Worked examples — Radiation safety — units (Bq, Gy, Sv), shielding

This is the drill page for Radiation safety — units & shielding. The parent note built the three units and the two exponential laws. Here we hammer every kind of question they can produce, one worked example per case, so that no exam scenario is new to you.

Before we start, we name every symbol that reappears below, in plain words, so nothing is used unearned:

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The scenario matrix

Every question in this topic lands in exactly one of these cells. Each cell gets its own worked example.

#	Case class	What makes it different	Example
C1	Straight forward activity → atoms	plug into $A=\lambda N$	Ex 1
C2	Decay in time (multiple half-lives)	$A=A_0{e}^{-\lambda t}$, whole number of $t_{1/2}$	Ex 2
C3	Non-integer time	$t$ not a whole multiple of $t_{1/2}$ → need $\ln$/exp	Ex 3
C4	Gy → Sv, harmful radiation	$w_R>1$ (alpha)	Ex 4
C5	Degenerate weighting	$w_R=1$ (gamma) → Gy $=$ Sv, the trap	Ex 5
C6	Shielding, integer HVL	intensity ratio is a power of $1/2$	Ex 6
C7	Shielding, non-integer HVL	ratio not a power of $1/2$ → need $\log$	Ex 7
C8	Distance only (inverse-square)	no shielding, $I\propto 1/r^2$	Ex 8
C9	Limiting / zero cases	$t=0$, $x=0$, $x\to \infty$, $t\to \infty$	Ex 9
C10	Real-world combined exam twist	Time + Distance + Shielding together	Ex 10

Constants used throughout: $\ln 2=0.693$, $1\text{ yr}=3.156\times 10^7\text{ s}$.

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Worked examples

Ex 1 — Cell C1: activity to number of atoms

Forecast: $A$ (decays per second) and $N$ (a count of atoms) have different units, so don't compare them directly. Instead guess the order of magnitude of $N$: since only a tiny fraction $\lambda \approx 10^{-5}$ of the atoms decay each second, do you expect $N$ to be far larger than the $2\times 10^9$ figure, or far smaller? Sketch it: picture a huge pile of atoms with only a thin sliver flaking off per second — that sliver is $A$, the whole pile is $N$.

1. Convert the half-life to seconds. $t_{1/2}=6.00\times 3600=2.16\times 10^4\text{ s}$. Why this step? Bq is per second, so $\lambda$ must also be per second — units must match before dividing.
2. Find $\lambda$. $\lambda =\frac{\ln 2}{t_{1/2}}=\frac{0.693}{2.16\times 10^4}=3.21\times 10^{-5}{\text{ s}}^{-1}$. Why this step? We only have activity and half-life; $\lambda$ is the bridge that lets us use $A=\lambda N$.
3. Solve for $N$. $N=\frac{A}{\lambda }=\frac{2.00\times 10^9}{3.21\times 10^{-5}}=6.23\times 10^{13}$ atoms. Why this step? Rearrange $A=\lambda N$ — the one relation linking a rate to a count.

Verify: units $\frac{\text{decay/s}}{1/\text{s}}=\text{decay}=$ atoms ✓. Forecast confirmed — because only a $10^{-5}$ sliver decays per second, the whole pile $N$ ($6.23\times 10^{13}$) is about $10^4$ times bigger than the per-second rate; check by re-multiplying: $\lambda N=3.21\times 10^{-5}\times 6.23\times 10^{13}=2.0\times 10^9$ Bq ✓.

See Radioactive decay law and Half-life for the underlying law.

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Ex 2 — Cell C2: decay over a whole number of half-lives

Forecast: 24 h with $t_{1/2}=6$ h. Sketch a staircase going down: each 6-hour step drops the height to half. Count the steps in your head, picture how low the last stair sits, and guess the final activity before computing.

1. Count the half-lives. $n=\frac{24}{6}=4$ half-lives. Why this step? When time is a clean multiple of $t_{1/2}$, we skip exponentials — each half-life just halves the number.
2. Halve four times. $A=A_0{\left(\frac{1}{2}\right)}^4=2.00\times 10^9\times \frac{1}{16}=1.25\times 10^8$ Bq. Why this step? ${\left(\frac{1}{2}\right)}^n$ IS $e^{-\lambda t}$ when $t=n{t}_{1/2}$; using powers of a half is the shortcut.

Verify: via the exponential to be sure — $\lambda t=3.21\times 10^{-5}\times (24\times 3600)=2.77$, and $e^{-2.77}=0.0625=\frac{1}{16}$ ✓. Same answer, so the shortcut is valid.

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Ex 3 — Cell C3: non-integer time (must use exp/ln)

Forecast: 8 yr is between 1 and 2 half-lives, so the answer must land between $A_0/2$ and $A_0/4$. Sketch a smooth decay curve and put your finger on it a bit past the first halving — that finger height is your guess. Is it nearer $A_0/2$ or $A_0/4$?

1. Get $\lambda$ in per-year first (time is given in years, so keep years). $\lambda =\frac{0.693}{5.27}=0.1315{\text{ yr}}^{-1}$. Why this step? $\lambda$ and $t$ must share a unit inside the exponent; both in years keeps it clean.
2. Form the exponent. $\lambda t=0.1315\times 8.0=1.052$. Why this step? $8$ yr is not a whole number of half-lives, so ${\left(\frac{1}{2}\right)}^n$ won't give a tidy fraction — we need the true exponential.
3. Evaluate. $A=A_0{e}^{-1.052}=3.7\times 10^{10}\times 0.349=1.29\times 10^{10}$ Bq. Why this step? $e^{-\lambda t}$ is the exact surviving fraction for any $t$.

Verify: $8/5.27=1.52$ half-lives, so fraction should be $(\frac{1}{2}{)}^{1.52}=0.349$ ✓, and $1.29\times 10^{10}$ lies between $A_0/4=0.93\times 10^{10}$ and $A_0/2=1.85\times 10^{10}$ — exactly where the forecast said. ✓

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Ex 4 — Cell C4: Gy → Sv with damaging radiation

Forecast: alpha is the "sharp pin". Picture the same amount of energy delivered as one dense stab versus a soft glow — will the sievert number be bigger or smaller than the $0.25$ gray?

1. Pick the weighting factor. Alpha ⇒ $w_R=20$ (from Types of radiation (alpha beta gamma)). Why this step? Sv exists precisely because a joule of alpha rips DNA far worse than a joule of gamma — the factor encodes that (see Biological effects of radiation).
2. Apply $H=w_{R}D$. $H=20\times 0.25=5.0$ Sv. Why this step? This is the definition of equivalent dose — energy-per-kg re-weighted by biological punch.

Verify: units $[]\times \text{J/kg}=\text{J/kg (weighted)}=$ Sv ✓. Sanity: 5 Sv is a lethal-range whole-body dose from just $0.25$ Gy — that 20× amplification is the whole reason Sv is not Gy. ✓

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Ex 5 — Cell C5: degenerate weighting (the Gy = Sv trap)

Forecast: same 0.25 Gy as Ex 4, but now a soft glow instead of a stab. Picture the weighting bar shrinking down to length 1 — will the harm match or beat the gray number?

1. Weighting for gamma. $w_R=1$. Why this step? Gamma spreads its ionization thinly; it is the reference radiation, so its factor is exactly 1.
2. Apply $H=w_{R}D$. $H=1\times 0.25=0.25$ Sv. Why this step? With $w_R=1$ the number of sievert equals the number of gray — a coincidence, not a law.

Verify: compare cells — same $0.25$ Gy gives $0.25$ Sv (gamma) but $5.0$ Sv (alpha), a 20× gap for identical absorbed energy. This is the degenerate case that makes people wrongly write "Gy = Sv". ✓

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Ex 6 — Cell C6: shielding, integer number of HVLs

Forecast: $1/16$ is a clean power of a half. Look at the figure below and imagine climbing down the magenta dots, each one at half the previous height — how many dots down is $1/16$?

1. Express the ratio as a power of a half. $\frac{1}{16}={\left(\frac{1}{2}\right)}^4$, so $n=4$ HVL. Why this step? Each HVL halves the beam (same "constant chance per slab" logic as decay), so integer halvings map straight to a power of $\frac{1}{2}$.
2. Multiply by the HVL. $x=n\times \text{HVL}=4\times 8.0=32\text{ mm}$. Why this step? Thickness is just the number of half-value layers stacked.

Verify: via the law — using $\mu =\frac{\ln 2}{x_{1/2}}=\frac{\ln 2}{8.0\text{mm}}=0.0866{\text{ mm}}^{-1}$ (the HVL↔$\mu$ link, with $\mu$ in mm${}^{-1}$ because the HVL is in mm), $I/I_0=e^{-\mu x}=e^{-0.0866\times 32}=e^{-2.77}=0.0625=\frac{1}{16}$ ✓. Note $\mu x=0.0866{\text{mm}}^{-1}\times 32\text{mm}$ is dimensionless, as it must be. See the fall-off in the figure below.

Look at the magenta dots on the figure: they mark each successive HVL, and each one sits at half the height of the one before — four dots bring you to $1/16$.

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Ex 7 — Cell C7: shielding, non-integer HVL (need logarithms)

Forecast: $1/100$ is NOT a power of a half. On the figure, find the height $1/100$ and slide across to the curve — does it land between the 6th and 7th magenta dots? Guess which two.

1. Set up the ratio. We need ${\left(\frac{1}{2}\right)}^n=\frac{1}{100}$, i.e. $2^n=100$. Why this step? $1/100$ can't be reached by whole halvings, so $n$ will be a fraction — we must solve, not count.
2. Solve with a logarithm. $n={\log }_{2}100=\frac{\ln 100}{\ln 2}=\frac{4.605}{0.693}=6.64$ HVL. Why this step? The log is the tool that answers "which power gives this ratio?" — exactly what $\arctan$ does for angles, $\log$ does for exponents.
3. Convert to thickness. $x=6.64\times 8.0=53.1\text{ mm}$. Why this step? Same "HVLs → thickness" conversion as Ex 6.

Verify: using $\mu =\frac{\ln 2}{8.0\text{mm}}=0.0866{\text{ mm}}^{-1}$ (mm${}^{-1}$ to match the mm thickness), $e^{-\mu x}=e^{-0.0866\times 53.1}=e^{-4.60}=0.01000=\frac{1}{100}$ ✓. And $6.64$ sits between the 6-HVL ($1/64$) and 7-HVL ($1/128$) marks — right where the forecast pointed. On the figure, this thickness lands between two magenta HVL dots, confirming the non-integer answer.

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Ex 8 — Cell C8: distance only (inverse-square law)

Forecast: distance ×4. On the orange curve in the figure below, picture the same energy smeared over an ever-bigger sphere. Slide from $r=1$ to $r=4$ — does the rate drop by 4, 8, or 16? Guess before you read on.

1. Invoke the inverse-square law. $I\propto \frac{1}{r^2}$ (from Inverse-square law). Here $I$ is the dose rate arriving at distance $r$. Why this step? A point source's energy spreads over a sphere of area $4\pi r^2$; four times the radius means the same energy is smeared over $4^2=16$ times the area.
2. Scale the rate. $I_2=I_1{\left(\frac{r_1}{r_2}\right)}^2=400\times {\left(\frac{1.0}{4.0}\right)}^2=400\times \frac{1}{16}=25\mu \text{Sv/h}$. Why this step? Ratios of intensities equal the inverse ratio of squared distances — no absolute source strength needed.

Verify: $25\times 16=400$ ✓. Distance quadrupled ⇒ rate cut by 16, not 4 — the square is doing the work. This is the cheapest of the three protection pillars: back away.

The figure shows the orange $1/r^2$ curve: doubling $r$ quarters the rate, quadrupling it drops to a sixteenth — the drop is far steeper than a straight line.

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Ex 9 — Cell C9: limiting and degenerate cases

Forecast: picture two decay curves side by side, one for $I$ vs thickness and one for $A$ vs time. Both start at full height and slide toward the floor. Guess: which of the four edge cases sit exactly at the start value, which hit exactly zero, and which only approach zero without ever touching?

1. Case (a), $x=0$: $I=I_0{e}^{-\mu \cdot 0}=I_0{e}^0=I_0$. Why this step? No shield ⇒ beam unchanged; $e^0=1$ encodes "nothing removed."
2. Case (b), $x\to \infty$: $I=I_0{e}^{-\mu x}\to I_0\cdot 0=0$, but only in the limit — for any finite lead, $I>0$. Why this step? Exponentials approach zero but never touch it; this is WHY gamma is "attenuated, never fully stopped."
3. Case (c), $t=0$: $A=A_0{e}^{-\lambda \cdot 0}=A_0{e}^0=A_0$. Why this step? Clock start ⇒ full activity; the same $e^0=1$ logic as case (a), just with time in place of thickness. The source hasn't decayed at all yet.
4. Case (d), $t\to \infty$: $A=A_0{e}^{-\lambda t}\to A_0\cdot 0=0$, again only in the limit — for any finite wait, $A>0$. Why this step? Given infinite time every nucleus eventually decays, so the count drains to zero; but like case (b), it approaches zero asymptotically and never exactly reaches it in finite time.

Verify: all four follow from $e^0=1$ and $e^{-\infty }=0$. The crucial safety takeaway: cases (b) and (d) reach zero only at infinity — no finite shield and no finite wait removes 100% of gamma or activity. Cases (a) and (c) sit exactly at the start value $I_0$ / $A_0$, because $e^0=1$ leaves the starting amount untouched. ✓

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Ex 10 — Cell C10: real-world combined exam twist

Forecast: three independent factors multiply. Picture three sliders — Distance, Shielding, Time — each pulling the dose down by its own amount. Roughly, is the combined cut around 10×, 100×, or 1000×? Guess before stacking them.

1. Distance factor (inverse-square). ${\left(\frac{1.0}{2.0}\right)}^2=\frac{1}{4}$, so rate $\to 800\times \frac{1}{4}=200\mu \text{Sv/h}$. Why this step? Pillar 1 — Distance; from Ex 8 the square law rules a point source.
2. Shielding factor. $2$ HVL ⇒ ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$, so rate $\to 200\times \frac{1}{4}=50\mu \text{Sv/h}$. Why this step? Pillar 2 — Shielding; each HVL halves (Ex 6). Distance and shielding factors multiply because they act on the same beam in sequence.
3. Time factor (dose = rate × exposure time). Dose $=50\mu \text{Sv/h}\times t_{\text{exp}}=50\times 0.5=25\mu \text{Sv}$. Why this step? Pillar 3 — Time; dose is the steady rate multiplied by how long she stands there, $t_{\text{exp}}$ (not a decay time).
4. Overall cut. Un-mitigated dose $=800\times 2=1600\mu \text{Sv}$; mitigated $=25\mu \text{Sv}$; factor $=\frac{1600}{25}=64$. Why this step? The three factors compose: $4\times 4\times 4=64$, matching the ratio directly.

Verify: $4\times 4\times 4=64$ ✓ and $1600/64=25\mu \text{Sv}$ ✓. Every pillar (Time–Distance–Shielding) contributed a factor of 4 here — the multiplicative power of stacking all three is the whole point of practical safety. ✓

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Recall Which cell is each question?

Match the question type to its master relation before computing. Activity ↔ atom count uses which formula? ::: $A=\lambda N$. "After 3 half-lives" — shortcut? ::: Multiply by ${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$; no exponential needed. "After 7 years, $t_{1/2}=5.27$ yr" — which tool? ::: Non-integer time ⇒ use $A=A_0{e}^{-\lambda t}$. "Reduce to 1/100" through lead — which tool? ::: Non-integer HVL ⇒ $n={\log }_{2}100\approx 6.64$. Move from 1 m to 5 m — factor on rate? ::: ${\left(\frac{1}{5}\right)}^2=\frac{1}{25}$ (inverse-square). Why can no finite shield give zero intensity? ::: $e^{-\mu x}>0$ for all finite $x$; it reaches 0 only as $x\to \infty$.