# Physics AQA Synoptic QuestionWatch

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#1
The F isotope is used to obtain medical images in a PET (positron emission
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.
0
11 years ago
#2
o you should halve it i thought it said have my bad, dunno not on my syllabus!
0
11 years ago
#3
Just done this question - this is how I did it:
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 =]
0
11 years ago
#4
What the OP has calculated is the difference in distances between the event and the detectors.
0
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