A full lesson: a cartoon movie to start, then why an orbiting electron should spiral into the nucleus, Bohr's three postulates, quantised angular momentum, the radius, speed and energy of the nth orbit, the n-trend machine, hydrogen-like ions, NEET shortcuts, problems to solve yourself, and 12 MCQs.
A ball on a string that winds round a pole always ends up at the pole. Classical physics says an orbiting electron should do the same within about 10⁻¹¹ s. Atoms do not collapse, so Bohr made new rules: only certain orbits are allowed, and light comes out only when the electron jumps between them.
In the school ground, Arjun ties a ball to a rope fixed at the top of a pole and hits it round and round. With every turn the rope wraps around the pole, the circle gets smaller, the ball goes faster, and finally it thuds into the pole. His cousin Riya laughs: "It always ends at the pole!"
In Rutherford's atom the electron goes round the nucleus, and classical physics says it should behave like Arjun's ball: lose energy, spiral inwards and crash into the nucleus. Yet the hydrogen in a star or in a discharge tube is perfectly stable and glows with sharp lines. In 1913 Niels Bohr solved this puzzle with three bold rules.
"When a theory predicts something that never happens, the theory has to change. Here, classical electromagnetism predicts that every atom should collapse. Bohr kept the nuclear atom but added quantum rules."
"As Arjun's ball winds inwards, it goes round faster and faster. If the ball were a charge giving out light at its rotation frequency, what colours would come out?"
Every colour in turn, sliding smoothly from one frequency to the next: a continuous spectrum. Real atoms give only sharp lines, which is the second failure of the classical picture.
An electron in a circular orbit is always accelerating towards the centre. Classical electromagnetic theory says an accelerated charge radiates energy. So the orbit should shrink, the electron should spiral into the nucleus, and the frequency of the emitted light should change smoothly. Run the classical prediction, then compare it with real hydrogen.
Losing energy, the electron spirals in. A classical estimate gives a lifetime of about 10⁻¹¹ s.
A smooth change of frequency means a continuous spectrum, not sharp lines.
Each gas emits a line fingerprint; white light through the same gas shows dark lines at the same wavelengths.
"Two predictions, two failures: the atom should collapse, and the spectrum should be continuous. Neither happens. So classical ideas cannot be applied unchanged inside the atom."
In 1913 Bohr combined classical ideas with the early quantum hypothesis. Step through the three postulates. In postulate 2, slide the orbit radius and see that only special radii have an angular momentum that is a whole number of h/2π.
An electron can revolve in certain stable orbits without emitting energy. Each orbit has a definite total energy.
Only orbits whose angular momentum is an integer multiple of h/2π are allowed. n is the principal quantum number.
Light is emitted only when the electron jumps from a higher orbit to a lower one, as one photon. Absorbing a photon of exactly the right energy lifts it up.
"Postulate 1 says no radiation in an allowed orbit. Postulate 2 tells you which orbits are allowed. Postulate 3 says light comes only from jumps. Emission needs a drop to lower energy; going up needs absorption."
Combine the force equation of Lesson 2 with Bohr's angular momentum rule and everything about the orbit follows. Pick the orbit number n and watch the radius, the speed and the energy level.
Force balance (classical) and angular momentum rule (quantum).
Divide the first by the second. The speed falls as 1/n.
a₀ is the Bohr radius. Second orbit 4a₀, third 9a₀.
Lowest (most negative) for n = 1, rising towards 0 as n grows. The minus sign means the electron is bound.
"Radius n squared, speed one over n, energy minus one over n squared. Remember also: minus 0.85 eV is a higher energy than minus 3.4 eV, even though 0.85 is the smaller number."
Most NEET questions on Bohr's model are ratio questions. Choose a quantity and two orbits; the machine plots the quantity for n = 1 to 6 and computes the ratio for you. Then try to beat it in your head.
| Quantity | Depends on n as | n = 1 value (hydrogen) |
|---|---|---|
| radius r | n² | 0.53 Å |
| speed v | 1/n | 2.2 × 10⁶ m/s |
| energy E | −1/n² | −13.6 eV |
| ang. momentum L | n | h/2π = 1.05 × 10⁻³⁴ J s |
| time period T = 2πr/v | n³ | 1.5 × 10⁻¹⁶ s |
"Write the power of n first, then the ratio is just arithmetic. Radius from n equals 1 to 3: three squared, nine times. Energy: minus 13.6 becomes minus 13.6 over 9, closer to zero."
"Is the energy needed to take the electron from n = 1 to n = 2 bigger or smaller than from n = 2 to n = 3?"
Bigger. 1 → 2 needs 13.6 − 3.4 = 10.2 eV; 2 → 3 needs 3.4 − 1.51 = 1.89 eV. The levels crowd together as n grows. These gaps give the lines of the hydrogen spectrum.
Bohr's model works for any atom with a single electron: hydrogen, singly ionised helium He⁺ (Z = 2), doubly ionised lithium Li²⁺ (Z = 3). A bigger nuclear charge pulls the electron closer and binds it more tightly. Very hot stars contain He⁺, and astronomers see its lines in their spectra.
He⁺ ground state: 0.265 Å, half of hydrogen's.
He⁺ ground state: −54.4 eV. Li²⁺: −122.4 eV.
The electron in Li²⁺ (n = 1) moves at 6.6 × 10⁶ m/s.
Also L ∝ n, T ∝ n³/Z², frequency of revolution ∝ Z²/n³.
For n = 1: K = 13.6 eV, U = −27.2 eV, E = −13.6 eV.
Third orbit: 3h/2π, not 9h/2π. L grows as n, not n².
3 → 2 gives 1.89 eV (656 nm, red); 2 → 1 gives 10.2 eV (122 nm, ultraviolet).
Hydrogen from n = 1: 13.6 eV; from n = 2: 3.4 eV.
Closer to zero means higher. The ground state has the lowest energy.
"Learn the power table and you can answer any Bohr ratio in ten seconds. And never forget the sign: the energy is negative, and it increases towards zero as n increases."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute. Use a₀ = 0.53 Å, h = 6.63 × 10⁻³⁴ J s.
For the electron in the n = 3 orbit of hydrogen, find r, E, v and L.
r = n²a₀, E = −13.6/n², v = 2.2 × 10⁶/n, L = nh/2π.
In which orbit of hydrogen is the radius 8.48 Å?
n² = r / a₀.
How much energy is needed to lift the electron of hydrogen from n = 1 to n = 2? What photon is emitted when it falls back?
ΔE = E₂ − E₁; λ (nm) ≈ 1240 / E (eV).
Find the radius and energy of the ground state of He⁺ (Z = 2).
r = 0.53 n²/Z, E = −13.6 Z²/n².
Classically, an electron at r = 0.53 Å with v = 2.2 × 10⁶ m/s would radiate at its frequency of revolution. Find it, and explain why the classical spectrum would be continuous.
ν = v / 2πr.
Write the power of n first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "classical theory predicts" | accelerated charge radiates → spiral in, continuous spectrum | Saying the orbit is stable |
| "angular momentum in nth orbit" | L = nh/2π | Writing n²h/2π or nh |
| "radius of nth orbit" | r = 0.53 n²/Z Å | Using n instead of n² |
| "speed in nth orbit" | v = 2.2 × 10⁶ Z/n m/s | Thinking outer electrons move faster |
| "energy of nth level" | E = −13.6 Z²/n² eV | Dropping the minus sign or the Z² |
| "which energy is higher" | closer to zero = higher | Picking −13.6 eV because 13.6 is bigger |
| "photon emitted when" | hν = E_i − E_f, E_i > E_f | Saying emission happens on jumping up |
| "ratio of time periods" | T ∝ n³ | Using n² because r ∝ n² |
Lesson 4 turns the energy ladder into the hydrogen spectrum: every line is a jump between two levels, grouped into the Lyman, Balmer and Paschen series. Lesson 5 explains Bohr's strange angular momentum rule with de Broglie's electron waves.
Values used: a₀ = 0.53 Å, E₁ = −13.6 eV, v₁ = 2.2 × 10⁶ m/s, h = 6.63 × 10⁻³⁴ J s, h/2π = 1.05 × 10⁻³⁴ J s, hc ≈ 1240 eV nm.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.