Isotopes of hydrogen — protium, deuterium, tritium: Exercises

This page is a self-testing ladder. Each problem builds on the parent note the isotopes topic. Work each one before opening its solution. The levels climb: L1 Recognition → L2 Application → L3 Analysis → L4 Synthesis → L5 Mastery.

A quick reminder of symbols we use throughout, so nothing appears unearned:

L1 — Recognition

Problem 1.1

State the number of protons, neutrons, and electrons in a neutral atom of protium ( $^{1}$ H), deuterium ( $^{2}$ H), and tritium ( $^{3}$ H).

Recall Solution

Hydrogen means $Z = 1$ , so 1 proton in all three. Neutral means electrons = protons = 1 electron in all three. Neutrons come from $N = A - Z$ :

Protium: $N = 1 - 1 = 0$ neutrons.
Deuterium: $N = 2 - 1 = 1$ neutron.
Tritium: $N = 3 - 1 = 2$ neutrons.

Same $Z$ , same electron count → same chemistry; different $N$ → different mass. See Atomic Structure and Isotopes.

Problem 1.2

Which of the three hydrogen isotopes is radioactive, and what type of decay does it undergo?

Recall Solution

Tritium ( $^{3}$ H) is radioactive. It undergoes beta-minus decay: $_{1}^{3} H \to_{2}^{3} He + e^{-} + \overset{ν}{ˉ}_{e}$ A neutron inside the nucleus turns into a proton, spitting out an electron and an antineutrino. Protium and deuterium are both stable. See Beta Decay.

L2 — Application

Problem 2.1

At the same temperature, compare the root-mean-square speeds of $H_{2}$ gas ( $M = 2$ g/mol) and $D_{2}$ gas ( $M = 4$ g/mol). By what factor is $D_{2}$ slower?

Recall Solution

Why this tool? Boiling-point and effusion differences between the isotopes come straight from how fast their molecules move, and $v_{r m s}$ is the clean handle on that from Kinetic Theory of Gases: $v_{r m s} = \frac{3 R T}{M}$ Same $T$ , same $R$ , so those cancel in a ratio and only mass survives: $\frac{v _{D_{2}}}{v _{H_{2}}} = \frac{M _{H_{2}}}{M _{D_{2}}} = \frac{2}{4} = \frac{1}{2} \approx 0.707$ So $D_{2}$ moves at about 70.7% of the $H_{2}$ speed — i.e. roughly 30% slower. Slower molecules linger near each other, boosting effective attraction, which is why $D_{2}$ (23.67 K) boils higher than $H_{2}$ (20.28 K).

Problem 2.2

Compute the reduced mass $μ$ of a C–H bond and a C–D bond, treating carbon as effectively infinitely heavy. Take $m_{H} = 1$ u, $m_{D} = 2$ u.

Recall Solution

Why reduced mass? A vibrating bond is a mass-on-a-spring; the mass that actually matters is $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ , not either atom alone. With carbon huge ( $m_{C} \to \infty$ ): $μ \approx lim_{m_{C} \to \infty} \frac{m _{C} m _{H}}{m _{C} + m _{H}} = m_{H} .$ So:

C–H: $μ_{H} \approx 1$ u.
C–D: $μ_{D} \approx 2$ u.

The ratio $μ_{D} / μ_{H} = 2$ is the seed of the whole kinetic isotope effect: the heavier reduced mass lowers the vibration frequency and the zero-point energy.

L3 — Analysis

Problem 3.1

Tritium has half-life $t_{1/2} = 12.32$ yr. (a) Find its decay constant $λ$ . (b) What fraction of a tritium sample remains after 50 years? (c) Roughly how many half-lives is that?

Recall Solution

Why the exponential? Radioactive decay has a fixed per-atom-per-second chance of decaying, which mathematically forces an exponential law $N (t) = N_{0} e^{- λ t}$ (see Beta Decay).

(a) Half-life means $N = N_{0} /2$ at $t = t_{1/2}$ , so $e^{- λ t_{1/2}} = \frac{1}{2}$ , giving $λ = \frac{l n 2}{t _{1/2}} = \frac{0.6931}{12.32 yr} = 0.05626 yr^{- 1} .$

(b) After 50 years: $\frac{N}{N _{0}} = e^{- λ t} = e^{- 0.05626 \times 50} = e^{- 2.813} = 0.0600.$ About 6.0% remains.

(c) Number of half-lives $= 50/12.32 = 4.06$ . And indeed $(1/2)^{4.06} \approx 0.060$ — the two methods agree.

Problem 3.2

Estimate the kinetic isotope effect $k_{H} / k_{D}$ at $T = 298$ K, given the C–H stretch frequency $\tilde{ν}_{H} = 2900 cm^{- 1}$ . Assume the rate difference comes entirely from the difference in zero-point energy, and that $\tilde{ν}_{D} = \tilde{ν}_{H} / 2$ .

Recall Solution

Why zero-point energy? Even at absolute zero a bond keeps vibrating with energy $E_{0} = \frac{1}{2} h ν$ . The heavier C–D bond vibrates slower, so it sits in a lower energy well. To reach the transition state (bond mostly broken, ZPE nearly gone), C–H starts higher and has less to climb — it reacts faster.

Convert wavenumbers to energy with $E = h c \tilde{ν}$ ; the extra activation energy for D is the ZPE gap: $Δ E_{Z P E} = \frac{1}{2} h c (\tilde{ν}_{H} - \tilde{ν}_{D}) = \frac{1}{2} h c \tilde{ν}_{H} (1 - \frac{1}{2}) .$ Numerically $\tilde{ν}_{H} - \tilde{ν}_{D} = 2900 (1 - 0.7071) = 849.6 cm^{- 1}$ , so half of that is $424.8 cm^{- 1}$ . Using $h c / k_{B} = 1.4388 cm\cdotK$ : $\frac{Δ E _{Z P E}}{k _{B} T} = \frac{1.4388 \times 424.8}{298} = 2.051.$ Then (from Arrhenius, Chemical Kinetics): $\frac{k _{H}}{k _{D}} = e^{Δ E_{Z P E} / k_{B} T} = e^{2.051} = 7.78.$ So $k_{H} / k_{D} \approx 7 - 8$ , matching the textbook "6–7" range for real C–H bonds (which lose a little ZPE in the transition state, shaving the number down). See the energy-well picture below.

Figure — Isotopes of hydrogen — protium, deuterium, tritium

L4 — Synthesis

Problem 4.1

Compute the Q-value (energy released) of tritium beta decay from atomic masses, and explain why using atomic masses (which include electrons) still gives the correct answer even though a free electron is emitted.

Given: $m (^{3} H) = 3.016049 u$ , $m (^{3} He) = 3.016029 u$ , $1 u = 931.5 MeV / c^{2}$ .

Recall Solution

Why it works with atomic masses: $^{3}$ H atom has 1 electron; $^{3}$ He atom has 2 electrons. When the nucleus emits a $β^{-}$ , the daughter $^{3}$ He nucleus needs one more orbital electron to become neutral — and that is exactly the emitted electron. So counting whole atoms, the emitted $e^{-}$ mass is already booked inside the $^{3}$ He atomic mass. We do not subtract $m_{e}$ separately: $Q = [m (^{3} H) - m (^{3} He)] c^{2} .$ $Q = (3.016049 - 3.016029) \times 931.5 MeV = 0.000020 \times 931.5 = 0.01863 MeV = 18.6 keV .$ This tiny 18.6 keV is why tritium's betas are so gentle they don't pierce skin — see Nuclear Stability and Binding Energy.

Problem 4.2

A CANDU-type reactor uses heavy water $D_{2} O$ instead of ordinary water $H_{2} O$ as a neutron moderator. Given neutron-absorption cross-sections $σ (H_{2} O) = 0.664$ barns and $σ (D_{2} O) = 0.00133$ barns per molecule, compute the factor by which $H_{2} O$ absorbs more neutrons, and state the physical consequence.

Recall Solution

Why cross-section? A cross-section is the effective "target area" a nucleus presents to a passing neutron — bigger area, more likely absorption. The ratio is: $\frac{σ ( H _{2} O )}{σ ( D _{2} O )} = \frac{0.664}{0.00133} = 499 \approx 500.$ So ordinary water swallows neutrons about 500× more greedily than heavy water. Protium's nucleus readily grabs a neutron (forming deuterium); deuterium is far more reluctant to grab a second. Consequence: in $D_{2} O$ neutrons survive to keep the chain reaction going, which is why heavy-water reactors can run on natural (un-enriched) uranium. See Heavy Water and Nuclear Reactors.

L5 — Mastery

Problem 5.1

The deuterium-tritium fusion reaction powering ITER is $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n + Q .$ Verify that charge and nucleon number balance, then compute $Q$ (in MeV) from these nuclear masses: $m (^{2} H) = 2.014102 u$ , $m (^{3} H) = 3.016049 u$ , $m (^{4} He) = 4.002602 u$ , $m (n) = 1.008665 u$ , $1 u = 931.5 MeV$ . (Use atomic masses; electrons balance since total charge $Z = 2$ on each side.)

Recall Solution

Conservation check:

Nucleons: $2 + 3 = 5$ on the left; $4 + 1 = 5$ on the right. ✓
Charge (protons): $1 + 1 = 2$ on the left; $2 + 0 = 2$ on the right. ✓ (Two electrons on each side, so atomic masses are consistent.)

Why mass defect → energy: the products weigh slightly less than the reactants; the missing mass is released as kinetic energy via $E = Δ m c^{2}$ . $Δ m = (2.014102 + 3.016049) - (4.002602 + 1.008665)$ $Δ m = 5.030151 - 5.011267 = 0.018884 u .$ $Q = 0.018884 \times 931.5 = 17.59 MeV \approx 17.6 MeV .$ This matches the quoted D–T energy of 17.6 MeV — the reaction with the lowest ignition temperature, which is why it is the front-runner for Nuclear Fusion.

Problem 5.2

An NMR chemist dissolves a sample in $CDCl_{3}$ (deuterated chloroform) rather than $CHCl_{3}$ . Explain quantitatively-in-spirit why the deuterium does not clutter the $^{1}$ H spectrum, referencing nuclear spin, and connect this to why deuterium and protium behave identically chemically but differently magnetically and kinetically.

Recall Solution

The reasoning chain:

$^{1}$ H-NMR listens only to nuclei with spin $I = \frac{1}{2}$ at the proton's resonance frequency. Protium has $I = \frac{1}{2}$ ; deuterium has $I = 1$ and a different gyromagnetic ratio, so it resonates far away and is invisible in the $^{1}$ H window. Hence $CDCl_{3}$ contributes no solvent peak to crowd your sample's protons. See NMR Spectroscopy.
Chemically identical: both isotopes have electron configuration $1 s^{1}$ — same valence, same bonds, same compounds. Chemistry is decided by electrons, and the isotopes share them exactly.
Magnetically different: chemistry ignores the nucleus's spin, but NMR is built on it — so swapping $I = \frac{1}{2}$ for $I = 1$ changes everything for NMR while changing nothing for bonding.
Kinetically different: the heavier nucleus lowers zero-point energy (Problem 3.2), so bond-breaking rates differ (the kinetic isotope effect) even though bond identities do not.

One clean sentence: isotopes differ only in the nucleus; anything that reads the nucleus (mass, spin, decay) changes, anything that reads the electrons (chemistry) does not.

Recall Self-test summary (cover the right side)

Speed ratio $v_{D_{2}} / v_{H_{2}}$ ::: $1/ 2 \approx 0.707$ Reduced-mass ratio $μ_{D} / μ_{H}$ for C–D vs C–H ::: $2$ Tritium $λ$ from $t_{1/2} = 12.32$ yr ::: $0.0563 yr^{- 1}$ Fraction of tritium left after 50 yr ::: $\approx 6%$ Estimated $k_{H} / k_{D}$ at 298 K ::: $\approx 7 - 8$ Tritium beta-decay Q-value ::: $18.6 keV$ $σ (H_{2} O) / σ (D_{2} O)$ ::: $\approx 500$ D–T fusion energy $Q$ ::: $17.6 MeV$