satisverborum2003
Badges: 0
Rep:
?
#1
Report Thread starter 11 years ago
#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.
Attached files
0
reply
chr15chr15
Badges: 1
Rep:
?
#2
Report 11 years ago
#2
o you should halve it i thought it said have my bad, dunno not on my syllabus!
0
reply
chidona
Badges: 15
Rep:
?
#3
Report 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; c=\frac{PR}{T}  (1)
similarily;  c = \frac{1.4 - PR}{T + 0.4 nm} (2) 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) -

 C(T + 0.4nm) = 1.4 - PR

 CT + (3x10^8)(0.4 x 10^{-9}) = 1.4 -PR

 PR + (3x10^8)(0.4x10^{-9}) = 1.4 - PR

 2PR = 1.28m

 PR = 0.64 m

Hope it helps =]
0
reply
teachercol
Badges: 9
Rep:
?
#4
Report 11 years ago
#4
What the OP has calculated is the difference in distances between the event and the detectors.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Bristol
    Undergraduate Open Afternoon Undergraduate
    Wed, 23 Oct '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Wed, 23 Oct '19
  • University of Nottingham
    Mini Open Day Undergraduate
    Wed, 23 Oct '19

Have you made up your mind on your five uni choices?

Yes I know where I'm applying (116)
66.29%
No I haven't decided yet (35)
20%
Yes but I might change my mind (24)
13.71%

Watched Threads

View All