Certain cancers, especially those of the liver, can be treated by injecting small glass spheres (30 to 35 μm in diameter) contai
ning radioactive 90Y into the tumor. The spheres become lodged in the small capillaries of the tumor, both cutting off its blood supply and delivering a high dose of radiation which damages the tumor, hope-fully sparing surrounding healthy tissue. The spheres stay in place until the radioactivity has completely de-cayed. 90Y undergoes beta-minus decay with a half-life of 64 h; the energy released in the decay is carried away by the beta particle, which has a kinetic energy 0.94 MeV.
a) What is the daughter nucleus for the 90Y decay?
b) The activity is measured when the spheres are prepared; the patient is to be injected 12 days later. If the patient is to be injected with spheres of total activity 1.0 GBq, what should be the approximate activity of the spheres at the time of preparation?
c) Spheres are injected into a liver with a 500 g tumor mass. If the total activity of the spheres administered is 1.0 GBq, and all of the energy deposited ends up in the tumor, what is the dose equivalent in Sv once all of the nuclei have decayed?
a) What is the daughter nucleus for the 90Y decay?
b) The activity is measured when the spheres are prepared; the patient is to be injected 12 days later. If the patient is to be injected with spheres of total activity 1.0 GBq, what should be the approximate activity of the spheres at the time of preparation?
c) Spheres are injected into a liver with a 500 g tumor mass. If the total activity of the spheres administered is 1.0 GBq, and all of the energy deposited ends up in the tumor, what is the dose equivalent in Sv once all of the nuclei have decayed?
1 answer:
You might be interested in
Answer:
0.08 ft/min
Explanation:
To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.
So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:
since the difference between the upper and lower base is the increase in the base and we are only at halft the height.
Now we can calculate the longitudinal section <em>A</em> at that point:
And the raising speed <em>v </em>of the water is given by:
where <em>q</em> is the water flow (1 cubic foot per minute).