Hydrogen emission spectrum — Lyman, Balmer, Paschen, Brackett, Pfund: Question bank

Before we start, one reminder of the master relation so every answer stays anchored:

True or false — justify

The spectrum is discrete because the atom is small

False — size is irrelevant; discreteness comes from quantized energy levels (Quantization of Energy). Only fixed

Δ E

jumps exist, so only fixed

λ

photons appear.

Every line in the Balmer series is visible to the human eye

False — only the first few (H-α red, H-β blue, H-γ violet) are visible; as

n_{2}

climbs the lines crowd toward the series limit at 365 nm, which is UV, not visible.

The Lyman series limit equals the ionization energy of hydrogen from the ground state

True — the limit is the

n = 1 \to \infty

transition run backwards, so its photon energy equals

13.6 eV

, exactly the Ionization Energy of Hydrogen.

A single hydrogen atom emits all lines of all series at once

False — one atom makes one photon per fall. A whole glowing sample contains billions of atoms falling by different routes, and the spectrum is the collective record of all of them.

The first line of a series has the largest wavelength within that series

True — the first line is the

n_{2} = n_{1} + 1

jump, the smallest energy drop, hence the longest

λ

; the series limit (

n_{2} \to \infty

) is the shortest.

Paschen, Brackett and Pfund all lie in the infrared

True — they land on

n_{1} = 3, 4, 5

, shallow parts of the energy well, so their drops are small, giving small

Δ E

and long (IR) wavelengths (Electromagnetic Spectrum).

Doubling

n_{2}

roughly halves the emitted wavelength

False —

\overset{ν}{ˉ}

depends on

1/ n_{2}^{2}

, not

1/ n_{2}

, and the

1/ n_{1}^{2}

term dominates once

n_{2}

is large, so

λ

barely changes near the series limit.

Wavenumber

\overset{ν}{ˉ}

is directly proportional to photon energy

True — since

Δ E = h c / λ = h c \overset{ν}{ˉ}

, and

h c

is constant,

Δ E \propto \overset{ν}{ˉ}

; that's why chemists compare energies using

\overset{ν}{ˉ}

directly.

Spot the error

"Balmer means the electron starts at level 2"

Wrong — the series is named by the destination

n_{1}

, so Balmer means the electron lands on level 2, starting from any

n_{2} > 2

\overset{ν}{ˉ} = R_{H} (\frac{1}{n _{2}^{2}} - \frac{1}{n _{1}^{2}})

for emission"

Sign flipped — with the larger

n_{2}

first the bracket is negative, giving negative

\overset{ν}{ˉ}

. Emission needs the ==smaller

n

first==:

\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}} > 0

"A bigger

n_{2} \to n_{1}

jump gives a bigger wavelength"

Backwards — bigger jump means bigger

Δ E

, and since

λ \propto 1/Δ E

, the wavelength gets smaller. This is the most common trap in the whole topic.

"The Rydberg formula gives energy in joules directly"

No — it gives

1/ λ

(a wavenumber in

m^{- 1}

). To get energy you multiply by

h c

Δ E = h c \overset{ν}{ˉ}

(Photon Energy and Planck Relation).

"Since

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

is negative, the emitted photon has negative energy"

The levels are negative (bound), but the photon carries the difference

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

, which is positive because

n_{2} > n_{1}

makes

E_{n_{2}}

less negative. Emitted photons always have positive energy.

"The series limit is where the first line of the next series begins"

No — the series limit is

n_{2} \to \infty

landing on the same

n_{1}

; the next series has a different

n_{1}

entirely. Their wavelength ranges can even overlap but they are separate families.

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

works for helium and lithium ions too"

Only after scaling by

Z^{2}

(nuclear charge). Bare

R_{H}

is for hydrogen; Rydberg Constant and Spectra of Hydrogen-like Ions shows

\overset{ν}{ˉ} = R_{H} Z^{2} (\dots)

for hydrogen-like ions.

Why questions

Why does hydrogen glow in specific colours and not a smooth rainbow?

Because electron energies are quantized (Bohr Model of the Atom), only fixed energy differences exist, so only fixed-

λ

photons come out — sharp lines instead of a continuum.

Why does the Lyman series have the shortest wavelengths of all series?

It lands on

n_{1} = 1

, the deepest point of the energy well, so every drop into it is the largest possible

Δ E

, giving the smallest

λ

(UV).

Why do lines within a series bunch up as

n_{2}

grows?

Because energy levels get closer together as

n

increases (

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

flattens out), so successive jumps differ less and less, crowding the lines toward the series limit.

Why can we cancel

h c

from both sides when deriving the Rydberg formula?

Both

Δ E

(as

h c / λ

) and the Bohr energy expression carry a common factor

h c

; dividing it out leaves a pure geometric relation between

1/ λ

and the

n

-values.

Why is the ionization energy tied to the Lyman limit and not, say, the Balmer limit?

Ionization from the ground state is a

1 \to \infty

transition, which is precisely the Lyman series limit (

n_{1} = 1

). The Balmer limit corresponds to removing an electron already sitting at

n = 2

, which costs only

13.6/4 = 3.4 eV

Why is emission a "fall" while absorption is a "climb"?

Emission releases energy as the electron drops to lower

n

(photon out); absorption takes in a photon of exactly the right energy to lift the electron to higher

n

. Same energy gaps, opposite direction.

Edge cases

What happens to

\overset{ν}{ˉ}

n_{2} \to \infty

for a fixed

n_{1}

The

1/ n_{2}^{2}

term vanishes, so

\overset{ν}{ˉ} \to R_{H} / n_{1}^{2}

— the series limit, the shortest wavelength of that series and a hard boundary the lines approach but never cross.

Is a transition from

n_{2} = 4

n_{1} = 4

allowed?

No — that is no jump at all (

Δ E = 0

), so no photon. A real transition needs

n_{2} > n_{1}

strictly; equal levels are degenerate cases with nothing emitted.

Can the Rydberg formula ever give a negative wavelength?

Only if you wrongly put the larger

n

first. Physically

λ > 0

always; a negative result is a sign-error signal, not a real photon.

What does a transition with

n_{2} < n_{1}

describe?

That is absorption written backwards — the electron would be climbing, not falling, so it absorbs rather than emits. The emission formula assumes

n_{2} > n_{1}

; reversing them describes the reverse process.

Beyond the series limit, is the spectrum empty?

Not quite — past the limit the electron is free (unbound), and it can carry any leftover kinetic energy, so a faint continuous absorption/emission edge appears there rather than discrete lines.

Does the Paschen series ever overlap the Balmer wavelength range?

Their ranges are distinct in practice (Paschen is IR, Balmer visible/near-UV), but the formula allows nearby values near limits; the key is they are different families defined by different

n_{1}

, never merged.

Connections

Bohr Model of the Atom — the quantized levels behind every trap here.
Quantization of Energy — the reason the spectrum is discrete.
Photon Energy and Planck Relation — converts $\overset{ν}{ˉ}$ to $Δ E$ .
Rydberg Constant and Spectra of Hydrogen-like Ions — the $Z^{2}$ generalisation.
Ionization Energy of Hydrogen — equals the Lyman series limit.
Electromagnetic Spectrum — places each series in UV/visible/IR.