Heavy water D₂O, hydrogen peroxide H₂O₂ — structure, preparation, reactions: Visual walkthrough

This page builds ONE result from nothing: why heavy water gets richer in deuterium the more you electrolyse the ordinary water, and how much richer. We do it slowly, with a picture at every step. Everything traces back to the parent note Heavy water & H₂O₂.

Step 1 — What "H" and "D" even are

WHAT. Hydrogen's nucleus is one tiny particle: a proton. Deuterium's nucleus is a proton plus a neutron — a neutral particle of almost the same weight. Both have exactly one electron, so chemically they behave almost identically. The only real difference is mass: D weighs about twice as much as H.

WHY start here. Every later step is really about mass. If you don't see that D is "H wearing a backpack," nothing else makes sense. The chemistry (bonds, shape) is the same; only the speed of vibration and the ease of breaking a bond changes — and that is a pure mass effect.

PICTURE. On the left, plain hydrogen: one proton, one electron. On the right, deuterium: proton + neutron, one electron. Same electron cloud, heavier core.

Step 2 — A bond is a spring, and a spring's speed depends on mass

WHAT. Picture the O–H bond not as a stiff stick but as a spring joining oxygen to hydrogen. The atoms jiggle back and forth. How fast they jiggle is the vibration frequency, written $ν$ (Greek "nu"). A light ball on a spring buzzes fast; a heavy ball on the same spring wobbles slowly.

WHY this tool. We need a way to say "why does D behave differently if the chemistry is the same?" The answer is vibration. The formula for a spring's frequency is the cleanest place where mass alone changes the answer — exactly what we need.

PICTURE. Two identical springs. Top: light H buzzing fast (tight wiggles). Bottom: heavy D wobbling slowly (loose wiggles). Same spring, different speeds.

The one thing to carry forward: $ν \propto \frac{1}{μ}$ — frequency falls as mass rises.

Step 3 — Reduced mass: whose weight actually matters?

WHAT. When two atoms of mass $m_{1}$ and $m_{2}$ are joined, the vibration doesn't feel either mass alone — it feels a blend called the reduced mass: $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$

WHY this exact combination. Both atoms move when the bond vibrates, so "how heavy is the wobble?" is a shared question. The reduced mass is the single effective weight that makes the two-body wobble look like one ball on a spring. If one partner is huge (oxygen, 16), $μ$ leans toward the light partner — so swapping H for D still shifts $μ$ a lot.

PICTURE. Oxygen (big) tied to H, then to D. The little arrow shows the "effective weight" of the wobble jumping from $0.94$ to $1.78$ when we swap H for D.

Plug in the numbers (oxygen $= 16$ ): $μ_{O H} = \frac{16 \times 1}{16 + 1} = \frac{16}{17} \approx 0.94 amu$ $μ_{O D} = \frac{16 \times 2}{16 + 2} = \frac{32}{18} \approx 1.78 amu$

Numerator $m_{1} m_{2}$ — the product of the two weights.
Denominator $m_{1} + m_{2}$ — their sum.
Result: O–D is nearly twice as "heavy to wobble" as O–H.

So by Step 2, $ν_{O D} = \frac{μ _{O H}}{μ _{O D}} ν_{O H} = \frac{0.94}{1.78} ν_{O H} \approx 0.73 ν_{O H}$ — the D bond buzzes at about 73% of H's speed.

Step 4 — Slower wobble ⇒ deeper resting energy ⇒ harder to break

WHAT. Even at absolute-zero cold, a spring can never sit perfectly still — quantum rules force a leftover jiggle called zero-point energy (ZPE): $E_{Z P E} = \frac{1}{2} h ν$

WHY this matters for breaking bonds. To break a bond you must climb from where the atom sits up to the "escape" energy. If the atom already sits higher (big ZPE, fast buzz = H), it has a shorter climb — easier to break. If it sits lower (small ZPE, slow buzz = D), it has a longer climb — harder to break. That climb-difference is the whole reason H leaves first.

PICTURE. One energy "valley" (the bond). Two floors inside it: the higher H floor (short red climb to the rim) and the lower D floor (tall red climb). Same rim, different starting heights.

Step 5 — Turn "H leaves faster" into a separation number $α$

WHAT. Define the separation factor: $α = \frac{rate H is removed}{rate D is removed}$ Experiment for electrolysis gives $α \approx 6$ : H leaves about 6× faster than D.

WHY define this. We need ONE number that packages all the physics of Steps 1–4 (mass → frequency → ZPE → break-rate) so we can do arithmetic. $α$ is that number. $α = 1$ would mean "no separation"; $α = 6$ means "strong separation."

PICTURE. A doorway (electrolysis). Six H molecules stream out for every one D molecule — the crowd left inside gets visibly darker (more D) as it drains.

Step 6 — Rayleigh's law: the leftover puddle gets richer

WHAT. Let $L$ = the fraction of the original liquid still remaining as a puddle. Start full: $L = 1$ . Electrolyse away half: $L = 0.5$ . Down to 1% left: $L = 0.01$ . The enrichment of D in the puddle is: $\frac{R _{f ina l}}{R _{ini t ia l}} = L^{(1/ α) - 1}$

Here $R$ = the deuterium ratio (how much D per unit H) in the liquid. $R_{f ina l} / R_{ini t ia l}$ tells us "how many times richer the puddle now is."

WHY this exact shape. As the puddle shrinks, each moment removes mostly H (rate 1) and a little D (rate $1/ α$ ). Summing that continuous, ever-changing removal is exactly what gives a power law in $L$ . The exponent is $(1/ α) - 1$ : with $α = 6$ it is $\frac{1}{6} - 1 = - 0.833$ , a negative number. Negative exponent means: as $L$ shrinks toward 0, the ratio blows up — the puddle keeps getting richer.

PICTURE. A curve of enrichment vs how much liquid is left. Reading right-to-left (draining the tank), the enrichment climbs steeply — flat at first, then rocketing as the last drops remain.

Let's put $α = 6$ and drain to $L = 0.01$ (99% electrolysed): $\frac{R _{f ina l}}{R _{ini t ia l}} = (0.01)^{- 0.833} = 1 0^{2 \times 0.833} = 1 0^{1.667} \approx 46$

$(0.01) = 1 0^{- 2}$ , and $(- 2) \times (- 0.833) = + 1.667$ , so the answer is $1 0^{1.667} \approx 46$ .

Enriched about 46×. Starting D content $0.0156%$ becomes $0.0156% \times 46 \approx 0.72%$ .

Step 7 — The degenerate cases (never leave a scenario unshown)

WHAT & WHY. A good derivation must survive its extreme inputs. Let's test the corners of $L^{(1/ α) - 1}$ .

PICTURE. The same curve with three flags planted: the start ( $L = 1$ ), the "no separation" line ( $α = 1$ , dead flat), and the runaway tail ( $L \to 0$ ).

Case	What we plug in	Result	Meaning
Nothing electrolysed	$L = 1$	$1^{anything} = 1$	No enrichment — makes sense, we did nothing.
No isotope effect	$α = 1$	exponent $= 1 - 1 = 0$ , so $L^{0} = 1$	If H and D left at equal rates, the puddle never changes. Enrichment needs $α > 1$ .
Drain to nearly dry	$L \to 0$	$L^{- 0.833} \to \infty$	Ratio explodes — but you're left with a drop, and it's still mostly H. Enrichment is huge per unit, tiny in total.
Pure D leaves	$α \to \infty$	exponent $\to - 1$ , so $R \propto 1/ L$	If only H ever left, D would concentrate exactly like $1/ L$ — the strongest possible enrichment.

The one-picture summary

The chain, left to right: extra neutron (mass ↑) → slower spring buzz ( $ν \propto 1/ μ$ ) → lower resting floor (smaller ZPE) → harder to break → H exits ~6× faster ( $α = 6$ ) → puddle enriches as $L^{(1/ α) - 1}$ → ~46× in one deep stage, so cascade to reach 99.8%.

Recall Feynman retelling — say it like a story

Imagine a room packed with people; almost everyone is light and quick (H), but a rare few wear heavy boots (D). We open a door. The quick people rush out first; the boot-wearers dawdle. As the room empties, the fraction of boot-wearers among those left keeps rising. That's the whole trick.

Why do the boot-wearers dawdle? Their extra neutron makes the bond-spring wobble slower, which parks them at a lower energy floor, so they need a longer climb to escape. Package all that into one number — $α = 6$ , "H leaves six times faster." Then a bit of arithmetic (Rayleigh's law) says: if you drain the room to 1% full, the heavy fraction is about 46 times what you started with. Not pure yet — so you repeat the whole thing in stage after stage until the water is truly heavy.

Recall Quick self-check

Why is O–D harder to break than O–H if the chemistry is identical? ::: Same spring stiffness $k$ , but larger reduced mass $μ$ ⇒ lower frequency $ν$ ⇒ lower zero-point energy ⇒ longer climb to break. What does $α = 1$ mean physically? ::: H and D leave at equal rates — no enrichment ever happens. Why is the Rayleigh exponent negative? ::: $(1/ α) - 1 < 0$ because $α > 1$ ; a negative exponent makes the ratio grow as $L$ shrinks. Does draining to $L \to 0$ give lots of D₂O? ::: No — the ratio explodes but the amount vanishes; the leftover is a tiny, still-mostly-H drop.