off: radius grows as ==n2====n^2== and shrinks as ZZ increases (a bigger nuclear charge pulls the electron in tighter).

Derivation of the energy EnE_n

Step 1 — Write total energy. E=KE+PEE = KE + PE.

  • Kinetic: KE=12mv2KE = \tfrac12 m v^2.
  • Potential (Coulomb, attractive so negative): PE=−14πε0Ze2rPE = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r}.

Step 2 — Use (A) to relate KE and PE. From (A), mv2=14πε0Ze2rmv^2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r}, so KE=12mv2=12⋅14πε0Ze2r=−12 PEKE = \tfrac12 m v^2 = \frac{1}{2}\cdot\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r} = -\frac{1}{2}\,PE Why this step? This is the virial result for a 1/r1/r force — it lets us write everything in terms of PEPE.

Step 3 — Total energy. E=KE+PE=−12PE+PE=12PE=−12⋅14πε0Ze2rE = KE + PE = -\tfrac12 PE + PE = \tfrac12 PE = -\frac{1}{2}\cdot\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r} E=−Ze28πε0rE = -\frac{Ze^2}{8\pi\varepsilon_0 r} Why this step? Energy is half the potential energy — and it's negative, meaning the electron is bound.

Step 4 — Plug in rnr_n. Substitute rn=ε0n2h2πmZe2r_n=\dfrac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}: En=−Ze28πε0⋅πmZe2ε0n2h2=−mZ2e48ε02n2h2E_n = -\frac{Ze^2}{8\pi\varepsilon_0}\cdot\frac{\pi m Z e^2}{\varepsilon_0 n^2 h^2} = -\frac{m Z^2 e^4}{8\varepsilon_0^2 n^2 h^2}

Read it off: energy is negative (bound), scales as ==Z2/n2====Z^2/n^2==, and gets closer to 00 (less bound) as nn grows.

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

Worked examples

Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine a ball on a string you're whirling around your head. The string's pull keeps it circling — for the atom, that "string" is the electric attraction of the tiny positive middle (nucleus) on the electron. But nature has a weird rule: the electron can only whirl at certain special sizes, like steps on a staircase, never in between. From those two facts alone — "the pull equals the whirl-force" and "only special step-sizes are allowed" — you can calculate exactly how big each orbit is and how much energy each one has. That's all Bohr did.

Active-recall flashcards

What two physical statements does Bohr's derivation combine?
Coulomb attraction = centripetal force (electrostatics) AND angular momentum mvr=nh/2πmvr=nh/2\pi (quantization).
Write the force-balance equation.
14πε0Ze2r2=mv2r\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2}=\frac{mv^2}{r}
State Bohr's quantization condition.
mvr=nh2π=nℏmvr = \frac{nh}{2\pi}=n\hbar, with n=1,2,3,…n=1,2,3,\dots
Formula for the nnth Bohr radius?
rn=ε0n2h2πmZe2=a0 n2/Zr_n=\frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}=a_0\,n^2/Z, with a0=0.529a_0=0.529 Å.
Formula for the nnth energy level?
En=−me48ε02h2Z2n2=−13.6 Z2/n2E_n=-\frac{me^4}{8\varepsilon_0^2 h^2}\frac{Z^2}{n^2}=-13.6\,Z^2/n^2 eV.
How does radius depend on nn and ZZ?
r∝n2/Zr\propto n^2/Z (grows with n2n^2, shrinks with ZZ).
How does energy depend on nn and ZZ?
E∝−Z2/n2E\propto -Z^2/n^2 (negative; Z2Z^2 tighter, n2n^2 looser).
Relationship between KE, PE and E for a 1/r1/r force?
KE=−12PEKE=-\tfrac12 PE, so E=12PE=−KEE=\tfrac12 PE=-KE.
Ionization energy of hydrogen ground state?
0−(−13.6)=13.60-(-13.6)=13.6 eV.
Ground-state energy of He+^+?
−13.6×22=−54.4-13.6\times2^2=-54.4 eV.
Why is EnE_n negative?
The electron is bound; energy is measured relative to the free electron at E=0E=0 (n=∞n=\infty).
Electron speed scaling in nnth orbit?
vn∝Z/nv_n\propto Z/n; v1≈2.18×106v_1\approx2.18\times10^6 m/s for H.

Connections

Concept Map

orbit governed by

Bohr postulate

multiply by r

solve for v

substitute into

solve for r

n=1, Z=1 gives

feeds into

fixes

produces

scales as n2/Z

Hydrogen-like atom charge +Ze

Coulomb = centripetal force

Angular momentum quantized mvr=nh/2pi

Relation A: Ze2/4pe0r = mv2

v = nh/2pi m r

Radius r_n = e0 n2 h2 / pi m Z e2

Bohr radius a0 = 0.529 Angstrom

Energy E_n = KE + PE

Discrete energy ladder

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Bohr ka pura idea sirf do baaton pe khada hai. Pehli baat — electron nucleus ke around ghoom raha hai, aur usko orbit me rakhne ke liye jo andar ki taraf khinchav (centripetal force) chahiye, wo aata hai Coulomb ke electrostatic attraction se. Matlab "Coulomb force = centripetal force". Ye pure classical physics hai, koi jaadu nahi. Dusri baat — Bohr ne ek naya rule daala: electron sirf un hi orbits me ghoom sakta hai jahan uska angular momentum mvrmvr ek poora multiple ho h/2πh/2\pi ka. Isko quantization kehte hain, aur yahi cheez orbits ko discrete "seedhi (staircase)" me convert kar deti hai.

In dono equations ko mila do, vv ko eliminate kar do, to seedha radius nikal aata hai: rn=a0n2/Zr_n = a_0 n^2/Z. Yaad rakho — radius n2n^2 ke saath badhta hai aur ZZ badhne pe chhota hota hai (zyada charge, zyada khinchav, chhoti orbit). Energy ke liye total energy =KE+PE= KE + PE likho. Ek trick: 1/r1/r force me KE=−12PEKE = -\tfrac12 PE, isliye total E=12PEE = \tfrac12 PE, jo negative aata hai. Radius daal do to En=−13.6 Z2/n2E_n = -13.6\,Z^2/n^2 eV.

Do cheezein exam me kaam aati hain. Ek — energy me Z2Z^2 hota hai, sirf ZZ nahi (ek ZZ Coulomb se, ek ZZ radius ke 1/Z1/Z se). Do — energy negative hai kyunki electron "bound" hai; jaise-jaise nn badhta hai, energy zero ke kareeb jaati hai, matlab electron kam bandha hua hai aur nikalna aasaan. Hydrogen ki ionization energy isiliye exactly 13.6 eV hai — n=1n=1 se n=∞n=\infty tak ka gap.

Bas isi do-rule ki jodi se saara Bohr model derive ho jaata hai. Ratne ki zarurat nahi — force balance aur quantization yaad rakho, baaki khud nikal aayega.

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Connections

Bohr model of hydrogen — postulates, radius rₙ = 0.529 n² - Z Å, energy Eₙ = −13.6 Z² - n² eVChemistry · 1.2.7Coulomb's law — force, comparison with gravityPhysics · 1.8.2Centripetal force — what provides it in various situationsPhysics · 1.2.16Conservation of angular momentum — conditionsPhysics · 1.5.12Ionization energy — first, second, …; trends and anomalies (e.g. B - Be)Chemistry · 2.2.4