Exercises — Flame tests — characteristic colours

• Na⁺ → golden/persistent yellow
• K⁺ → lilac / violet
• Ca²⁺ → brick red
• Ba²⁺ → apple/pale green
• Cu²⁺ → blue–green
• Li⁺ → crimson red
• Sr²⁺ → crimson / blood red

Anchor with the mnemonic: "Na is yellow mellow, K wears violet pillow."

Ba²⁺ and Cu²⁺. Both sit near $\lambda \approx 510\text{–}525\text{nm}$, the green region. Flame test alone cannot separate them — that needs a wet test (see L3.2).

WHAT: plug into $\Delta E=hc/\lambda$. WHY: we are given wavelength (colour) and asked for the energy gap — the equation runs exactly that way.

Convert: $589\text{nm}=589\times 10^{-9}\text{m}=5.89\times 10^{-7}\text{m}$. $\Delta E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{5.89\times 10^{-7}}$ Numerator: $6.626\times 10^{-34}\times 3\times 10^8=1.9878\times 10^{-25}$. $\Delta E=\frac{1.9878\times 10^{-25}}{5.89\times 10^{-7}}\approx 3.37\times 10^{-19}\text{J}$ So each sodium photon carries about $3.37\times 10^{-19}\text{J}$.

WHAT: compare energies from wavelengths alone. WHY this reasoning: $\Delta E=hc/\lambda$ — since $h,c$ are constants, $\Delta E$ is inversely proportional to $\lambda$. Bigger $\lambda$ ⇒ smaller $\Delta E$.

$766\text{nm}>589\text{nm}$, so K's photon has the smaller energy gap; Na's photon (shorter wavelength) carries more energy. This matches the picture: the redder end (longer $\lambda$) is the low-energy end.

WHAT: invert the formula to get $\lambda$. WHY: we now have energy and want colour, so use $\lambda =hc/\Delta E$. $\lambda =\frac{(6.626\times 10^{-34})(3\times 10^8)}{3.05\times 10^{-19}}=\frac{1.9878\times 10^{-25}}{3.05\times 10^{-19}}$ $\lambda \approx 6.52\times 10^{-7}\text{m}=652\text{nm}$ Drop $652\text{nm}$ onto the spectrum strip (Figure 2): it lands in the deep red → consistent with Sr²⁺ (crimson, ~650–700 nm). Li (671 nm) is also plausible; a wet test would decide.

WHAT: add dilute H₂SO₄ to a solution of the salt (a wet confirmatory test). WHY: barium sulphate is highly insoluble; copper sulphate is soluble.

• Ba²⁺: ${\text{Ba}}^{2+}+{\text{SO}}_4^{2-}\to {\text{BaSO}}_4\downarrow$ — a white precipitate.
• Cu²⁺: no precipitate (soluble sulphate; solution may stay pale blue).

Conclusion: white precipitate ⇒ Ba²⁺; no precipitate ⇒ Cu²⁺. This is the Wet Tests for Cations step that flame screening cannot replace.

• Salt A → Na⁺. The yellow is persistent (lasts seconds), the hallmark of genuine sodium.
• Salt B → K⁺. The brief yellow flash is sodium contamination; blue cobalt glass absorbs that yellow (see the definition above), unmasking the potassium lilac. Persistent lilac through the glass = K⁺.
• Salt C → Ba²⁺. Green flame narrows to Ba²⁺/Cu²⁺; the white BaSO₄ precipitate with dilute H₂SO₄ confirms Ba²⁺ (Cu²⁺ would give none).

This chains three ideas: persistence rule (Na), cobalt-glass filter (K), and confirmatory precipitation (Ba vs Cu).

(a) WHAT: convert the violet wavelength to its energy gap. WHY use $\Delta E=hc/\lambda$ this way: we are given a wavelength (a colour) and want the energy of the jump that produced it, so we run the formula in the $\lambda \to \Delta E$ direction. $405\text{nm}=4.05\times 10^{-7}\text{m}$. $\Delta E_{\text{violet}}=\frac{1.9878\times 10^{-25}}{4.05\times 10^{-7}}\approx 4.91\times 10^{-19}\text{J}$ That is a large gap — consistent with violet being the high-energy end of the strip in Figure 2.

(b) A single ion produces several emission lines, not one, because its electron can fall between many different pairs of levels — each pair has its own gap, its own $\lambda$, its own colour. So "one gap = one colour" is only the first-approximation story; the perceived flame colour is the visual blend of whichever lines are bright enough to see. This is exactly why one ion shows a whole line pattern in Atomic Spectra and Emission Lines, and why the safe exam strategy is to memorise the observed colours rather than reason from a single gap. "Bigger atom = redder" is therefore not a reliable law.

(a) WHAT: single-photon energy via $\Delta E=hc/\lambda$. WHY this direction: we are handed the colour ($\lambda =515$ nm) and asked for the energy of one photon, so we feed wavelength into $\Delta E=hc/\lambda$. $515\text{nm}=5.15\times 10^{-7}\text{m}$. $\Delta E=\frac{1.9878\times 10^{-25}}{5.15\times 10^{-7}}\approx 3.86\times 10^{-19}\text{J}$

(b) WHY multiply by $N_A$: part (a) is the energy of one photon from one electron; a mole contains $6.022\times 10^{23}$ of them, so we scale the single-photon energy up by Avogadro's number to get energy per mole. $E_{\text{mol}}=(3.86\times 10^{-19})(6.022\times 10^{23})\approx 2.32\times 10^5\text{J/mol}$ $=232\text{kJ/mol}$ A tidy visible-light gap sits around a couple hundred kJ/mol — the right size for the loosely-held s-block electrons of Group 1 and Group 2 Elements (s-block).

WHAT: the largest energy comes from the shortest wavelength (400 nm), the smallest from the longest (700 nm). WHY: $\Delta E=hc/\lambda$ makes energy fall as wavelength grows, so the two band edges map to the two energy extremes. $\Delta E_{\max }=\frac{1.9878\times 10^{-25}}{4.00\times 10^{-7}}\approx 4.97\times 10^{-19}\text{J}(400\text{nm, violet})$ $\Delta E_{\min }=\frac{1.9878\times 10^{-25}}{7.00\times 10^{-7}}\approx 2.84\times 10^{-19}\text{J}(700\text{nm, red})$ So visible photons carry roughly $2.84\times 10^{-19}$ to $4.97\times 10^{-19}\text{J}$.

Why gaps outside this band are invisible: a gap smaller than $\Delta E_{\min }$ emits infrared ($\lambda >700$ nm) and a gap larger than $\Delta E_{\max }$ emits ultraviolet ($\lambda <400$ nm); the eye's photoreceptors respond to neither, so the flame looks colourless — the same energy window that governs the Photoelectric Effect.