IB · E · data analysis
Nuclear and quantum physics · Question 7
Nuclear and quantum physics · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 12
- Topics
- 1
- Answer
- Complete
A student measures the activity A of a radioactive source at intervals, correcting each reading for background. She expects A = A₀e^(−λt).
| t / minutes | 0 | 10 | 20 | 30 | 40 |
|---|---|---|---|---|---|
| A / Bq | 500 | 397 | 315 | 250 | 198 |
| ln (A / Bq) | 6.215 | 5.984 | 5.753 | 5.521 | 5.288 |
- (a)
Explain Explain why a graph of ln A against t is plotted, and state what its gradient represents.
3 marks - (b)
Determine Determine the decay constant and hence the half-life of the source.
4 marks - (c)
Outline Outline why the readings had to be corrected for background before being processed.
2 marks - (d)
Explain The student's individual readings scatter about the line even after correction. Explain why this scatter cannot be removed by taking more care with the apparatus.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Explain Explain why a graph of ln A against t is plotted, and state what its gradient represents.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
taking natural logarithms of A = A₀e^(−λt) gives ln A = ln A₀ − λt
- 2
this is linear in t, so the points lie on a straight line if the decay is exponential — which the plot therefore also tests
- 3
the gradient is −λ, the negative of the decay constant
(b)
Determine Determine the decay constant and hence the half-life of the source.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
gradient = (5.288 − 6.215)/(40 − 0) = −0.0232 min⁻¹
- 2
λ = 0.0232 min⁻¹
- 3
t½ = ln 2 / λ = 0.693/0.0232
- 4
t½ = 30 minutes
(c)
Outline Outline why the readings had to be corrected for background before being processed.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
the detector registers radiation from cosmic rays, rocks and other natural sources as well as from the sample
- 2
background adds a constant to every reading, which is a systematic error — it does not decay away, so an uncorrected plot of ln A against t would curve rather than being straight
(d)
Explain The student's individual readings scatter about the line even after correction. Explain why this scatter cannot be removed by taking more care with the apparatus.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
radioactive decay is a random process — which nucleus decays, and when, cannot be predicted
- 2
so the number of decays counted in any fixed interval fluctuates about a mean value, and this fluctuation is a property of the process, not of the instrument
- 3
it can only be reduced by counting for longer or over more nuclei, since the fractional fluctuation falls as the total count rises — not by improving the apparatus
Private on this device