Atomic, nuclear and quantum physics for Cambridge IGCSE Physics

Reading cue. The paper and the 5 mm aluminium are alternative absorbers put in the same place, one at a time and never stacked, so each reading tests one absorber only.

1. aState State the quantity that Fig. 2.1 deliberately keeps unchanged when the absorber is put in place, and state why it must be kept unchanged.

   routine2 marks

   Check answer · 2 marks

   1. the distance between the source and the window of the tube (1)
   2. so that any fall in count rate is caused by the absorber alone, and not by the radiation spreading out over a greater distance (1)

2. bDescribe Describe how the counter in Fig. 2.1 should be used so that the reading with the absorber can be fairly compared with the reading without it.

   routine3 marks

   Check answer · 3 marks

   1. first record the count with the source removed, to obtain the background (1)
   2. take every count over the same measured counting time, or convert each count to a count rate (1)
   3. subtract the background from each reading before the two are compared (1)

3. cExplain With no absorber the counter records 620 counts in 100 s. With the sheet of paper in place, at the same distance, it records 611 counts in 100 s. Explain whether this shows that the paper absorbs some of the radiation.

   demanding4 marks

   Check answer · 4 marks

   1. count rates are 620 / 100 = 6.2 counts/s and 611 / 100 = 6.1 counts/s (1)
   2. radioactive decay is random, so repeated counts vary even when nothing has changed (1)
   3. the difference of about 0.1 counts/s is no larger than this random variation, so it is not evidence of absorption (1)
   4. count for much longer, or repeat the readings, to decide (1)

4. dSuggest The paper is replaced by the 5 mm sheet of aluminium, at the same distance, and the counter records 33 counts in 100 s; with the source taken right away it records 30 counts in 100 s. From these readings a student writes that the aluminium stops all the radiation from the source. Suggest why that conclusion is not safe, and suggest one change to the arrangement in Fig. 2.1 that would test it.

   top of the paper5 marks

   Check answer · 5 marks

   1. corrected count rate with the aluminium = (33 − 30) / 100 = 0.03 counts/s, which is small but not zero (1)
   2. a difference of 3 counts is smaller than the random variation in counts of this size, so the readings cannot show whether anything is getting through (1)
   3. a weak, more penetrating emission such as gamma could be present as well and would still be passing through the aluminium (1)
   4. put lead absorbers of increasing thickness at the same position and look for any further fall in the corrected count rate (1)
   5. count for far longer at each thickness, so that the random variation is small compared with the fall being looked for (1)

Reading cue. The plates, the air gap, the cell and the alarm form one series loop, so whatever happens in the gap fixes the current everywhere; the arrow shows smoke entering, not the source moving.

1. aState State the part of the series circuit drawn in Fig. 6.1 that the radiation from the source must act on before any current can flow round the loop, and state why.

   recall2 marks

   Check answer · 2 marks

   1. the air gap between the two plates (1)
   2. air is an insulator unless it is ionised, and the radiation from the source produces the ions that carry the charge across the gap (1)

2. bExplain The source in Fig. 6.1 emits α-particles. Explain why the current in the circuit falls when smoke drifts sideways into the gap, as shown by the arrow.

   routine3 marks

   Check answer · 3 marks

   1. smoke particles absorb the α-particles before they have crossed the whole gap (1)
   2. fewer ions are produced in the air, and ions that do form attach to the smoke particles and move more slowly (1)
   3. so less charge passes between the plates each second, the current falls and the alarm circuit is triggered (1)

3. cExplain Explain why the α-source in Fig. 6.1 is fixed to the underside of the upper plate rather than mounted outside the chamber, and why the gap between the plates is only a few millimetres wide.

   demanding4 marks

   Check answer · 4 marks

   1. alpha particles travel only a few centimetres in air (1)
   2. mounted outside, they would be absorbed by the wall of the chamber and no ions would be made in the gap (1)
   3. a narrow gap means ions are produced right across it, all the way to the lower plate (1)
   4. so a steady, measurable current can be maintained between the plates (1)

4. dSuggest A manufacturer suggests replacing the α-source in Fig. 6.1 with a beta source of the same activity. Suggest the effect on the current between the plates and on how well the alarm works.

   top of the paper4 marks

   Check answer · 4 marks

   1. beta particles are much less strongly ionising than alpha particles (1)
   2. far fewer ions are produced in the gap each second, so the current is much smaller (1)
   3. beta is barely absorbed by smoke, so the current changes very little when smoke enters (1)
   4. the alarm becomes unreliable or fails to trigger, and the more penetrating beta also escapes through the casing (1)

Reading cue. The three lines are the paths of three different α-particles, not one particle bouncing about; each angle is measured from the original beam direction set by the channel in the lead.

1. aState State the purpose of the narrow channel cut through the lead block in Fig. 7.1.

   recall2 marks

   Check answer · 2 marks

   1. the lead absorbs alpha particles emitted in all other directions (1)
   2. so a narrow beam of known direction reaches the foil, against which deflection angles can be measured (1)

2. bDescribe Describe the force that acted on the α-particle following the middle path in Fig. 7.1, the one deflected upwards through a moderate angle, and describe where in the gold atom that force acted.

   routine3 marks

   Check answer · 3 marks

   1. an electrostatic force of repulsion (1)
   2. between the positively charged α-particle and the positively charged nucleus (1)
   3. acting as the particle passed close to a nucleus, but not directly at it (1)

3. cExplain Explain how Fig. 7.1 would have to be redrawn if the positive charge of each gold atom were spread evenly throughout the whole atom, as in the earlier model of the atom.

   demanding4 marks

   Check answer · 4 marks

   1. only the straight-through path, with at most very small deflections, would be drawn (1)
   2. no path turned back through more than 90° would appear (1)
   3. charge spread through the whole atom produces a much weaker repulsion at any point in it (1)
   4. this force is far too small to reverse the motion of a fast, massive α-particle (1)

4. dSuggest The gold foil in Fig. 7.1 is replaced by an aluminium foil of the same thickness. An aluminium nucleus holds 13 protons and a gold nucleus holds 79. Suggest how the three paths drawn would change.

   top of the paper4 marks

   Check answer · 4 marks

   1. most α-particles would still pass straight through, so that path is unchanged (1)
   2. the aluminium nucleus carries a much smaller positive charge than the gold nucleus (1)
   3. so the repulsive force at the same distance of approach is much smaller (1)
   4. fewer particles are deflected through large angles, and far fewer are turned back through more than 90° (1)

Reading cue. The window-to-source gap is fixed for every reading, and the counter shows a total number of counts, not a rate, so divide by the counting time before comparing any two readings.

1. aState State why the tube in Fig. 10.1 is clamped with its window pointing straight down at the source, and state one effect on the results if the retort stand were nudged part-way through the experiment.

   recall2 marks

   Check answer · 2 marks

   1. the same fraction of the radiation emitted by the source enters the window for every reading (1)
   2. nudging the stand changes the distance or direction, so the count rate would change for a reason other than the decay of the source (1)

2. bDetermine With the source taken away but the stand and tube left exactly as drawn, the counter records 300 counts in 12.0 minutes. The source is then placed at the fixed distance and the counter records 1085 counts in 5.0 minutes. Determine the count rate due to the source alone.

   routine3 marks

   Check answer · 3 marks

   1. background rate = 300 / 12.0 = 25 counts/min (1)
   2. measured rate with source = 1085 / 5.0 = 217 counts/min (1)
   3. corrected rate = 217 − 25 = 192 counts/min (1)

3. cDetermine The corrected count rate due to the source is 192 counts/min at the start, and the half-life of the source is 30 minutes. Determine the total number of counts the counter in Fig. 10.1 will record in one minute, 90 minutes after the start.

   demanding4 marks

   Check answer · 4 marks

   1. 90 minutes is 3 half-lives (1)
   2. 192 → 96 → 48 → 24, so the source alone gives 24 counts/min (1)
   3. the background must be added back, because the tube detects it as well as the source (1)
   4. total recorded = 24 + 25 = 49 counts in that minute (1)

4. dSuggest The student repeats the whole experiment with the tube clamped twice as far above the source, everything else unchanged. Suggest how her corrected count rates and her value of the half-life would each be affected.

   top of the paper4 marks

   Check answer · 4 marks

   1. every corrected count rate would be much smaller (1)
   2. the radiation spreads out in all directions, so a smaller fraction of it enters the window (1)
   3. the background and the random variation then form a larger proportion of each reading, so the results are less reliable (1)
   4. the half-life obtained would be the same, because it is a property of the source and does not depend on how much of the radiation is detected (1)