Bohr model of hydrogen — postulates, radius rₙ = 0.529 n² - Z Å, energy Eₙ = −13.6 Z² - n² eV: Question bank

Before we start, one reminder of the symbols so nothing here is a surprise:

$n$ = the whole-number label of the orbit (1, 2, 3, …); bigger $n$ = orbit farther out.
$Z$ = number of protons in the nucleus (the "pull strength").
$r_{n} = 0.529 \frac{n ^{2}}{Z}$ Å is the orbit size (note the division: $n^{2}$ over $Z$ ); $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}$ eV is its energy (again a division: $Z^{2}$ over $n^{2}$ ).
KE = kinetic energy (energy of motion, $\frac{1}{2} m v^{2}$ ); PE = potential energy (energy of position in the electric pull, here negative). Total energy $E = K E + P E$ .
A hydrogen-like species = an atom or ion with exactly one electron (H, He⁺, Li²⁺, …).

The two pictures below are worth carrying in your head while you answer — the first shows orbits shrinking as $Z$ grows, the second shows energy levels crowding toward zero as $n$ grows. The traps mostly punish forgetting one of these two shapes.

Figure — Bohr model of hydrogen — postulates, radius rₙ = 0.529 n² - Z Å, energy Eₙ = −13.6 Z² - n² eV

True or false — justify

TRUE or FALSE: As

n

increases, the electron's total energy

E_{n}

increases.

TRUE — but "increases" means moves toward zero (less negative), e.g.

- 13.6 \to - 3.4 \to - 1.51

eV. Larger numbers on the number line, even though the magnitude shrinks.

TRUE or FALSE: A larger nuclear charge

Z

makes the orbits bigger.

FALSE —

r_{n} \propto 1/ Z

, so a stronger positive pull drags the electron inward and shrinks every orbit. Only

n

(via

n^{2}

) makes orbits grow.

TRUE or FALSE: The electron in a stationary Bohr orbit is accelerating.

TRUE — it moves in a circle, and circular motion always has centripetal acceleration pointing to the nucleus. Bohr's radical move was to forbid this accelerating charge from radiating, contradicting classical electromagnetism.

TRUE or FALSE: An electron in the

n = 2

orbit is moving faster than one in

n = 1

FALSE — speed obeys

v_{n} \propto Z / n

, so higher orbits are slower. Why $Z / n$ ? From

m v r = n ℏ

we get

v = n ℏ/ (m r)

; substituting

r_{n} \propto n^{2} / Z

makes the

n^{2}

in the bottom overpower the

n

on top, leaving

v \propto Z / n

. So the

n = 2

electron moves at half the

n = 1

speed for the same

Z

TRUE or FALSE: The Bohr model correctly predicts the spectrum of helium (neutral, He).

FALSE — neutral He has two electrons, and their mutual repulsion is not in the model. Bohr works only for one-electron species. It does work for He⁺ (one electron).

TRUE or FALSE: The total energy equals minus the kinetic energy,

E = - K E

TRUE — this is the virial relation. Where the ½ comes from: force balance gives

K E = \frac{1}{2} k Z e^{2} / r

, while

P E = - k Z e^{2} / r

, so

P E = - 2 K E

. Adding them,

E = K E + P E = K E - 2 K E = - K E = \frac{1}{2} P E

. The negative sign shows the electron is bound.

TRUE or FALSE: The Bohr radius

a_{0} = 0.529

Å is the smallest possible orbit for any hydrogen-like ion.

FALSE — it's the smallest for hydrogen (

Z = 1

). For Li²⁺ (

Z = 3

) the

n = 1

radius is

0.529/3 \approx 0.176

Å, smaller still. Bigger

Z

shrinks the ground orbit.

TRUE or FALSE: Doubling

n

doubles the orbit radius.

FALSE — radius scales as

n^{2}

, so doubling

n

makes the radius four times larger, not twice.

TRUE or FALSE: The photon emitted in an

n_{2} \to n_{1}

jump carries exactly

E_{n_{2}} - E_{n_{1}}

of energy.

TRUE — energy conservation: the atom drops by

Δ E = E_{n_{2}} - E_{n_{1}}

(a positive number since it falls) and packages that into one photon of frequency

ν = Δ E / h

Spot the error

Spot the error: "

m v r = nh

, so angular momentum is quantised in units of

h

The correct condition is

m v r = nh /2 π = n ℏ

. Angular momentum comes in units of

ℏ = h /2 π

, not

h

. Dropping the

2 π

corrupts every derived constant (radius, energy, speed).

Spot the error: "Since

E_{3} = - 1.51

eV and

E_{1} = - 13.6

eV, and

1.51 < 13.6

, the

n = 3

electron has less energy."

It has smaller magnitude but greater actual energy. On the number line

- 1.51 > - 13.6

, so

n = 3

sits higher. Confusing

∣ E ∣

with

E

is the classic sign slip.

Spot the error: "PE is positive because the electron has energy stored in its orbit."

PE (potential energy) is negative for an attractive force:

P E = - k Z e^{2} / r

. A bound electron sits in a potential well below zero. We set PE = 0 at infinite separation, so being attracted means PE < 0.

Spot the error: "The electron radiates light continuously as it orbits, which is why hydrogen glows."

In a stationary orbit the electron radiates nothing (postulate 3). Light is emitted only during a jump between orbits, and it comes out as a single-colour photon — that's why the glow is sharp lines, not a smear.

Spot the error: "For He⁺, use

E_{n} = - 13.6/ n^{2}

eV like hydrogen."

You must include

Z^{2}

E_{n} = - 13.6 Z^{2} / n^{2}

. For He⁺,

Z = 2

, so every level is

4 \times

deeper:

E_{1} = - 54.4

eV, not

- 13.6

eV.

Spot the error: "As

n \to \infty

the energy goes to

- \infty

because the orbit is huge."

The opposite:

E_{n} = - 13.6 Z^{2} / n^{2} \to 0

n \to \infty

. The electron becomes barely bound and then free at

E = 0

. Large

n

means energy approaches zero from below, not diverges.

Why questions

Why is the total energy of a bound electron negative rather than positive?

We measure energy relative to a free electron at rest infinitely far away (

E = 0

). A bound electron is trapped — you'd have to add energy to free it — so it must sit below zero.

Why does the Bohr model predict discrete spectral lines instead of a continuous spectrum?

Because

n

can only be whole numbers, only certain energies

E_{n}

exist, so only certain differences

Δ E

(hence certain photon colours) are possible. Discrete levels → discrete jumps → discrete lines.

Why did Bohr need postulate 3 (no radiation in stationary states)?

Classical physics says an accelerating charge (the orbiting electron) must radiate energy, spiral in, and crash within nanoseconds. Postulate 3 simply forbids this in allowed orbits, saving the atom from collapse.

Why does the quantisation of angular momentum (

m v r = n ℏ

) lead to quantised radii and energies?

It supplies a second equation alongside the force balance. Two equations pin down both

v

and

r

uniquely for each

n

, and since

n

is discrete, the resulting

r_{n}

and

E_{n}

are discrete too.

Why does He⁺ in

n = 2

have the same energy as H in

n = 1

E \propto Z^{2} / n^{2}

. For He⁺:

Z^{2} / n^{2} = 4/4 = 1

; for H:

1/1 = 1

. The fourfold

Z^{2}

deepening exactly cancels the fourfold

n^{2}

spreading, giving identical

- 13.6

eV.

Why does the same

Δ E

formula also give the Rydberg spectral formula?

Dividing the emitted energy

Δ E = h ν

h c

converts energy into wavenumber

\overset{ν}{ˉ} = 1/ λ

. The constant bundle

13.6 eV / (h c)

is exactly the Rydberg constant

R_{H} = 1.097 \times 1 0^{7}

m⁻¹. See Hydrogen spectrum & Rydberg formula.

Why is the ground-state ionisation energy of hydrogen equal to

+ 13.6

eV?

Ionisation moves the electron from

n = 1

(

E = - 13.6

eV) to

n = \infty

(

E = 0

). The energy needed is

0 - (- 13.6) = + 13.6

eV. See Ionisation energy.

Edge cases

Edge case: What happens to

r_{n}

and

E_{n}

in the limit

n \to \infty

r_{n} \to \infty

(orbit becomes infinitely large) and

E_{n} \to 0^{-}

(approaches zero from below). This is the boundary between bound and free — the ionisation threshold.

Edge case: Is there an

n = 0

orbit?

No. Setting

n = 0

gives

r_{0} = 0

(electron on the nucleus) and

E_{0} = - \infty

, both unphysical. Quantisation starts at

n = 1

; there is no lower level to fall into, which is why hydrogen is stable.

Edge case: What is the smallest possible energy gap for a given hydrogen-like ion?

Gaps between adjacent high-

n

levels: as

n \to \infty

, consecutive

E_{n}

values crowd together toward zero, so

Δ E \to 0

. Lines bunch up near the series limit before merging into a continuum.

Edge case: If

Z

could be zero, what would the model predict?

With

Z = 0

the radius formula

r_{n} = 0.529 n^{2} / Z

blows up:

r_{n} \to \infty

(division by zero). Meanwhile

E_{n} = - 13.6 Z^{2} / n^{2} \to 0

. No nuclear charge means no Coulomb pull, no bound orbit — the model degenerates because its one force vanishes.

Edge case: For a jump within the same level (

n_{2} = n_{1}

), what photon is emitted?

None —

Δ E = 13.6 Z^{2} (1/ n_{1}^{2} - 1/ n_{1}^{2}) = 0

. No energy difference means no photon. Emission requires a drop to a strictly lower

n

Edge case: Does a heavier ion (larger

Z

) always have a smaller ground-state radius but a larger energy magnitude?

Yes.

r_{1} \propto 1/ Z

shrinks while

∣ E_{1} ∣ \propto Z^{2}

grows. Stronger pull means the electron sits tighter and deeper — smaller orbit, more tightly bound.

Recall Quick self-test before you leave

Cover the answers: (1) Sign of $E_{n}$ ? (2) Radius scaling with $n$ ? (3) Energy scaling with $Z$ ? (4) Which atoms does Bohr work for? Sign of $E_{n}$ ? ::: Negative (bound), approaching 0 from below as $n \to \infty$ . Radius scaling with $n$ ? ::: $r_{n} \propto n^{2}$ (quadruples when $n$ doubles). Energy scaling with $Z$ ? ::: The value $E_{n} = - 13.6 Z^{2} / n^{2}$ gets more negative as $Z$ grows (deeper well); its magnitude $∣ E_{n} ∣ \propto Z^{2}$ (four times deeper for He⁺). Which atoms does Bohr work for? ::: One-electron (hydrogen-like) species only: H, He⁺, Li²⁺, Be³⁺.

Connections

Parent: Bohr model — the formulas these traps test.
Rutherford model — the instability Bohr's postulate 3 patches.
Hydrogen spectrum & Rydberg formula — why discrete levels give sharp lines.
Quantum mechanical model of atom — where Bohr's picture ultimately fails.
Ionisation energy — the $n = 1 \to \infty$ edge case made physical.
de Broglie wavelength — the standing-wave reason behind $m v r = n ℏ$ .