Flame tests — characteristic colours: Worked examples

Figure — Flame tests — characteristic colours

Look at the figure above: the left ladder has widely-spaced rungs (big $Δ E$ ) → the falling electron shoots out a short, tightly-packed wave (blue). The right ladder has close rungs (small $Δ E$ ) → a long, stretched wave (red). This single picture is the engine behind every example below.

The scenario matrix

Every flame-test question is one of these cells. Each worked example is tagged with the cell(s) it covers.

Cell	Case class	What makes it tricky	Example
A	Given $λ$ → find $Δ E$	unit conversion nm→m	Ex 1
B	Given $Δ E$ → find $λ$ + name the colour	reverse direction	Ex 2
C	Compare two metals (small vs large gap)	which is redder/bluer	Ex 3
D	Degenerate: two metals, same colour	flame test alone fails	Ex 4
E	Zero / limiting input	$Δ E \to 0$ or $λ \to \infty$	Ex 5
F	Contamination twist	persistent vs flash yellow	Ex 6
G	Procedure error	wrong acid / wrong flame region	Ex 7
H	Real-world word problem	fireworks / lab unknown	Ex 8
I	Exam twist: frequency form + multiple lines	$E = h ν$ , two wavelengths	Ex 9

Cell A — given wavelength, find the energy gap

Worked example Ex 1 (Cell A)

Sodium's flame is golden-yellow at $λ = 589 nm$ . Find the energy gap $Δ E$ (in joules) of the transition responsible.

Forecast: guess first — will $Δ E$ be a big number or a tiny one? (Photon energies are minuscule, so expect something like $1 0^{- 19}$ J.)

Convert nm to m: $589 nm = 589 \times 1 0^{- 9} m = 5.89 \times 1 0^{- 7} m$ .
- Why this step? $h$ and $c$ are in SI units (J·s and m/s). Mixing nanometres in would give nonsense — always match units first.
Plug into $Δ E = \frac{h c}{λ}$ : $Δ E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{5.89 \times 1 0 ^{- 7}}$
- Why this step? The boxed relation directly maps a known $λ$ to its gap.
Compute the top: $6.626 \times 1 0^{- 34} \times 3 \times 1 0^{8} = 1.988 \times 1 0^{- 25} J\cdotm$ .
Divide: $\frac{1.988 \times 1 0 ^{- 25}}{5.89 \times 1 0 ^{- 7}} = 3.37 \times 1 0^{- 19} J$ .

Verify: Units: $\frac{J\cdotm}{m} = J$ ✓. Magnitude $\sim 1 0^{- 19}$ J, exactly the visible-light ballpark ✓. Our forecast (tiny) was right.

Cell B — given energy gap, find wavelength & name the colour

Worked example Ex 2 (Cell B)

A metal ion has an excitation gap of $Δ E = 3.06 \times 1 0^{- 19} J$ . Find $λ$ , and say what colour and (roughly) which metal.

Forecast: This gap is smaller than sodium's ( $3.37 \times 1 0^{- 19}$ ). Smaller gap → longer $λ$ → redder than yellow. Predict red-ish.

Use $λ = \frac{h c}{Δ E}$ : $λ = \frac{1.988 \times 1 0 ^{- 25}}{3.06 \times 1 0 ^{- 19}}$
- Why this step? We know the gap and want the colour's wavelength — the rearranged form isolates $λ$ .
Compute: $λ = 6.50 \times 1 0^{- 7} m = 650 nm$ .
Name it: $650 nm$ sits in the red band. From the parent table, ~650 nm crimson-red = Sr²⁺ (strontium).
- Why this step? Converting the physics answer back into the chemistry identity is the whole point of a flame test.

Verify: Feed $λ = 6.50 \times 1 0^{- 7}$ back into $Δ E = h c / λ$ : $1.988 \times 1 0^{- 25} /6.50 \times 1 0^{- 7} = 3.06 \times 1 0^{- 19}$ J ✓. And $650 > 589$ nm, so it is redder than sodium ✓ — forecast confirmed.

Cell C — compare two metals

Worked example Ex 3 (Cell C)

Barium glows apple-green ( $λ \approx 515 nm$ ); lithium glows crimson ( $λ \approx 671 nm$ ). Which metal's transition has the larger energy gap?

Forecast: $λ$ and $Δ E$ are inversely related, so the shorter wavelength wins the larger gap. Barium is shorter — predict Ba.

Recall the inverse law: $Δ E = \frac{h c}{λ}$ — bigger $λ$ downstairs makes $Δ E$ smaller.
- Why this step? We don't even need full numbers; the structure of the equation answers "which is bigger?".
Compute both (numerator $h c = 1.988 \times 1 0^{- 25}$ ):
- Ba: $\frac{1.988 \times 1 0 ^{- 25}}{515 \times 1 0 ^{- 9}} = 3.86 \times 1 0^{- 19} J$ .
- Li: $\frac{1.988 \times 1 0 ^{- 25}}{671 \times 1 0 ^{- 9}} = 2.96 \times 1 0^{- 19} J$ .
Compare: $3.86 > 2.96$ → barium has the larger gap.

Verify: The blue-ward colour (green 515) beat the red-ward colour (crimson 671) for gap size — consistent with figure s01 (blue = wide ladder) ✓. Forecast confirmed.

Worked example Ex 4 (Cell D)

An unknown salt gives a green flame. Name the two candidates and give the decisive follow-up test.

Forecast: Green is not unique — remember two metals live there.

List green emitters: from the parent table, green ⇒ Ba²⁺ (515 nm) or Cu²⁺ (~510–525 nm).
- Why this step? This is the degenerate cell: the flame test cannot separate them because their wavelengths overlap — nearly the same $Δ E$ , so nearly the same colour.
Escalate to a wet test (see Wet Tests for Cations): add dilute H₂SO₄.
- Why this step? When one tool saturates (both green), switch tools. Solubility differs even when colour doesn't.
Read the result: Ba²⁺ + SO₄²⁻ → white BaSO₄ precipitate; Cu²⁺ gives no such white precipitate.

Verify: Logic sanity check — a flame test is a screening tool; a degenerate colour is exactly when you must confirm. Behaviour matches the parent's "green → Ba or Cu" warning ✓.

Cell E — zero / limiting input

Worked example Ex 5 (Cell E)

Two thought-experiments. (a) As a transition's gap $Δ E \to 0$ , what happens to $λ$ and can you see the light? (b) What would $Δ E = 0$ mean physically?

Forecast: $λ = h c /Δ E$ has $Δ E$ in the denominator — shrinking it must blow $λ$ up.

Limit (a): as $Δ E \to 0^{+}$ , $λ = \frac{h c}{Δ E} \to \infty$ .
- Why this step? Dividing a fixed positive numerator by an ever-smaller positive number grows without bound — that's the definition of this limit.
Interpret the light: huge $λ$ means infrared → radio, far past the red edge (~700 nm) of vision. Your eye sees nothing.
- Why this step? This is why metals with tiny gaps (many transition/heavy metals) give faint or invisible flame colours — the parent note's "gaps in the UV/IR" remark, made precise.
Case (b) $Δ E = 0$ : the "high" and "low" levels coincide — there is no transition at all, so no photon, $λ$ undefined (division by zero).
- Why this step? Covers the truly degenerate zero-input: no gap, no emission.

Verify: Check the trend numerically at three shrinking gaps and confirm $λ$ strictly increases; the exact $Δ E = 0$ case is a division-by-zero, i.e. no physical emission ✓.

Cell F — contamination twist

Worked example Ex 6 (Cell F)

A student tests a salt. The flame flashes yellow for about half a second, then turns lilac and stays. What metal(s) are truly present?

Forecast: "Yellow = sodium" is the classic trap. Was the yellow persistent?

Judge the yellow: it lasted <1 s then vanished → this is the signature of sodium contamination (sweat/glass/dust), NOT the sample.
- Why this step? The parent rule: only persistent yellow (several seconds) counts as genuine Na.
Read the persistent colour: the steady lilac is the real signal → K⁺ (potassium).
- Why this step? The lasting colour reflects the bulk sample vaporising; the flash is trace impurity burning off first.
Confirm K: view through blue cobalt glass — it absorbs stray Na-yellow, and the lilac (often crimson-pink through the glass) persists.

Verify: Consistent with the parent's contamination rule and cobalt-glass logic ✓. Answer: the sample is a potassium salt; the yellow flash was contamination.

Cell G — procedure error

Worked example Ex 7 (Cell G)

A student makes the paste with concentrated H₂SO₄ instead of HCl and gets a barely-visible colour. Identify the two procedural faults if they also placed the wire in the luminous (yellow) flame tip.

Forecast: Two independent things can wash out a colour — the salt not vaporising, and the flame masking it.

Wrong acid: H₂SO₄ makes metal sulphates, which are non-volatile / high-boiling, so little vapour reaches the flame → weak colour.
- Why this step? Colour intensity depends on how much metal vaporises; volatility is set by the compound formed. HCl → volatile chlorides is the fix.
Wrong flame region: the luminous (yellow) part of a Bunsen flame is cooler and already coloured, so it both under-excites and masks the true colour.
- Why this step? You must use the hot, colourless non-luminous (blue) region so the emitted colour is neither swamped nor tinted.
Corrected procedure: paste with conc. HCl, introduce into the non-luminous flame.

Verify: Both fixes match the parent's procedure and its "sulphate mistake" callout ✓. Two faults, two fixes.

Cell H — real-world word problem

Worked example Ex 8 (Cell H)

A firework designer wants a deep-red star at $λ \approx 671 nm$ and separately a bright-yellow burst at $λ \approx 589 nm$ . Which metal salts should go in each, and how much more energy does each yellow photon carry than each red one?

Forecast: Red 671 nm → lithium; yellow 589 nm → sodium. Yellow is shorter $λ$ , so more energy per photon.

Match colours to metals (parent table): $671 nm$ crimson-red → Li⁺; $589 nm$ golden → Na⁺.
- Why this step? The word problem is just the flame table read forwards — colour tells the metal.
Energy per photon $= h c / λ$ :
- Red (Li): $\frac{1.988 \times 1 0 ^{- 25}}{671 \times 1 0 ^{- 9}} = 2.96 \times 1 0^{- 19} J$ .
- Yellow (Na): $\frac{1.988 \times 1 0 ^{- 25}}{589 \times 1 0 ^{- 9}} = 3.37 \times 1 0^{- 19} J$ .
- Why this step? "How much more energy" needs actual $Δ E$ values, not just wavelengths.
Take the ratio: $\frac{3.37 \times 1 0 ^{- 19}}{2.96 \times 1 0 ^{- 19}} = 1.14$ . Each yellow photon carries about 14% more energy than each red one.

Verify: Shorter $λ$ → higher energy, so ratio $> 1$ ✓; both energies are the ~ $1 0^{- 19}$ J visible-light scale ✓. Forecast confirmed.

Cell I — exam twist: frequency form + two lines

Worked example Ex 9 (Cell I)

Potassium famously shows two lines: ~766 nm and ~405 nm. (a) Find the frequency of the 766 nm line using $c = λ ν$ . (b) Which line is more energetic, and what colours are they?

Forecast: Frequency $= c / λ$ (rearranged wave equation). Shorter $λ$ (405 nm) → higher $ν$ → more energy.

Rearrange the wave equation: from $c = λ ν$ , $ν = \frac{c}{λ}$ .
- Why this step? The question hands us $λ$ and $c$ and asks for $ν$ — this isolates it. (This is the same $ν$ that plugs into $E = h ν$ , see Photoelectric Effect.)
Convert & compute (766 nm line): $λ = 766 \times 1 0^{- 9} m$ , $ν = \frac{3 \times 1 0 ^{8}}{766 \times 1 0 ^{- 9}} = 3.92 \times 1 0^{14} Hz .$
Compare the two lines (b): $405 nm$ is much shorter than $766 nm$ , so it has the larger $ν$ and larger $E = h ν$ → more energetic. Colours: $766 nm$ is deep red/near-infrared (dim to the eye); $405 nm$ is violet. Their blend is what we call the lilac/violet potassium flame.
- Why this step? Two emission lines mixing explains why K looks pale-lilac rather than a single pure colour (see Atomic Spectra and Emission Lines).

Verify: Check $ν λ = c$ : $3.92 \times 1 0^{14} \times 766 \times 1 0^{- 9} = 3.00 \times 1 0^{8}$ m/s ✓. And $405 < 766$ nm so the violet line is the energetic one ✓.

Recall Did every cell get covered?

A(Ex1) · B(Ex2) · C(Ex3) · D(Ex4) · E(Ex5) · F(Ex6) · G(Ex7) · H(Ex8) · I(Ex9). No blank cells.

Connections

Flame Tests — Characteristic Colours — the parent topic
3.5.03 Flame tests — characteristic colours (Hinglish)
Planck's Quantum Theory — $E = h ν$ , source of $Δ E = h c / λ$
Bohr Model of the Atom — the energy-level ladder
Atomic Spectra and Emission Lines — why K shows two lines
Group 1 and Group 2 Elements (s-block) — the vivid-colour metals
Wet Tests for Cations — confirmatory step for Cell D
Photoelectric Effect — same $ν = c / λ$ , opposite direction
Inorganic Qualitative Analysis