Radioactivity Worked ExamplesIGCSE Physics 0625

Exposure is reduced in exactly three ways: less time near the source, more distance between the source and the body, and shielding that absorbs the radiation. Test each action in each list against these three.

B: a lead-lined box is shielding; working quickly reduces time; long tongs at arm's length increase distance. Every action reduces exposure, so B is correct.

A is wrong because short tongs bring the source closer to the body. C is wrong because choosing a longer half-life is not a precaution: it changes none of time, distance or shielding, and the source simply stays radioactive for longer. D is wrong because holding the source in a hand puts it in contact with the body (a glove does not stop gamma rays), and working slowly increases the time of exposure.

(a) The detector also records background radiation, so subtract $30$ counts/min from every reading.

(b) Paper reduces the corrected count rate by only $10$ counts/min (from $1820$ to $1810$), which is within random variation. Alpha particles are stopped by paper, so the source emits no α-particles.

$5$ mm of aluminium causes a large fall (from $1810$ to $300$). Beta particles pass through paper but are stopped by a few millimetres of aluminium, so the source emits β-particles.

Even behind $20$ mm of lead the corrected count rate is $40$ counts/min, which is still above zero, so some radiation passes through aluminium and is only partly absorbed by thick lead. That is γ-radiation.

(c) Radioactive emission is random, so the count in a short time varies a lot. Counting over a longer time (or repeating and averaging) gives a more reliable count rate.

A β-particle is a fast-moving electron emitted from the nucleus. Its nucleon number is $0$ and its charge (proton number) is $-1$, so it is written:

Nucleon numbers balance on the top row: $131 = A + 0$, so the xenon nucleus has $A = 131$.

Proton numbers balance on the bottom row: $53 = Z + (-1)$, so $Z = 54$. The proton number goes up by one, which is why the element changes.

The complete equation:

(b) One neutron in the nucleus changes into a proton and an electron; the electron is emitted as the β-particle. So the nucleus has one neutron fewer and one proton more.

(a) An α-particle is a helium nucleus: $2$ protons and $2$ neutrons, written $^{4}_{2}\alpha$. Subtract these from the parent nucleus.

Write the equation and check both rows balance ($211 = 207 + 4$, $83 = 81 + 2$).

(b) Two marking points, each a comparison: α-particles have more kinetic energy than β-particles, and α-particles have a greater charge ($+2$ compared with $-1$). So each α-particle removes electrons from many more atoms along its path.

(c) α-particles have a very short range: they are stopped by a few centimetres of air or by paper, so from outside the body they would not penetrate the skin. Placed inside the tumour, all their ionisation happens in the cancer cells, killing them.

Track the two numbers one decay at a time. β-emission: nucleon number unchanged, proton number up by $1$. α-emission: nucleon number down by $4$, proton number down by $2$.

Apply the three decays in order.

So the answer is B. A ignores both β-decays. C wrongly takes each β-decay as lowering the proton number. D ignores the α-decay. (The final nucleus has proton number $82$ again, so it is lead-208, a different isotope of the starting element.)

(a) γ-radiation is electromagnetic and has no charge, so it continues in a straight line, undeflected.

α-particles are positive, so they curve towards the negative (top) plate. The curve starts as soon as they enter the field, and it is a smooth curve, not a sharp bend.

β-particles are negative, so they curve towards the positive (bottom) plate, in the opposite direction. They are deflected more than the α-particles because their mass is so much smaller.

(b) Marking points: β-particles are negatively charged; a charged particle in an electric field experiences a force; opposite charges attract, so the force is towards the positive plate and the path curves that way.

(c) The paper absorbs the α-particles, so the α path disappears. β-particles and γ-radiation pass through paper, so their paths are unchanged.

The force on a moving charge in a magnetic field is perpendicular to both the field and the motion, so the field cannot point along the deflection. That rules out C (and D), which are in the plane of the page.

β-particles are electrons (negative), so the conventional current is in the opposite direction to their motion: towards the left.

Use Fleming's left-hand rule: thuMb = force (towards the top of the page), seCond finger = current (to the left). The First finger (field) then points out of the page.

So the answer is A. B is what you get if you forget that the β-particles are negative and point the current to the right.

(a) Only the count rate from the sample halves. At $t = 0$ the corrected count rate is:

Half of this is $400$ counts/min. Add the background back on to find where to read the measured graph:

Read across from $500$ to the curve and down: this happens at $t = 10$ min.

Check with a second half-life: corrected $400 \rightarrow 200$, measured $200 + 100 = 300$, which the graph reaches at $t = 20$ min, another $10$ min later. ✓

(b) After many half-lives the sample's contribution becomes almost zero, so the measured count rate levels off at the background value.

(a) Background radiation is the ionising radiation that is always present in the environment, from natural (and some artificial) sources, even when no source is being used. Sources include: radon gas in the air, rocks and buildings, food and drink, cosmic rays (any two).

(b)(i) Subtract background to get the count rate due to the source alone.

Number of half-lives in $9.0$ hours:

Halve the source's count rate three times.

The background is still there, so add it back on to give the detector reading.

(ii) Each half-life halves the number of undecayed nuclei.

(a) Link the choice to what happens to the count rate. α-particles would be completely stopped by $1$ mm of plastic, so the detector would read only background whatever the thickness. γ-radiation would pass through almost unaffected, so the count rate would hardly change when the thickness changes.

β-particles are partly absorbed by $1$ mm of plastic, so the count rate changes noticeably when the thickness changes: a thicker sheet absorbs more and the count rate falls.

(b) More β-particles are getting through, so the sheet has become thinner. The controller should increase the gap between the rollers (reduce the pressure) to make the sheet thicker again.

(c) With a long half-life the activity (and so the count rate for a given thickness) stays almost constant over months, so a change in count rate is caused by a change in thickness, not by the source decaying, and the source does not need frequent replacement.

(a) The half-life is the time for the activity to fall to half: read across from $50\%$ to the curve and down.

(b) Read across from $80\%$ to the curve and down to the age axis. The curve crosses $80\%$ just before $10$ years.

So the source must be replaced after about $10$ years, well before one half-life has passed. Note how the decision comes from the graph reading at $80\%$, not from the half-life.

(c) With a lower activity the count rate is lower, so a small change in thickness produces a smaller change in count rate. Because emission is random, the count rate fluctuates anyway; the small change becomes hard to distinguish from these random fluctuations.

(a) The half-life of an isotope is the time taken for half the nuclei of that isotope in any sample to decay.

(b) Number of half-lives in $24$ days:

Undecayed nuclei after three half-lives:

The question asks how many have decayed, so subtract from the starting number.

(c) Halve repeatedly and compare with $1\%$ ($0.01$): after $6$ half-lives $\tfrac{1}{64} \approx 1.6\%$ remains, which is still too much; after $7$ half-lives $\tfrac{1}{128} \approx 0.78\%$ remains, which is below $1\%$.

So $7$ half-lives are needed.

(d) Radioactive decay is spontaneous and random: every undecayed nucleus has the same chance of decaying in a given time, and nothing (such as temperature) changes it, so the half-life describes a large number of nuclei, not one.