CLASS 12 · CHAPTER 12 · MODERN PHYSICS

Atoms

Climb the Bohr energy ladder one rung at a time. Every rung up should feel like it's getting further from the one below, right? Watch the gaps instead, they collapse, not grow, the higher you climb.

01 · See it

Watch it happen

Energy level n: 1

This is the ground state, n=1, the lowest rung of the ladder. There is no level below it to fall to.

Level n

-13.6000 eV

Energy Eₙ

—

Gap to n−1

0.529 Å

Radius rₙ

Drag the slider from n=1 up to n=6. The energy levels you’re looking at aren’t evenly spaced, they’re drawn at their real E_n positions, and the gap between each level and the one below it keeps shrinking: 10.2 eV, then 1.89 eV, then 0.66 eV, then smaller still. The ladder’s rungs crowd together as you go up, converging toward E = 0 (full ionisation) without ever quite reaching it at any finite n.

02 · Derive it

Where the formula comes from

Bohr rescued the hydrogen atom from classical instability with two postulates. First, electrons move in circular orbits around the nucleus without radiating energy, so long as their angular momentum is quantised:

Quantisation of angular momentum

L = m v r = \dfrac{nh}{2\pi}, \quad n = 1, 2, 3, \ldots

Second, an atom emits or absorbs a photon only when an electron jumps between two such allowed orbits, with the photon’s energy equal to the energy difference between them, $h\nu = E_i - E_f$ .

Combining the quantisation condition with Coulomb attraction supplying the centripetal force (the full algebra isn’t needed at this level, only the results) gives the radius and energy of the n-th allowed orbit:

Orbit radius

\textcolor{#e08a1e}{r_n} = n^2 \times 0.529\ \text{\AA}

Orbit energy

\textcolor{#e08a1e}{E_n} = -\dfrac{13.6}{n^2}\ \text{eV}

The negative sign simply means the electron is bound, it takes energy to pull it free. As $n \to \infty$ , $E_n \to 0$ : the electron is just barely unbound, at rest infinitely far from the nucleus. That is the ionisation limit.

03 · Break it

Where the shortcut stops working

Intuition built on springs and ladders says equal steps should feel equally far apart, or that things get more spread out as they get bigger. Energy levels in an atom do the opposite: because $E_n \propto -1/n^2$ , the gap between consecutive levels is

\Delta E_{n \to n+1} = 13.6\left(\dfrac{1}{n^2} - \dfrac{1}{(n+1)^2}\right)\ \text{eV}

which gets smaller every time n increases, because both $1/n^2$ and $1/(n+1)^2$ are themselves shrinking, and shrinking numbers that are close together have even smaller differences. Concretely: n=1→2 releases 10.2000 eV, n=2→3 only 1.8889 eV, n=3→4 just 0.6611 eV, n=4→5 a mere 0.3060 eV, and n=5→6 only 0.1662 eV. The rungs of the ladder crowd together as you climb, converging toward (but never touching) E = 0.

This shrinking-gap behaviour is exactly why atomic spectra look the way they do. Within the Lyman series (every transition ending at n=1), the Balmer series (ending at n=2, the one with visible lines), and the Paschen series (ending at n=3), the individual spectral lines are spaced far apart for small starting n, then bunch closer and closer together as the starting level rises, piling up at a sharply defined series limit, the wavelength corresponding to ionisation from that lower level. A spectroscope pointed at hydrogen gas doesn’t show evenly spaced lines, it shows this exact crowding pattern, and that crowding is direct experimental evidence for $E_n \propto -1/n^2$ .

Students who assume “higher energy means more spread out” (true for, say, a particle in an infinite square well, where gaps actually grow with n) import that assumption here and get the direction backwards. In the Coulomb potential of an atom, it is specifically the opposite.

04 · Master it

Apply it under exam conditions

Q1. An electron in a hydrogen atom falls from n=3 to n=2. Find the energy of the emitted photon and its wavelength.

E = E_3 - E_2 = (-1.5111) - (-3.4000) = \textcolor{#e08a1e}{1.8889\text{ eV}}

Using the shortcut $E(\text{eV}) = 1240/\lambda(\text{nm})$ :

\lambda = \dfrac{1240}{1.8889} \approx \textcolor{#e08a1e}{656.6\text{ nm}}

This is the famous H-alpha line, the first (longest wavelength) line of the Balmer series, and it falls right in the red part of the visible spectrum.

Q2. Find the series limit (minimum photon energy and maximum wavelength needed to just ionise the atom) for the Balmer series, i.e. for an electron starting at n=2.

The series limit corresponds to a transition all the way up to $n=\infty$ , where $E_\infty = 0$ :

E_{\text{limit}} = E_\infty - E_2 = 0 - (-3.4000) = \textcolor{#e08a1e}{3.4000\text{ eV}}

\lambda_{\text{limit}} = \dfrac{1240}{3.4000} \approx \textcolor{#e08a1e}{364.7\text{ nm}}

Every Balmer line (n=3→2, n=4→2, n=5→2, …) has a shorter wavelength than this, and they crowd closer and closer to 364.7 nm as the starting level increases, exactly the shrinking-gap pattern from the simulation above.

05 · FAQs

Quick answers

Don't higher energy levels get further apart, not closer? Energy itself is bigger up there, shouldn't the gaps be bigger too?+

This is the single biggest trap in this chapter. It feels natural because the levels are climbing toward higher (less negative) energy, so you'd expect the spacing to climb too. But E_n = -13.6/n² eV, and the DIFFERENCE between consecutive levels shrinks as n grows: 10.2 eV from n=1 to n=2, then 1.89 eV, then 0.66 eV, then 0.31 eV, and so on. The levels bunch up and converge toward E=0 as n→∞, they never spread apart.

Why does the electron not spiral into the nucleus, if it's constantly accelerating in a circular orbit?+

Classical electromagnetism says an accelerating charge should radiate energy continuously and spiral inward, which would make every atom unstable in a tiny fraction of a second. Bohr's fix was to simply postulate that electrons in certain special orbits (quantised angular momentum, L = nh/2π) don't radiate at all, they're stable 'stationary states'. Radiation only happens when the electron jumps between these orbits, not while it sits in one.

What decides whether a transition shows up in the Lyman, Balmer, or Paschen series?+

Only the final (lower) level the electron lands on. Any transition that ends at n=1 is part of the Lyman series (all in the ultraviolet), any that ends at n=2 is Balmer (this is the one with visible lines), and any ending at n=3 is Paschen (infrared). The starting level can be anything higher than the final level.

Does the Bohr model work for any atom, or just hydrogen?+

Only hydrogen and other hydrogen-like single-electron ions (He⁺, Li²⁺, and so on) are described correctly by this model. The moment there's more than one electron, electron-electron repulsion complicates the energy levels in ways the simple Bohr picture can't capture, which is exactly why it was eventually replaced by full quantum mechanics.

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