A technician investigates how the radiation from a radioactive source changes over time. The results from the detector are: at t=0t = 0t=0 min, the count rate is 510 counts per minute510\text{ counts per minute}510 counts per minute; at t=3.0t = 3.0t=3.0 min, it is 270 counts per minute270\text{ counts per minute}270 counts per minute; at t=6.0t = 6.0t=6.0 min, it is 150 counts per minute150\text{ counts per minute}150 counts per minute; at t=9.0t = 9.0t=9.0 min, it is 90 counts per minute90\text{ counts per minute}90 counts per minute; and at t=12.0t = 12.0t=12.0 min, it is 60 counts per minute60\text{ counts per minute}60 counts per minute. The half-life of the source is known to be exactly 3.0 minutes. Explain why the measured count rate is always greater than half of the previous reading, and calculate the background count rate in counts per minute.
484 exam-style questions on CIE IGCSE Physics 5.2 Radioactivity, covering 5.2.1 Background radiation and measuring count rate, 5.2.2a Alpha, beta and gamma emission, 5.2.2b Deflection and ionising effects of nuclear radiation, 5.2.3a Radioactive decay, 5.2.3b Effects of decay on the nucleus and decay equations, 5.2.4a Half-life, 5.2.4b Choosing isotopes for applications, and 5.2.5 Effects of ionising radiation and safety precautions. Each one has a worked solution and a mark scheme showing where the marks go.