Worked examples — Isotopes of hydrogen — protium, deuterium, tritium

This page is the practice arena for the parent topic. We do not re-teach the theory; we exercise every corner of it. Before each example you will Forecast (guess the answer), then we grind through it, then we Verify by plugging back and checking units.

A quick note on formatting: key terms below appear highlighted like this (Obsidian renders them with a coloured background). If you are reading in plain text, just treat the doubled equals signs as emphasis.

A quick reminder of the symbols we will reuse (all earned in the parent note):

Recall The four tools we keep pulling off the shelf

• $A=Z+N$ — mass number = protons + neutrons. This is just counting the heavy particles in the nucleus.
• $v_{rms}=\sqrt{\frac{3RT}{M}}$ — the "typical speed" of a gas molecule. Heavier $M$ ⇒ slower. Comes from Kinetic Theory of Gases.
• $k=A_0{e}^{-E_a/RT}$ — the Arrhenius rate law. Bigger activation energy $E_a$ ⇒ exponentially slower reaction. From Chemical Kinetics.
• $N(t)=N_0{e}^{-\lambda t}$, with $\lambda =\frac{\ln 2}{t_{1/2}}$ — radioactive decay. From Beta Decay.

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The scenario matrix

Think of every problem this topic can throw at you as one cell in this grid. Our job: fill every cell with at least one solved example so you never meet a scenario cold.

Cell	Case class	What makes it tricky	Covered by
C1	Counting / identity — given $A$ and $Z$, name the isotope	zero-neutron edge (protium)	Ex 1
C2	Degenerate mass ratio — the lightest possible mass doubling	relative mass change is 100 %, not tiny	Ex 2
C3	Gas speed / effusion — compare $v_{rms}$ across isotopes	which is faster, and by exactly how much	Ex 3
C4	Kinetic isotope effect — rate ratio $k_H/k_D$	quantum zero-point energy, sign of $\Delta E$	Ex 4
C5	Decay over time — how much tritium is left after $t$	non-integer number of half-lives	Ex 5
C6	Limiting behaviour — $t\to 0$, $t\to \infty$	check the formula does not blow up	Ex 5b
C7	Q-value / mass–energy — sign of released energy	tiny mass difference, unit conversion	Ex 6
C8	Real-world word problem — heavy-water moderator choice	ratio of cross-sections	Ex 7
C9	Exam twist — the "deuterium is radioactive?" trap	conceptual, no decay to compute	Ex 8

Every numeric answer below is machine-checked in the verify block.

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Ex 1 — Cell C1: Name that nucleus (the zero-neutron edge)

Step 1 — Use $N=A-Z$. Why this step? Mass number counts all nucleons; subtract the protons to isolate neutrons. $N=A-Z=1-1=0.$

Step 2 — Interpret. Why this step? An isotope's name is decided by neutron count for hydrogen (parent note). Zero neutrons ⇒ protium, symbol ${}_1^1\text{H}$. This is the degenerate edge case: a nucleus that is a bare single proton — the only common nucleus with no neutrons at all.

Step 3 — Repeat for $A=3$. $N=3-1=2\Rightarrow \text{tritium},{}_1^3\text{H (or T)}.$

Verify: Rebuild mass numbers: protium $A=Z+N=1+0=1$ ✓; tritium $A=1+2=3$ ✓. Units: $A$, $Z$, $N$ are pure counts (dimensionless) — consistent. See the nucleon pictures in Atomic Structure and Isotopes.

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Ex 2 — Cell C2: The doubling nobody else gets

Step 1 — Percentage change formula. Why this step? "Relative mass difference" tells us how much a change matters compared with what we started with. Let $m_{\text{start}}$ be the mass before adding the neutron, and let $\Delta m$ be the mass added (here always one neutron, ≈ 1 u). Then: $\text{percent increase}=\frac{\Delta m}{m_{\text{start}}}\times 100\%.$ For hydrogen $m_{\text{start}}=1$ u and $\Delta m=1$ u: $\text{H}\to \text{D}:\frac{1}{1}\times 100\%=100\%.$

Step 2 — Carbon comparison. Now $m_{\text{start}}=12$ u, still $\Delta m=1$ u: ${}^{12}\text{C}\to {}^{13}\text{C}:\frac{1}{12}\times 100\%\approx 8.33\%.$

Step 3 — Interpret. Why this step? This is the reason hydrogen isotopes get their own names (parent note's opening intuition). A 100 % mass change shifts speeds, boiling points and reaction rates measurably; an 8 % change barely does.

The bar chart below drives the point home: the same single neutron produces a towering 100 % bar for hydrogen but a stubby ~8 % bar for carbon. Take-away — read the red annotation: the hydrogen effect is a full 12× taller than the carbon effect, which is why only hydrogen's isotopes earn distinct names.

Verify: $\frac{1}{1}=1.00$ and $\frac{1}{12}=0.0833$; ratio of the two effects $=12$, i.e. hydrogen's relative jump is 12× larger than carbon's. ✓

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Ex 3 — Cell C3: Who is faster, and by how much?

Step 1 — Why $v_{rms}=\sqrt{3RT/M}$? Why this tool and not another? We need to compare speeds, and kinetic theory ties speed to mass through exactly this relation. At the same $T$, the $\sqrt{3RT}$ factor is shared, so it cancels in a ratio — leaving only the masses.

Step 2 — Take the ratio. $\frac{v_{D_2}}{v_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{D_2}}}=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}\approx \mathrm{0.707.}$ So D₂ is ~29.3 % slower ($1-0.707$).

Step 3 — Tritium. $\frac{v_{T_2}}{v_{H_2}}=\sqrt{\frac{2}{6}}=\frac{1}{\sqrt{3}}\approx 0.577,$ i.e. T₂ is about 42 % slower.

The curve below plots "speed relative to ${\text{H}}_2$" against molar mass; the three coloured dots mark H₂, D₂ and T₂. Take-away: the curve falls as mass grows — heavier means slower — and it falls steeply at first, so the jump from H₂ to D₂ (a 30 % drop) is bigger per neutron than the later D₂→T₂ step. This gentle $1/\sqrt{M}$ shape is why isotope speed differences shrink for heavier elements.

Verify: $0.707^2=0.500=2/4$ ✓ and $0.577^2=0.333=2/6$ ✓. Units: masses in the same units cancel, leaving a dimensionless ratio — correct for a speed ratio. This slower motion is exactly why D₂ boils higher (parent note). See Kinetic Theory of Gases.

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Ex 4 — Cell C4: Kinetic isotope effect, with the sign nailed down

Step 1 — Zero-point energy (ZPE). Why this tool? Even at absolute zero a bond vibrates; that leftover energy is $E_0=\frac{1}{2}h\nu$. The lighter H sits higher in its well, so it needs less extra push to break — that is the whole effect. $E_{0,H}=\frac{1}{2}h\nu ,E_{0,D}=\frac{1}{2}h\nu \sqrt{\frac{{\mu }_H}{{\mu }_D}}=\frac{1}{2}h\nu \cdot \frac{1}{\sqrt{2}}.$ Why the $\sqrt{\mu }$? Frequency $\nu \propto \sqrt{k/\mu }$; doubling reduced mass drops $\nu$ by $\sqrt{2}$.

Step 2 — Difference feeds activation energy. Why this step? In the transition state the bond is essentially broken, so its ZPE vanishes for both. The reactant ZPE difference becomes an activation-energy difference: $\Delta E_{ZPE}=E_{0,H}-E_{0,D}=\frac{1}{2}h\nu \left(1-\frac{1}{\sqrt{2}}\right).$ Numerically, $\Delta E_{ZPE}=\frac{1}{2}(6.626\times 10^{-34})(9.0\times 10^{13})(0.2929)\approx 8.74\times 10^{-21}\text{J per bond}$.

Step 3 — Arrhenius ratio. Why exponential? Rates depend on $e^{-E_a/RT}$; a difference in $E_a$ becomes a ratio of rates. The pre-exponential factor $A_0$ is essentially the same for both isotopes, so it cancels. Per molecule use $k_B$ instead of $R$: $\frac{k_H}{k_D}=\exp \left(\frac{\Delta E_{ZPE}}{k_{B}T}\right)=\exp \left(\frac{8.74\times 10^{-21}}{(1.381\times 10^{-23})(298)}\right)\approx e^{2.12}\approx \mathrm{8.3.}$

Verify: Sign check — C–H has more ZPE ⇒ smaller barrier ⇒ H breaks faster ⇒ $k_H/k_D>1$ ✓, matching the parent note's "6–7-ish" range (our idealised $\nu$ pushes it a touch higher). Units: J / (J/K · K) = dimensionless in the exponent ✓. Deuterated drugs exploit exactly this to slow metabolism (parent note). See Chemical Kinetics.

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Ex 5 — Cell C5: How much tritium survives?

Step 1 — Decay constant. Why this step? The clean exponential $N=N_0{e}^{-\lambda t}$ needs $\lambda$, and $\lambda =\ln 2/t_{1/2}$ because after one half-life $e^{-\lambda t_{1/2}}=\frac{1}{2}$. $\lambda =\frac{0.6931}{12.32}=0.05627{\text{yr}}^{-1}.$

Step 2 — Evaluate at $t=50$ yr. $\frac{N}{N_0}=e^{-\lambda t}=e^{-0.05627\times 50}=e^{-2.814}\approx \mathrm{0.0600.}$

Step 3 — Cross-check via half-lives. Why this step? Sanity: $50/12.32=4.06$ half-lives, so fraction $=(1/2{)}^{4.06}\approx 0.060$. Both routes agree.

The decay curve below shows the fraction remaining sliding down as time passes; the dashed vertical lines mark successive half-lives (each one halves what is left), and the red dot marks our 50-year answer. Take-away: notice the curve is steepest early and flattens later — the first few years strip away most of the sample, and by 50 years (just past the 4th dashed line) only the red-marked 6 % survives.

Verify: $e^{-2.814}=0.0600$ and $2^{-4.06}=0.0600$ ✓ — ≈ 6 % remains. Units: $\lambda t={\text{yr}}^{-1}\cdot \text{yr}$ dimensionless ✓. See Beta Decay.

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Ex 5b — Cell C6: The limits (does the formula misbehave?)

Step 1 — $t=0$. $N/N_0=e^0=1$ — the full sample is present. ✓ (No decay has happened yet.)

Step 2 — $t\to \infty$. $N/N_0=e^{-\infty }=0$ — everything has decayed. The curve approaches zero but never reaches it (asymptote), which is why "how many atoms are left" is only ever statistically zero.

Verify: The exponential is finite and positive for all finite $t$, so the model never gives a negative or >1 population — a physically sane function everywhere. This is the degenerate/limiting cell C6. ✓

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Ex 6 — Cell C7: Q-value of tritium beta decay (mind the tiny signs)

Step 1 — Atomic-mass shortcut. Why this step? Atomic masses include the electrons; the emitted ${\beta }^{-}$ is captured by using the neutral daughter's electron count. So the extra $m_e$ term cancels: $Q=\left[m{(}^3\text{H})-m{(}^3\text{He})\right]c^2.$

Step 2 — Subtract, then convert. Why this tool? $1\text{u}\leftrightarrow 931.5\text{MeV}$ via $E=m{c}^2$ — the mass defect becomes kinetic energy of the electron + antineutrino. $\Delta m=3.016049-3.016029=2.0\times 10^{-5}\text{u},$ $Q=2.0\times 10^{-5}\times 931.5\text{MeV}=0.01863\text{MeV}=18.6\text{keV}.$

Step 3 — Interpret the sign. Why this step? $Q>0$ ⇒ decay is energetically allowed (mass is lost, energy released). A positive $Q$ is exactly what makes tritium unstable while a hypothetical energy-costing decay would be forbidden.

Verify: $2.0\times 10^{-5}\times 931.5=0.018630\text{MeV}=18.6\text{keV}$ ✓. Units: u × (MeV/u) = MeV ✓. This low 18.6 keV is why tritium betas don't pierce skin (parent note). See Nuclear Stability and Binding Energy.

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Ex 7 — Cell C8: Why heavy water moderates better (word problem)

Step 1 — Take the ratio. Why this step? Absorption loss scales with the cross-section (the target area a molecule presents to a neutron). The reduction factor when we swap H₂O for D₂O is simply the ratio of the two areas. $\frac{{\sigma }_{{\text{H}}_2\text{O}}}{{\sigma }_{{\text{D}}_2\text{O}}}=\frac{0.664}{0.00123}\approx 540.$

Step 2 — Interpret. Why this step? Heavy water absorbs roughly 540× fewer neutrons. Fewer swallowed neutrons ⇒ more survive to sustain the chain reaction ⇒ a reactor can run on natural (un-enriched) uranium. That is precisely the CANDU design choice, and it matches the parent note's "≈500×" figure — the parent was comparing the same molecular absorption, not a per-atom number.

Verify: $0.664/0.00123=540$ (to 2 sig figs) ✓ — heavy water absorbs about 540× less, consistent with the parent note's "≈500×." Units: barns/barns = dimensionless ✓. See Heavy Water and Nuclear Reactors and Nuclear Fusion for the D–T fuel angle.

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Ex 8 — Cell C9: The exam twist ("is deuterium radioactive?")

Step 1 — Test the stated logic. Why this step? The claim assumes a monotone rule: "more mass ⇒ more instability." We test it against the neutron-to-proton ratio $N/Z$, the quantity that actually governs light-nucleus stability (parent note):

• Protium: $N/Z=0/1=0$ — stable.
• Deuterium: $N/Z=1/1=1$ — a balanced, stable light nucleus.
• Tritium: $N/Z=2/1=2$ — too neutron-rich, so it ${\beta }^{-}$-decays.

Step 2 — Show the claimed decay is energetically forbidden. Why this step? A conclusive "it never decays" needs more than a ratio — it needs a negative $Q$. The only candidate decay for deuterium would be ${\beta }^{-}$ to ${}^2\text{He}$ (diproton), but ${}^2\text{He}$ is unbound (it does not exist as a bound nucleus), so there is no lower-energy nucleus for deuterium to fall into. With no accessible product of lower mass, $Q<0$ for every decay channel ⇒ decay is impossible.

Step 3 — Verdict. The mass-monotone logic is wrong: stability tracks $N/Z$ and the availability of a lower-mass product, not raw mass. Deuterium is perfectly stable. The statement is False.

Verify: Conceptual check — a truly monotone "heavier = more radioactive" rule would also make ordinary ${}^{13}\text{C}$ radioactive, which it is not. And numerically the $N/Z$ values are $0,1,2$ for H, D, T, with only the largest ($N/Z=2$, tritium) unstable — non-monotone in a way that singles out tritium alone. ✓ See Nuclear Stability and Binding Energy.

Recall Self-test

Which isotope moves fastest at fixed T, and why? ::: Protium — lowest molar mass, so largest $v_{rms}=\sqrt{3RT/M}$. After ~4 half-lives, roughly what fraction of tritium remains? ::: About 6 % ($2^{-4}=1/16\approx 0.0625$). Is $k_H/k_D$ greater or less than 1, and why? ::: Greater than 1 — C–H has higher zero-point energy, so a lower barrier, so it breaks faster. Is deuterium radioactive? ::: No — it is perfectly stable; $N/Z=1$ and there is no lower-mass nucleus for it to decay into.