91Ó°ÊÓ

Because photons follow the principles of quantum physics, their behaviour is wave-like and particle-like. A photon is registered as a single, particulate unit when it is detected by measuring equipment.

(a)

We look at two annihilated systems in this problem, \(\gamma \) rays that are from a PET scan. These two \(\gamma \)rays are traveling to the different sides of the two detectors.

For part a, we do a calculation. What is the change in photon arrival times if the place of origin is \({\rm{9}}{\rm{.00\;cm}}\) closer to one of the detectors?

First, we must remember that speed is defined as a change in distance travelled in a certain amount of time.

\(v = \frac{s}{t}\)

To get time \({\rm{\gamma }}\)rays have traveled we have:

\(t = \frac{s}{v}\)where is exactly the distance (path) traveled by annihilated \(\gamma \)rays. So if the origin is \(9\;cm\)closer \(\left( {{\rm{0}}{\rm{.09\;m}}} \right)\)we have:

\(t = \frac{{0.09}}{{3 \times {{10}^8}}}\)

\(t = 3 \times {10^{ - 10}}\;s\)

Therefore, the difference in arrival times of the photons is \(t = 3 \times {10^{ - 10}}\;s\).

(b)

We have the identical situation in part b, only we are seeking for time if we wish to have a resolution of

\(1\;\,mm\), so we got:

\(t = \frac{{0.001}}{{3 \times {{10}^8}}}\)

And we have a result of:

\({\rm{t = 3}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 12}}}}{\rm{\;s}}\)

So we can see that the closer we wish to measure (smaller route), the more precision and accuracy we'll require with the detection.