Radiation safety — units (Bq, Gy, Sv), shielding: Exercises

Intuition How to use this page

Each problem is tagged L1 → L5 (Recognition → Application → Analysis → Synthesis → Mastery). Read the problem, cover the solution, work it on paper, then open the collapsible [!recall]- block to check every step. If you never hit a step you couldn't do, level up. Everything traces back to the parent: Radiation safety — units & shielding.

The whole page rests on five formulas you already met there: $A = λ N, λ = \frac{l n 2}{t _{1/2}}, D = \frac{E}{m}, H = w_{R} D, I = I_{0} e^{- μx} .$ Every symbol below is defined the first time it appears. In particular, in the attenuation law $I = I_{0} e^{- μx}$ : $I_{0}$ is the incident intensity (the beam strength before it enters the shield), $I$ is the surviving intensity (what is left after passing through thickness $x$ ), $μ$ is the linear attenuation coefficient (probability of removal per unit thickness), and $x$ is the shield thickness — nothing here is assumed.

Level 1 — Recognition

Can you pick the right unit and the right factor without any real calculation?

Recall Solution

Source strength → becquerel (Bq). "Bq" answers how much is the source decaying? — it counts disintegrations per second and says nothing about any patient.
Energy per kg deposited → gray (Gy). Recall $1 Gy = 1 J/kg$ — pure physics, energy landing in matter.
Biological-harm estimate → sievert (Sv). Same J/kg but weighted by radiation type. The number happening to equal the Gy value is a hint the radiation here is $γ$ or $β$ (where $w_{R} = 1$ ), so $H = 1 \times D$ .

Recall Solution

$w_{R}$ compares biological damage of a radiation type to a "1 Gy of gamma" baseline.

(a) $γ$ : $w_{R} = 1$ (the reference — sparse ionization).
(b) $α$ : $w_{R} = 20$ (dumps energy in a tiny dense track, shredding DNA).
(c) neutrons: their $w_{R}$ ranges from about 5 up to 20 depending on the neutron's energy; the worst-case value is $w_{R} \approx 20$ . (Here "5 to 20" describes the range of the weighting factor $w_{R}$ , not the neutron energy.)

Level 2 — Application

Plug into one formula, mind the units.

Recall Solution

Use the equivalent-dose formula $H = w_{R} D$ , where $D$ is the absorbed dose in Gy. For $α$ , $w_{R} = 20$ , so $H = 20 \times 0.30 Gy = 6.0 Sv .$ Why ×20? The same joules delivered by alphas cause 20 times the biological harm of gammas — that is exactly what $w_{R}$ encodes.

Recall Solution

The decay constant is the probability per second that a given nucleus decays, and it links to half-life by $λ = \frac{ln 2}{t _{1/2}}$ . First convert the half-life to seconds so the answer comes out "per second" (matching Bq): $t_{1/2} = 5.27 \times 3.156 \times 1 0^{7} = 1.663 \times 1 0^{8} s .$ Then $λ = \frac{0.6931}{1.663 \times 1 0 ^{8}} = 4.17 \times 1 0^{- 9} s^{- 1} .$ Why convert first? Bq means "per second"; mixing years and seconds would give a nonsense unit.

Recall Solution

Absorbed dose is energy per unit mass: $D = \frac{E}{m}$ . $D = \frac{0.10 J}{2.0 kg} = 0.050 Gy = 50 mGy .$ Why divide by mass, not volume? Bond-breaking depends on energy shared among the atoms actually present — per-kilogram is the fair normaliser.

Level 3 — Analysis

Combine two formulas, or reason about behaviour.

Recall Solution

Activity, decay constant and atom count are tied by $A = λ N$ . Solve for $N$ : $N = \frac{A}{λ} = \frac{3.7 \times 1 0 ^{10}}{4.17 \times 1 0 ^{- 9}} = 8.87 \times 1 0^{18} atoms .$ Sanity check: that is under $1 0^{- 5}$ mol of cobalt — a speck — yet it decays ten billion times a second. Small mass, huge activity because $t_{1/2}$ is only a few years, so $λ$ is comparatively large. See Radioactive decay law and Half-life.

Recall Solution

Each HVL multiplies the surviving intensity by $\frac{1}{2}$ . To reach $\frac{1}{8}$ we need $n$ halvings with $(\frac{1}{2})^{n} = \frac{1}{8} = (\frac{1}{2})^{3} \Rightarrow n = 3.$ So thickness $= 3 \times 12 mm = 36 mm$ of lead.

Where does $μ = \frac{ln 2}{HVL}$ come from? The HVL is defined as the thickness $x_{1/2}$ at which the surviving fraction is exactly one half. Put that into the attenuation law: $\frac{I}{I _{0}} = \frac{1}{2} = e^{- μ x_{1/2}} .$ Take the natural log of both sides (the log is what "undoes" the exponential to free $μ$ ): $ln \frac{1}{2} = - μ x_{1/2} \Rightarrow - ln 2 = - μ x_{1/2} \Rightarrow μ = \frac{l n 2}{x _{1/2}} = \frac{l n 2}{HVL} .$ So the HVL and $μ$ are two ways of stating the same physical fact ("how fast the beam dies"). Substituting back, the thickness giving $\frac{1}{8}$ is $x = \frac{ln 8}{μ} = \frac{ln 8}{ln 2} HVL = 3 HVL$ — matching the halving count.

Read the figure below (Fig s01): the horizontal axis is thickness measured in whole HVLs $(0, 1, 2, 3)$ ; the vertical axis is the surviving fraction $I / I_{0}$ . The cyan curve is the smooth exponential $e^{- μx}$ . The amber dots sit on that curve at each whole HVL and are labelled $I_{0}, I_{0} /2, I_{0} /4, I_{0} /8$ — each dot exactly half the height of the one before. The white vertical arrow drops at $x = 3$ HVL to the $I_{0} /8$ level, showing visually that three halvings — not more, not fewer — land the beam on one-eighth.

Figure — Radiation safety — units (Bq, Gy, Sv), shielding

Recall Solution

We need $(\frac{1}{2})^{n} = \frac{1}{100}$ , i.e. $2^{n} = 100$ . Take logs: $n = \frac{l n 100}{l n 2} = \frac{4.605}{0.6931} = 6.64 HVL .$ Thickness $= 6.64 \times 12 mm = 79.7 mm \approx 8.0 cm$ of lead. Why non-integer is fine: attenuation is a smooth exponential ( $I = I_{0} e^{- μx}$ ), not a staircase — HVL is just a convenient ruler for it.

Level 4 — Synthesis

Chain three or more ideas across topics.

Recall Solution

Two independent reducers multiply together. (a) Distance — inverse-square law. From Inverse-square law, intensity falls as $1/ r^{2}$ : $\frac{H _{dist}}{H _{1}} = (\frac{r _{1}}{r _{2}})^{2} = (\frac{1}{4})^{2} = \frac{1}{16} .$ (b) Shielding. $2$ HVL multiplies by $(\frac{1}{2})^{2} = \frac{1}{4}$ . (c) Combine: $H_{2} = 400 \times \frac{1}{16} \times \frac{1}{4} = 400 \times \frac{1}{64} = 6.25 μ Sv/h .$ Why multiply? Distance and shielding act on the beam in sequence and independently, so their reduction factors compose — this is the "Distance + Shielding" pillars of protection working at once.

Read the figure below (Fig s02): the amber dot at the far left is the source (400 µSv/h). A cyan arrow measures $r_{1} = 1 m$ and a dashed white arrow measures the full $r_{2} = 4 m$ along the beam axis. Just in front of the worker sits the hatched cyan slab — the lead shield of $2$ HVL. The white dot at right is the worker, receiving 6.25 µSv/h. The amber caption spells out the two composing factors — distance $\frac{1}{16}$ and shield $\frac{1}{4}$ — multiplying to the total $\frac{1}{64}$ .

Recall Solution

Step by step, watching every unit. Absorbed dose: dose rate $\times$ activity $\times$ time. $D = (5.0 \times 1 0^{- 9} \frac{Gy}{h \cdot Bq}) \times (2.0 \times 1 0^{7} Bq) \times (3.0 h) .$ $D = 5.0 \times 1 0^{- 9} \times 2.0 \times 1 0^{7} \times 3.0 = 0.30 Gy .$ Equivalent dose: $H = w_{R} D = 1 \times 0.30 = 0.30 Sv = 300 mSv$ . Why $w_{R} = 1$ ? It is gamma, so Gy and Sv coincide numerically — but we still write the $w_{R}$ to keep the harm interpretation explicit. This one exposure exceeds a typical annual occupational limit — the Time pillar (cut the 3 h) would be the first lever to pull.

Level 5 — Mastery

Design, invert, or reason about limiting/degenerate cases.

Recall Solution

(a) Start from the attenuation law $I = I_{0} e^{- μx}$ and set $\frac{I}{I _{0}} = 0.005$ : $0.005 = e^{- μx} \Rightarrow ln (0.005) = - μx \Rightarrow x = \frac{- l n ( 0.005 )}{μ} .$ $x = \frac{5.298}{0.080 cm ^{- 1}} = 66.2 cm of concrete .$ (b) One HVL $= \frac{ln 2}{μ} = \frac{0.6931}{0.080} = 8.66 cm$ . Number of HVLs: $\frac{66.2}{8.66} = 7.64 HVL (check: 2^{7.64} \approx 200 = \frac{1}{0.005}) .$ Why invert the exponential with a log? The unknown $x$ sits inside $e^{- μx}$ ; the natural log is the tool that "undoes" the exponential, freeing $x$ .

Recall Solution

(a) $x = 0$ : $I = I_{0} e^{0} = I_{0}$ . No shield $\Rightarrow$ full beam — correct sanity check.
(b) $x \to \infty$ : $e^{- μx} \to 0$ , so $I \to 0$ but never exactly reaches 0 for any finite thickness. Physically: γ rays are attenuated, never fully stopped — there is always some (astronomically tiny) leak. This is why γ shielding is quoted as HVLs "to an acceptable level", not "complete blocking". Contrast with $α$ , which has a hard finite range (stopped by paper) — see Types of radiation (alpha beta gamma).

Recall Solution

(a) Harm scales with $w_{R}$ : $α$ gives $H = 20 \times 1 = 20 Sv$ ; $γ$ gives $H = 1 \times 1 = 1 Sv$ . So alpha is 20× more damaging per gray absorbed — its dense ionization track breaks many bonds close together, overwhelming repair (see Biological effects of radiation). (b) That factor of 20 only matters if the energy is actually deposited in living tissue. Alphas are stopped by the dead outer skin layer or a sheet of paper — so an external α source deposits almost nothing in vital tissue ( $D \approx 0$ , hence $H \approx 0$ ). The danger flips only when the emitter is inhaled or ingested, placing the α track right inside living cells, where that same $\times 20$ harm factor now acts on DNA at full strength. Gamma, by contrast, penetrates deeply from outside, so it delivers dose to internal organs whether the source is inside you or across the room. Reconciliation & lesson: the two facts are about two different quantities. Fact (a) is about harm per unit absorbed dose ( $w_{R}$ ); fact (b) is about whether the dose is absorbed at all (penetration/range). Total risk $=$ (does the radiation reach the tissue?) $\times$ (how nasty per gray once there?). Alpha scores near-zero on the first factor externally but maximal on the second internally — which is exactly why the rule of thumb is: alpha is harmless in your hand but deadly in your lungs. Always judge both penetration and $w_{R}$ , never one alone.

Recall Self-test scoreboard

Level cleared cleanly? ::: Move up. Stuck on a step? Reread that formula's derivation in the parent note before advancing. The one habit that makes all of this automatic ::: For every number, name its unit and ask "energy (Gy) or harm (Sv)?", and for every reduction, multiply factors — never add.