A full lesson: a cartoon movie to start, then standing waves on a string, an electron wave that must close around its orbit, de Broglie's two-line proof of L = nh/2π, scaling laws for hydrogen and hydrogen-like ions, where the Bohr model fails, NEET shortcuts, problems to solve yourself, and 12 MCQs.
A string fixed at both ends keeps only waves that fit exactly. Bend the wave into a circle and the rule becomes: a whole number of wavelengths around the loop. For the electron that rule, 2πr = nλ, is exactly Bohr's L = nh/2π.
Bohr's second postulate says the electron may move only in orbits where its angular momentum is a whole-number multiple of h/2π: L = m v r = n h / 2π. It worked beautifully for hydrogen, but nobody could say why. It looked like a rule pulled out of a hat.
In 1923, ten years later, Louis de Broglie found the answer in something every Indian home has heard: a plucked string. A sitar or veena string is tied at both ends, so it can keep only those waves that fit exactly, with a node at each end. That is why it rings with clear notes instead of noise. De Broglie said the electron is also a wave, and its orbit is a string bent into a circle.
Try at home. Tie one end of a long rope (or a skipping rope) to a door handle. Shake the free end gently up and down. Most speeds give a messy wiggle. At certain speeds the rope suddenly forms one big steady loop, then two, then three. Only whole loops stand still.
"Today's single idea: a wave that must join itself can only have whole numbers of wavelengths. Put that together with λ = h/mv and Bohr's rule appears in two lines."
"Can you make the rope form two and a half loops and stay steady?"
No. The fixed end forces a node there. With two and a half loops the reflected wave meets the incoming wave out of step, and they cancel. Only whole numbers of half-wavelengths fit between two fixed ends.
A string of length L is fixed at both ends. A wave runs to one end, reflects and runs back. The forward and backward waves add. Slide the wavelength. When L is a whole number of half-wavelengths, the waves reinforce every trip: a strong standing wave. Otherwise the many reflections arrive out of step and wipe each other out.
A node at each end. Only these wavelengths: λ = 2L/n, n = 1, 2, 3, …
Nodes never move. Antinodes swing the most. The pattern stands still in space.
Nobody forces the string to choose: the misfit waves simply cancel. That is natural quantisation.
"Quantisation is not magic. Any wave trapped in a fixed region keeps only certain wavelengths. The sitar string does it, the air in a flute does it, and the electron in an atom does it."
De Broglie: a moving electron is a wave of wavelength λ = h / m v. In a circular orbit the wave has no ends; it runs round and round. After one lap it must arrive in step with itself, so the circumference must hold a whole number of wavelengths. Slide the number of waves around the orbit and watch the join (the yellow marker).
n = 1, 2, 3, … The nth orbit holds exactly n de Broglie wavelengths.
For the electron in the first Bohr orbit, v ≈ 2.19 × 10⁶ m/s, so λ ≈ 0.33 nm = 2π × 0.53 Å.
Lap after lap the wave arrives with a different phase, and the sum averages to almost nothing.
"The circle is just a string with its two ends glued together. Instead of 'a node at each end', the rule becomes 'the wave must meet itself in step'. Whole number of wavelengths, every time."
"How many de Broglie wavelengths fit in the third orbit of hydrogen? And in the ground state?"
Three in the third orbit and exactly one in the ground state. The orbit number n is the number of waves around it.
Choose an orbit. The canvas draws it to scale (r ∝ n²) with its n wavelengths, and computes everything from de Broglie's condition. Then reveal the proof one step at a time.
The circumference of the nth orbit holds n wavelengths.
Valid for speeds well below c, where momentum p = m v.
Replace λ in the first line.
Angular momentum L = m v r is quantised in units of h/2π. Bohr's rule is no longer a guess: it is the condition for a standing electron wave.
"Notice that λ = 2πr/n grows as n, because r grows as n². So the outer orbits hold more waves, and each wave is longer."
From r ∝ n² and v ∝ 1/n everything else follows. For a hydrogen-like ion (nucleus +Ze with just one electron) replace e² by Ze². Pick a quantity and a nucleus; the bars compare levels n = 1 to 5 with the hydrogen ground state.
| Quantity | Depends on n, Z | Hydrogen, n = 1 |
|---|---|---|
| Radius r | ∝ n² / Z | 0.529 Å |
| Speed v | ∝ Z / n | 2.19 × 10⁶ m/s (≈ c/137) |
| Energy E | ∝ −Z² / n² | −13.6 eV |
| Period T | ∝ n³ / Z² | 1.52 × 10⁻¹⁶ s |
| Ang. momentum L | ∝ n (no Z) | h/2π = 1.055 × 10⁻³⁴ J s |
| de Broglie λ | ∝ n / Z | 0.333 nm |
"For He⁺ the nucleus pulls twice as hard: the orbit is half as big and the energy is four times deeper, −54.4 eV. For Li²⁺ it is −122.4 eV. Angular momentum does not care about Z at all."
"Which level of He⁺ has the same energy as the ground state of hydrogen?"
E = −13.6 × 4 / n² = −13.6 eV gives n = 2. In general the level n = Z of a hydrogen-like ion matches hydrogen's ground-state energy.
The Bohr model is built for one electron around one nucleus. Choose an atom or ion. Count its electrons: if there is just one, Bohr's formulas work. If there are more, the electrons repel each other, and the model has no way to include that.
Works for hydrogen-like species. Fails for helium and every heavier neutral atom: electron–electron repulsion is ignored.
It predicts where the lines are, but not why some lines are bright and others faint.
Orbits are classical paths with a quantum rule added on. A full account of atoms needs quantum mechanics.
"Trap alert: Li⁺ looks like an ion, but it has two electrons, so Bohr fails. Li²⁺ has one electron, so Bohr works. Always count the electrons: Z minus the charge."
2πr_n = nλ. The ground state holds exactly one wavelength.
λ ∝ n/Z, not n². For hydrogen n = 2: λ = 0.67 nm.
Write these three first. Everything else is built from them.
T = 2πr/v. T₂/T₁ = 8 for hydrogen.
Jump from n₁ to n₂ changes L by (n₂ − n₁) h/2π.
Bohr is valid only if this is 1: He⁺, Li²⁺, Be³⁺. Not He, Li⁺, Ne⁺.
He⁺ n = 2 and Li²⁺ n = 3 have −13.6 eV, the same as hydrogen's ground state; their radii are 2a₀ and 3a₀.
Small compared with c, so the non-relativistic p = mv in λ = h/mv is fine.
"In NEET these are ratio questions. Write the proportionality with both n and Z, cancel, and you are done in fifteen seconds."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute. Use r₁ = 0.53 Å for hydrogen and h/2π = 1.055 × 10⁻³⁴ J s.
Find the de Broglie wavelength of the electron in the ground state of hydrogen, using de Broglie's standing-wave condition.
n = 1, so one wavelength fills the whole circumference.
Find the radius of the n = 3 orbit of hydrogen and the de Broglie wavelength of the electron in it. How does it compare with the ground state?
r_n = n² r₁, then λ = 2π r_n / n.
Find the angular momentum of the electron in the n = 3 orbit of hydrogen. By how much does it change when the electron jumps from n = 4 to n = 2?
L = n h/2π. Z does not appear.
Find the radius of the first orbit and the ionisation energy of (a) He⁺, (b) Li²⁺.
r₁ = 0.529/Z Å and E₁ = −13.6 Z² eV.
For the electron in hydrogen, compare the n = 2 orbit with the n = 1 orbit: find the ratios of (a) speed, (b) time period, (c) kinetic energy, (d) de Broglie wavelength.
v ∝ 1/n, T ∝ n³, K ∝ 1/n², λ ∝ n.
Take r₁ = 0.53 Å and E₁ = −13.6 eV for hydrogen. Write the proportionality first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "de Broglie's explanation of Bohr's postulate" | 2π r_n = n λ, λ = h/mv | Writing 2πr = λ/n or n²λ |
| "number of de Broglie waves in the nth orbit" | n | Answering n² because r ∝ n² |
| "wavelength of the electron in orbit n" | λ = 2π r_n / n ∝ n/Z | Using λ ∝ n² (that is the radius) |
| "time period / frequency of revolution" | T ∝ n³/Z², f ∝ Z²/n³ | Using n² for the period |
| "angular momentum" in a hydrogen-like ion | L = n h/2π | Multiplying by Z |
| "energy of He⁺, Li²⁺" | E = −13.6 Z²/n² eV | Using Z instead of Z² |
| "Bohr model is valid for" | one electron: Z − charge = 1 | Choosing He, Li⁺ or Ne⁺ |
| "Bohr model cannot explain" | many-electron atoms, line intensities | Thinking it fails for the hydrogen line positions |
This completes the Atoms chapter. It used the de Broglie relation λ = h/p from Dual Nature of Radiation and Matter, and the energy levels you used here power the photon calculations in Semiconductors and Nuclei. Now test yourself with the Atoms Challenge.
Values used: r₁ = 0.529 Å (0.53 Å in problems), E₁ = −13.6 eV, v₁ = 2.19 × 10⁶ m/s, h = 6.63 × 10⁻³⁴ J s, h/2π = 1.055 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg. Back to the chapter hub.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.