Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²): Question bank

The symbols this page uses — defined once, up front

Every trap below leans on the same handful of symbols. Read these before you begin so nothing is ambiguous.

Definition Symbols used on this page

For plain hydrogen the formula is $\frac{1}{λ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}), n_{2} > n_{1}$ and the general hydrogen-like version (any one-electron ion of nuclear charge $Z$ ) is $\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) .$

$λ$ = the wavelength of the light emitted or absorbed (metres).
$\frac{1}{λ}$ = the wavenumber (units $m^{- 1}$ ); it is proportional to the photon's energy.
$R_{H}$ = the Rydberg constant for hydrogen, $R_{H} \approx 1.097 \times 1 0^{7} m^{- 1}$ . It is the measured value that uses the reduced mass (see below). For hydrogen the general constant $R$ equals $R_{H}$ , and $Z = 1$ , so the $Z^{2}$ factor is just $1$ .
$R_{\infty}$ = the idealised Rydberg constant assuming an infinitely heavy nucleus, $R_{\infty} \approx 1.0974 \times 1 0^{7} m^{- 1}$ . Related by $R_{H} = R_{\infty} \cdot \frac{μ}{m _{e}}$ , so $R_{H}$ is very slightly smaller.
$R$ = the generic Rydberg constant for whichever atom you are treating; for hydrogen $R = R_{H}$ , and for other one-electron nuclei it shifts slightly because the reduced mass changes. Use it with the $Z^{2}$ form above.
$Z$ = the atomic number (number of protons in the nucleus): $Z = 1$ for hydrogen, $Z = 2$ for He⁺, $Z = 3$ for Li²⁺.
$n_{1}$ = quantum number of the lower (final, smaller) level; $n_{2}$ = quantum number of the higher level.
$μ$ = the reduced mass of the electron–proton pair, $μ = \frac{m _{e} m _{p}}{m _{e} + m _{p}}$ , where $m_{e}$ is the electron mass and $m_{p}$ the proton mass.

The picture to keep in your head

Figure — Rydberg formula 1 - λ = R(1 - n₁² − 1 - n₂²)

How to read this figure. The vertical axis is energy $E_{n} = - 13.6/ n^{2}$ in electron-volts (eV); the horizontal cyan lines are the allowed orbits $n = 1, 2, 3, \dots$ , and the dashed white line at $E = 0$ is the ionisation limit ( $n = \infty$ , electron just free). Notice the lines crowd together as you climb — the top steps of the staircase are nearly level. The amber arrows are the electron falling: the tall arrow ( $3 \to 2$ ) is a big energy drop that produces a shorter-wavelength (bluer) photon; the short arrow ( $6 \to 5$ ) is a tiny drop producing a longer-wavelength (redder) photon. The whole formula just measures the length of one such arrow and reports it as a wavelength.

True or false — justify

Answer T or F, then give the real reason. A bare "true" earns nothing — the reasoning is the point.

The wavenumber

\frac{1}{λ}

is directly proportional to the energy of the emitted photon.

True. Since

E_{photon} = \frac{h c}{λ} = h c \cdot \frac{1}{λ}

, and

h c

is a constant, wavenumber and photon energy differ only by that fixed factor — they rise and fall together.

If two transitions have the same energy gap, they emit the same wavelength.

True. Wavelength is fixed entirely by

Δ E

via

λ = h c /Δ E

; the atom "forgets" which

n_{1}, n_{2}

produced the gap once the photon leaves.

Doubling the energy gap doubles the emitted wavelength.

False. It halves the wavelength, because

λ \propto 1/Δ E

. Bigger gap ⇒ more energetic (bluer) photon ⇒ shorter

λ

The Balmer series is the visible series because

n_{1} = 2

is a "special" number.

False. There is nothing magic about 2; it is a coincidence of scale that gaps landing on

n = 2

happen to match the 380–700 nm visible band. Lyman gaps are too big (UV), Paschen too small (IR).

Within a single series, lines get more widely spaced as

n_{2}

increases.

False. They crowd together and pile up at the series limit, because the

1/ n_{2}^{2}

term shrinks ever more slowly as

n_{2}

grows — the top steps of the staircase are nearly level.

The emission and absorption of the

2 \leftrightarrow 4

transition happen at exactly the same wavelength.

True. The formula uses the lower level as

n_{1}

regardless of direction; absorption and emission swap only the direction of energy flow, not the size of the gap, so

λ

is identical.

The Rydberg constant

R_{H}

is a pure mathematical number with no physical ingredients.

False.

R_{H} = \frac{μ e ^{4}}{8 ε _{0}^{2} h ^{3} c}

is built from measurable constants — reduced mass, electron charge, Planck's constant, light speed. Its agreement with spectra is a physics triumph, not a definition.

R_{H}

is exactly equal to

R_{\infty}

False.

R_{H} = R_{\infty} \cdot \frac{μ}{m _{e}}

is very slightly smaller (by about

0.05%

) because the proton is not infinitely heavy;

R_{\infty}

is the idealised limit where the nucleus never recoils.

Spot the error

Each line contains a mistake in reasoning or setup. Name it.

"For the line

n_{2} = 5 \to n_{1} = 3

, I'll write

\frac{1}{λ} = R_{H} (\frac{1}{5 ^{2}} - \frac{1}{3 ^{2}})

The levels are swapped. The lower level (3) must be

n_{1}

, so it should be

(\frac{1}{3 ^{2}} - \frac{1}{5 ^{2}})

. As written the bracket is negative, giving a nonsensical negative wavelength.

"To get the He⁺ spectrum I just use

R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

unchanged."

Missing the

Z^{2}

factor, where

Z

is the atomic number. For a hydrogen-like ion,

\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

; for He⁺,

Z = 2

scales the wavenumber up by

Z^{2} = 4

"The series limit is where

n_{2} = n_{1}

, so the bracket is zero."

Confuses limit with a null transition. The series limit is

n_{2} \to \infty

(smallest

λ

), not

n_{2} = n_{1}

. Setting

n_{2} = n_{1}

is no transition at all — no photon.

"Absorption from

n = 1

n = 3

releases a photon of wavelength given by Rydberg."

Absorption takes in a photon; it does not release one. The wavelength is still Rydberg-correct for

n_{1} = 1, n_{2} = 3

, but the energy flows into the atom, not out.

"The Rydberg formula predicts the sodium D-lines, since they are visible yellow."

The formula is only exact for one-electron systems (hydrogen and hydrogen-like ions). Sodium has many electrons and screening; its lines are not given by the simple

1/ n^{2}

formula.

"Since Lyman lines are UV, they carry less energy than Balmer's visible lines."

Backwards. UV photons are more energetic (shorter wavelength) than visible ones; Lyman gaps land on

n = 1

, the deepest well, so they are the biggest drops.

Why questions

Why must

n_{2} > n_{1}

in the formula?

So the bracket

(\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

stays positive, guaranteeing a positive wavenumber and a real, positive wavelength — a physical photon.

Why does the formula use the reduced mass

μ = m_{e} m_{p} / (m_{e} + m_{p})

rather than the electron mass

m_{e}

Because both the electron and the proton orbit their common centre of mass; the reduced mass correctly accounts for the finite (not infinite) proton mass so both particles' motion is captured.

Why do the emitted colours "crowd together" toward the blue/UV end of each series?

Because the energy levels themselves crowd (

E_{n} \propto - 1/ n^{2}

), so successive gaps shrink and pile up at the series limit — the spectral lines inherit this crowding.

Why is the shortest-wavelength line of a series called the "series limit"?

It comes from

n_{2} \to \infty

, giving the largest possible gap for that

n_{1}

and hence the smallest

λ

. Beyond it lies the continuous ionisation continuum — no more discrete lines.

Why does the same formula work for both emission and absorption?

The wavelength depends only on the size of the energy gap

Δ E

, which is fixed by

n_{1}

and

n_{2}

; direction of energy flow (out = emission, in = absorption) does not change the gap.

Why does replacing

m_{p}

with a heavier nucleus (deuteron) shift the spectrum slightly?

A heavier nucleus makes

μ

closer to

m_{e}

, raising

R

a touch toward

R_{\infty}

, so lines shift very slightly to shorter wavelengths — the observable "isotope shift".

Why does the Lyman series (

n_{1} = 1

) sit in the ultraviolet while Paschen (

n_{1} = 3

) sits in the infrared?

Landing on

n = 1

means falling into the deepest part of the well — big gaps, big photon energies, short (UV) wavelengths. Landing on

n = 3

means small gaps, low energies, long (IR) wavelengths.

Why is the series limit of the Lyman series numerically equal to

R_{H}

itself?

With

n_{1} = 1, n_{2} \to \infty

the bracket becomes

(1 - 0) = 1

, so

\frac{1}{λ} = R_{H} \cdot 1 = R_{H}

. This same energy also equals the ionisation energy of ground-state hydrogen expressed as a wavenumber.

Edge cases

What does the formula predict for

n_{2} = n_{1}

The bracket is zero, so

1/ λ = 0

, meaning

λ \to \infty

— i.e. no photon. There is no transition, so there is nothing to emit.

What happens to

λ

n_{2} \to \infty

for fixed

n_{1}

The

1/ n_{2}^{2}

term vanishes, the bracket reaches its maximum, and

λ

reaches its minimum for that series — the series limit, corresponding to the electron just barely escaping.

Can

n_{1} = 0

ever appear?

No.

n = 0

would give

1/ 0^{2}

(division by zero) and is not an allowed quantised orbit; the ground state is

n = 1

Is there an upper bound on wavelength within a given series?

Yes — the longest wavelength (smallest gap) comes from the smallest allowed jump,

n_{2} = n_{1} + 1

. Every other line in that series is bluer (shorter) than this one.

What would the spectrum look like if all levels were equally spaced instead of

\propto - 1/ n^{2}

The lines would be evenly spaced, not crowded — the pile-up at the series limit would vanish. The observed crowding is direct evidence of the

1/ n^{2}

law.

Does the formula distinguish a photon absorbed by the atom from one merely passing through?

No — the formula only fixes which wavelengths can interact. Whether a given photon is absorbed depends on it matching a gap and on an electron being available in level

n_{1}

For a hydrogen-like ion, does

R

stay exactly

R_{H}

No. Besides the explicit

Z^{2}

factor, the reduced mass changes because the nucleus differs, so the effective

R

shifts slightly from atom to atom.

Active Recall

Recall One-line summary of every trap on this page

Reveal after you've done the bank. Answer: Keep $n_{2} > n_{1}$ ; wavenumber tracks energy (so bigger gap ⇒ shorter $λ$ ); use $Z^{2}$ for ions and $μ$ (not $m_{e}$ ) for the constant; the series limit is $n_{2} \to \infty$ , not $n_{2} = n_{1}$ ; and the formula is exact only for one-electron systems.