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:
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Answer:
Explanation:
When the rock is immersed in unknown liquid the forces that act on it are shown as under
1) Tension T by the string
2) Weight W of the rock
3) Force of buoyancy due to displaced liquid B
For equilibrium we have
=
When the rock is suspended in air for equilibrium we have
When the rock is suspended in water for equilibrium we have
+
=
Using the given values of tension and solving α,β,γ simultaneously for we get
Solving for density of liquid we get
E=mcθ
where E is the energy added,m is the mass,c is the specfic heat capacity and θ is the change in temprature.
making m subject u get E/c<span>θ=m
</span><span>θ=(30-20)=10</span><span>
plugging the values we get :
</span>
<span>
solving it we get the answer that is 0.25kg or 250 grams
</span>
where E is the energy added,m is the mass,c is the specfic heat capacity and θ is the change in temprature.
making m subject u get E/c<span>θ=m
</span><span>θ=(30-20)=10</span><span>
plugging the values we get :
</span>
<span>
solving it we get the answer that is 0.25kg or 250 grams
</span>