Photochemistry — Stark-Einstein law, quantum yield, Jablonski diagram, fluorescence vs phosphorescence

1. Stark–Einstein Law (Law of Photochemical Equivalence)

WHY does this law exist? Light absorption is a quantum event. A molecule cannot absorb "half a photon"; absorption is the disappearance of one whole photon and the simultaneous jump of one molecule to an excited state. So the primary event is strictly 1:1.

WHAT it does NOT say: It says nothing about how many product molecules form. After the primary act, secondary (chemical) steps can multiply or quench the result — that's why quantum yields can be huge or tiny (next section).

Worked example Energy per einstein at

$\lambda = 400$ nm Plug in $h = 6.626\times10^{-34}$ J s, $c = 3\times10^8$ m/s, $\lambda = 400\times10^{-9}$ m, $N_A = 6.022\times10^{23}$ . $E = \frac{(6.022\times10^{23})(6.626\times10^{-34})(3\times10^8)}{400\times10^{-9}} \approx 2.99\times10^5 \text{ J/mol} \approx 299\ \text{kJ/mol}$ Why this step? We multiplied the single-photon energy by $N_A$ to convert to a molar quantity comparable with bond energies. Note: ~299 kJ/mol is plenty to break many bonds — that's why violet light is photochemically active while red light often isn't.

2. Quantum Yield ( $\Phi$ )

WHY define it? The Stark–Einstein law fixes the primary step at 1:1, but real reactions have secondary chemistry. $\Phi$ measures the overall efficiency — how much chemistry one absorbed photon ultimately produces.

HOW to read the value:

$\Phi$ value	Meaning	Cause
$\Phi = 1$	ideal primary act, no extra steps	every excited molecule reacts once
$\Phi < 1$	inefficient	excited molecules de-excite (fluorescence, heat, collisional quenching) before reacting
$\Phi \gg 1$	chain reaction	one photon starts a chain that consumes many molecules

3. Jablonski Diagram — the map of fates

Figure — Photochemistry — Stark-Einstein law, quantum yield, Jablonski diagram, fluorescence vs phosphorescence

The processes (WHAT each arrow means):

Absorption ( $S_0 \to S_1$ ): photon kicks molecule up. Femtoseconds.
Vibrational relaxation / Internal Conversion (IC): molecule drops to the lowest vibrational level of $S_1$ as heat (wavy). Fast.
Fluorescence ( $S_1 \to S_0$ ): radiative emission, spin preserved (singlet→singlet). Allowed → fast ( $10^{-9}$ – $10^{-7}$ s).
Intersystem Crossing (ISC) ( $S_1 \to T_1$ ): non-radiative, spin flips to triplet (wavy). Spin-forbidden but happens via spin–orbit coupling.
Phosphorescence ( $T_1 \to S_0$ ): radiative, spin must flip back (triplet→singlet). Spin-forbidden → slow ( $10^{-3}$ s to minutes).

4. Fluorescence vs Phosphorescence

Property	Fluorescence	Phosphorescence
Transition	$S_1 \to S_0$	$T_1 \to S_0$
Spin change	none (allowed)	yes (forbidden)
Lifetime	$10^{-9}$ – $10^{-7}$ s	$10^{-3}$ s – minutes
Afterglow	stops at once	persists ("glow-in-dark")
Energy of emitted photon	higher	lower (since $T_1 < S_1$ )

5. Common Mistakes (Steel-manned)

6. Active Recall

Recall What decides fluorescence vs phosphorescence, and how does that decide lifetime?

Whether a spin flip is needed. Fluorescence ( $S_1\to S_0$ , no flip) is spin-allowed → fast ( $10^{-9}$ s). Phosphorescence ( $T_1\to S_0$ , flip needed) is spin-forbidden → slow (ms–min, afterglow).

Recall A reaction has

$\Phi = 2\times10^5$ . What does this tell you mechanistically? One absorbed photon produces ~ $2\times10^5$ product molecules ⇒ a chain reaction (e.g. radical chain). The primary act is still 1 photon → 1 activated molecule (Stark–Einstein intact).

Recall Explain Stokes shift in one breath.

Vibrational relaxation/IC dumps part of the absorbed energy as heat before emission, so emitted photons are lower energy (longer wavelength) than absorbed ones.

Recall (Feynman, explain to a 12-year-old)

Imagine a ball you throw up a staircase. Light is the throw — it must give exactly enough push to land on a step, never half a step (that's the photon rule). Once on a high step, the ball can: roll back down giving off a quick flash of light (fluorescence), or slip onto a secret slow ramp where it dribbles down for a long time, glowing softly even after you stop throwing (phosphorescence) — that's the glow-in-the-dark toy. And sometimes one push starts a row of dominoes (chain reaction), knocking down way more than one ball.

Flashcards

State the Stark–Einstein law of photochemical equivalence.

In the primary photochemical act, each absorbing molecule absorbs exactly one photon (quantum) of radiation; one photon → one excited molecule.

What is an "einstein" and its energy?

One mole of photons; energy

E = N_A h\nu = N_A hc/\lambda

Energy per einstein at 400 nm (approx)?

~299 kJ/mol.

Define quantum yield

\Phi

\Phi

= (number of molecules reacting / events) ÷ (number of photons absorbed).

Why can

\Phi \gg 1

A chain (e.g. radical) reaction: one photon initiates a chain that consumes many molecules; only the primary act is 1:1.

Why can

\Phi < 1

Competing de-excitation: fluorescence, internal conversion (heat), or collisional quenching remove excitation before reaction.

\Phi

for H₂ + Cl₂ → HCl, and why?

10^6

; long radical chain from Cl• and H• propagation.

What does a Jablonski diagram show?

Electronic/vibrational states (

S_0,S_1,T_1

) and all radiative (straight) and non-radiative (wavy) transitions.

Difference between singlet and triplet state?

Singlet: paired (antiparallel) spins; triplet: parallel spins.

Define internal conversion (IC).

Non-radiative transition between states of same multiplicity (e.g.

S_1\to S_0

), energy lost as heat.

Define intersystem crossing (ISC).

Non-radiative, spin-flipping transition

S_1\to T_1

(singlet to triplet) via spin–orbit coupling.

Fluorescence: transition, spin, lifetime?

S_1\to S_0

, no spin change (allowed),

10^{-9}

–

10^{-7}

Phosphorescence: transition, spin, lifetime?

T_1\to S_0

, spin change required (forbidden),

10^{-3}

s to minutes (afterglow).

Why does phosphorescence persist after light is removed?

T_1\to S_0

is spin-forbidden, so de-excitation is slow, leaking energy over a long time.

State Kasha's rule.

Emission occurs from the lowest excited state of a given multiplicity (

S_1

T_1

), independent of which higher state was excited.

What causes the Stokes shift?

Vibrational relaxation/IC loses energy as heat before emission, so emitted light is lower-energy (redder) than absorbed light.

Why is phosphorescence light redder than fluorescence?

T_1

lies below

S_1

, so

T_1\to S_0

emits a smaller energy gap (longer wavelength).

Connections

Planck's Quantum Theory & E=hν
Electronic Spectra & Selection Rules (spin-allowed/forbidden)
Spin Multiplicity & Singlet–Triplet States
Chain Reactions & Radical Mechanisms
Beer–Lambert Law (photons absorbed)
Fluorescence Spectroscopy / LASERs
Thermal vs Photochemical Reactions

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, photochemistry ka funda simple hai: normal reactions ko heat push karti hai, lekin yahan push light (photon) se aata hai. Stark–Einstein law kehta hai ki primary step mein har ek molecule sirf ek photon absorb karta hai — na aadha, na do. Ek photon, ek excited molecule. Photon ki energy $E = hc/\lambda$ hoti hai, aur ek mole photons ko "einstein" kehte hain, jiski energy $N_A hc/\lambda$ — 400 nm pe roughly 299 kJ/mol, jo bahut saare bonds tod sakti hai. Isliye violet/UV light reactive hoti hai, red light usually nahi.

Ab quantum yield $\Phi$ = (kitne molecules react hue) ÷ (kitne photons absorb hue). Agar $\Phi$ bahut bada hai (jaise H₂ + Cl₂ mein $10^6$ ), toh samajh jao chain reaction chal rahi hai — ek photon ne radical banaya aur woh chain laakhon molecules consume kar gayi. Agar $\Phi$ chhota hai, matlab excited molecule react karne se pehle hi energy lose kar deta hai (fluorescence, heat, ya quenching). Important: Stark–Einstein sirf primary act ke liye hai, products ke liye nahi — yeh confusion mat karna.

Jablonski diagram ek map hai jo dikhata hai excited molecule energy kaise wapas chhodta hai. Absorption se molecule $S_0

Test yourself — Physical Chemistry (Advanced)