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Zanzabum
3 years ago
5

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?

Physics
1 answer:
Semenov [28]3 years ago
8 0

Answer:

Explanation:

Base on the scenario been described in the question, let's use the method in the file attached below to solve the question

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Read 2 more answers
A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?
anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\  \end{gathered}

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}

now, replace

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}

hence, the answer is 0.495 m/s

3 0
1 year ago
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