Physics AQA Synoptic Question Watch
tomography) scanner. A solution containing the isotope is injected into the
patient. When a F nucleus in the body decays, the positron emitted travels no further
than a millimetre before it loses its kinetic energy and is annihilated by an electron. As
a result, two γ photons of the same energy are emitted in opposite directions.
A ring of detectors surrounding the patient is used to detect the photons.
A detector P was triggered by a γ photon from such an annihilation event at R, as
shown in Figure 7. A second detector Q at a distance of 1400 mm from P was
triggered 0.40 ns later by the other photon from this event.
Calculate the distance PR.
The trouble is I've already attempted this question. I got the first part right by using the equation speed = distance/time to find the distance from Q to R:
0.4* 10^-9 * 3* 10^8 = 0.12 m.
I then figured out that the next logical step to find PR is to subtract the overall distance from QR:
1400mm - 120mm =1280mm
but the mark scheme stated this was wrong and that I should halve 1280mm. Now, I'm asking why? And why was my method wrong. I am grateful for anyone who could provide any answer.
Assuming the gamma rays travel at the speed of light (let's call it 'c'), and the time for the gamma ray to get from P to R is 'T' , then -
speed = dist/time;
similarily; For the other photon hitting the other detector, which is a distance of 1.4 -PR away, as can be seen in the calculation.
From (1), we can deduce that TC = PR.
Now, considering (2) -
Hope it helps =]