The Classical Hydrogen Atom: Force Balance

Rutherford's model pictures the atom as an electrically neutral sphere: a tiny, massive, positive nucleus at the centre with electrons revolving in dynamically stable orbits. What keeps an electron in orbit? The electrostatic attraction to the nucleus supplies exactly the centripetal force required:

Fe=Fc⇒14πε0e2r2=mv2rF_e = F_c \quad\Rightarrow\quad \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} = \frac{mv^2}{r}

From this single equation (Eq. 12.2) the orbit's geometry follows:

r=e24πε0mv2r = \frac{e^2}{4\pi\varepsilon_0 mv^2}

Faster electrons orbit closer in; slower ones farther out — precisely like satellites.

Electron orbit force balance and energy relations

The Energy Bookkeeping: K, U and E

This is the most exam-productive algebra in the chapter. From the force balance, the kinetic energy:

K=12mv2=e28πε0rK = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r}

The electrostatic potential energy (attractive, hence negative):

U=−e24πε0rU = -\frac{e^2}{4\pi\varepsilon_0 r}

(The negative sign signifies the force is toward the nucleus, in the −r-r direction.) The total energy:

E=K+U=e28πε0r−e24πε0r=−e28πε0rE = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r}

The golden ratios

U=−2K,E=−K,E=U2\boxed{U = -2K, \qquad E = -K, \qquad E = \frac{U}{2}}

These hold for every circular Coulomb orbit — Rutherford's or Bohr's. If E = -13.6 eV, then instantly K = +13.6 eV and U = -27.2 eV.

Key Point: The total energy is negative, which means the electron is bound to the nucleus. If E were positive, the electron would not follow a closed orbit — it would escape. Energy must be supplied to ionise the atom.

[NEET Important] 'Kinetic, potential and total energy of the electron in hydrogen: which is negative? What are their ratios?' — asked nearly every year somewhere. K is always positive; U negative and twice as large in magnitude; E negative and equal in magnitude to K.

[JEE Tip] These are the Coulomb-force twins of the gravitational satellite relations from Class 11 — same virial structure, different force constant. Reuse the intuition.

NCERT's Worked Numbers: The Size and Speed of Hydrogen

Example 12.3's landmark calculation: experimentally, 13.6 eV is needed to separate a hydrogen atom into a proton and an electron. So E = -13.6 eV = −2.2×10−18-2.2 \times 10^{-18} J, and from E=−e28πε0rE = -\dfrac{e^2}{8\pi\varepsilon_0 r}:

r=−e28πε0E=(9×109)(1.6×10−19)22×2.2×10−18=5.3×10−11 mr = -\frac{e^2}{8\pi\varepsilon_0 E} = \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{2 \times 2.2 \times 10^{-18}} = 5.3 \times 10^{-11} \text{ m}

And from the force-balance relation:

v=e4πε0mr=2.2×106 m/sv = \frac{e}{\sqrt{4\pi\varepsilon_0 m r}} = 2.2 \times 10^6 \text{ m/s}

Two numbers worth carrying in your head: r ≈ 0.53 angstrom (this will turn out to be the Bohr radius a0a_0) and v ≈ 2.2×1062.2 \times 10^6 m/s ≈ c/137 (the fine-structure hint).

So far so good — but…

Everything above is classical and self-consistent for a frozen instant. The model even predicts sensible sizes and speeds. The problem — the fatal one — arrives the moment we remember that this orbiting electron is accelerating, and accelerating charges radiate. That catastrophe is the next section's story.

[JEE Tip] Given any ONE of E, K, U or r for a circular Coulomb orbit, you can generate all the others in two lines using E=−K=U/2=−e28πε0rE = -K = U/2 = -\dfrac{e^2}{8\pi\varepsilon_0 r}. Ratio questions are testing exactly this fluency.

Solved Examples

Example 1: Radius and speed from ionisation energy (NCERT Example 12.3)

Given that 13.6 eV separates hydrogen into a proton and electron, compute the orbital radius and electron speed.

Solution:

  1. Total energy: E = -13.6 eV = −2.2×10−18-2.2 \times 10^{-18} J.
  2. Radius: r=−e28πε0E=(9×109)(1.6×10−19)22×2.2×10−18=5.3×10−11r = -\dfrac{e^2}{8\pi\varepsilon_0 E} = \dfrac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{2 \times 2.2 \times 10^{-18}} = 5.3 \times 10^{-11} m.
  3. Speed: v=e4πε0mrv = \dfrac{e}{\sqrt{4\pi\varepsilon_0 mr}} with m = 9.1×10−319.1 \times 10^{-31} kg → v=2.2×106v = 2.2 \times 10^6 m/s.
  4. Takeaway: the experimentally known binding energy fixes both the atom's size and the electron's speed — no quantum theory needed yet.

Example 2: The K-U-E triple [NEET Rapid]

The total energy of the electron in a hydrogen atom is -13.6 eV. Write down K and U.

Solution:

  1. E = -K: K = +13.6 eV.
  2. U = 2E: U = -27.2 eV.
  3. Check: K + U = 13.6 - 27.2 = -13.6 eV = E. ✔
  4. Takeaway: this is NCERT Exercise 12.4 — the golden ratios answer it in five seconds.

Example 3: Deriving the golden ratios [Board Derivation]

Starting from the force balance, show that U = -2K and E = -K.

Solution:

  1. Force balance: e24πε0r2=mv2r⇒mv2=e24πε0r\dfrac{e^2}{4\pi\varepsilon_0 r^2} = \dfrac{mv^2}{r} \Rightarrow mv^2 = \dfrac{e^2}{4\pi\varepsilon_0 r}.
  2. Kinetic: K=12mv2=e28πε0rK = \frac{1}{2}mv^2 = \dfrac{e^2}{8\pi\varepsilon_0 r}.
  3. Potential: U=−e24πε0r=−2KU = -\dfrac{e^2}{4\pi\varepsilon_0 r} = -2K.
  4. Total: E=K+U=K−2K=−K=−e28πε0rE = K + U = K - 2K = -K = -\dfrac{e^2}{8\pi\varepsilon_0 r}. ∎

Example 4: What negative E means [Board Conceptual]

Why must the total energy of an orbiting electron be negative?

Solution:

  1. Negative total energy means the electron cannot escape to infinity — it is bound.
  2. If E were positive, NCERT notes, the electron would not follow a closed orbit around the nucleus — it would fly off.
  3. The magnitude |E| is exactly the energy needed to ionise (13.6 eV for hydrogen's ground state).
  4. Takeaway: bound ↔ negative E is a universal rule (gravity orbits included).

Example 5: Speed as a fraction of c [JEE Numerical]

Express the hydrogen electron's speed 2.2×1062.2 \times 10^6 m/s as a fraction of the speed of light.

Solution:

  1. vc=2.2×1063×108≈1137\dfrac{v}{c} = \dfrac{2.2 \times 10^6}{3 \times 10^8} \approx \dfrac{1}{137}.
  2. Answer: v ≈ c/137 — comfortably non-relativistic, justifying the classical (and later Bohr) treatment.
  3. Takeaway: 1/137 is the famous fine-structure constant — its appearance here is no accident, and JEE trivia occasionally pokes at it.

Example 6: If r doubles [JEE Scaling]

For a circular Coulomb orbit, how do K, U and E change if the orbital radius doubles?

Solution:

  1. All three scale as 1r\dfrac{1}{r}: K=e28πε0rK = \dfrac{e^2}{8\pi\varepsilon_0 r}, U=−e24πε0rU = -\dfrac{e^2}{4\pi\varepsilon_0 r}, E=−e28πε0rE = -\dfrac{e^2}{8\pi\varepsilon_0 r}.
  2. Doubling r halves each magnitude: K → K/2, U → U/2 (still negative), E → E/2 (still negative, i.e. E increases toward zero).
  3. Takeaway: larger orbits are less tightly bound — E rises toward zero as r grows, which is why outer Bohr levels crowd below E = 0.

Example 7: Orbit radius from speed [JEE Numerical]

An electron circles a proton at 1.1×1061.1 \times 10^6 m/s. Find the orbit radius.

Solution:

  1. Formula: r=e24πε0mv2r = \dfrac{e^2}{4\pi\varepsilon_0 mv^2}.
  2. Substitute: r=(9×109)(1.6×10−19)2(9.1×10−31)(1.1×106)2r = \dfrac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(9.1 \times 10^{-31})(1.1 \times 10^6)^2}.
  3. Calculate: r=2.3×10−281.1×10−18≈2.1×10−10r = \dfrac{2.3 \times 10^{-28}}{1.1 \times 10^{-18}} \approx 2.1 \times 10^{-10} m.
  4. Takeaway: halving the speed quadruples the radius (r∝1/v2r \propto 1/v^2) — check: this is 4× the 5.3 × 10⁻¹¹ m orbit of the 2.2×1062.2 \times 10^6 m/s electron. ✔

Example 8: Frequency of revolution [JEE Numerical]

Find the orbital frequency of the electron in the r = 5.3×10−115.3 \times 10^{-11} m orbit (v = 2.2×1062.2 \times 10^6 m/s).

Solution:

  1. Formula: ν=v2πr\nu = \dfrac{v}{2\pi r}.
  2. Substitute: ν=2.2×1062π×5.3×10−11\nu = \dfrac{2.2 \times 10^6}{2\pi \times 5.3 \times 10^{-11}}.
  3. Calculate: ν≈6.6×1015\nu \approx 6.6 \times 10^{15} Hz.
  4. Takeaway: file this number — the next section uses it as the classically predicted initial frequency of radiated light (NCERT Example 12.4).

Example 9: Energy to ionise from a given orbit [NEET Numerical]

An electron orbits with total energy -3.4 eV. How much energy is needed to ionise it, and what are its K and U?

Solution:

  1. Ionisation energy = |E| = 3.4 eV (raise E to zero).
  2. K = -E = +3.4 eV.
  3. U = 2E = -6.8 eV.
  4. Takeaway: -3.4 eV is hydrogen's n = 2 level; the ratios work at every level, and this exact set appears in NEET repeatedly.

Example 10: Which grows when the electron slows? [Conceptual]

In a circular Coulomb orbit, the electron moves to an orbit where its speed is smaller. What happens to r, K, U and E?

Solution:

  1. r=e24πε0mv2r = \dfrac{e^2}{4\pi\varepsilon_0 mv^2}: smaller v → larger r.
  2. K = 12mv2\frac{1}{2}mv^2 → decreases.
  3. U = -2K → increases (less negative).
  4. E = -K → increases (less negative, closer to zero — less tightly bound).
  5. Takeaway: slow, distant electrons are loosely bound; fast, close ones tightly bound. All four quantities move together through the golden ratios.