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Conceptual questions

Radiotherapy is more likely to be used to treat cancer in elderly patients than in young ones. Explain why. Why is radiotherapy used to treat young people at all?

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Problems&Exercises

A beam of 168-MeV nitrogen nuclei is used for cancer therapy. If this beam is directed onto a 0.200-kg tumor and gives it a 2.00-Sv dose, how many nitrogen nuclei were stopped? (Use an RBE of 20 for heavy ions.)

7.44 × 10 8 size 12{7 "." "44" times "10" rSup { size 8{8} } } {}

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(a) If the average molecular mass of compounds in food is 50.0 g, how many molecules are there in 1.00 kg of food? (b) How many ion pairs are created in 1.00 kg of food, if it is exposed to 1000 Sv and it takes 32.0 eV to create an ion pair? (c) Find the ratio of ion pairs to molecules. (d) If these ion pairs recombine into a distribution of 2000 new compounds, how many parts per billion is each?

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Calculate the dose in Sv to the chest of a patient given an x-ray under the following conditions. The x-ray beam intensity is 1.50 W /m 2 , the area of the chest exposed is 0.0750 m 2 , 35.0% of the x-rays are absorbed in 20.0 kg of tissue, and the exposure time is 0.250 s.

4.92 × 10 –4 Sv

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(a) A cancer patient is exposed to γ rays from a 5000-Ci 60 Co transillumination unit for 32.0 s. The γ rays are collimated in such a manner that only 1.00% of them strike the patient. Of those, 20.0% are absorbed in a tumor having a mass of 1.50 kg. What is the dose in rem to the tumor, if the average γ energy per decay is 1.25 MeV? None of the β size 12{β} {} s from the decay reach the patient. (b) Is the dose consistent with stated therapeutic doses?

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What is the mass of 60 Co in a cancer therapy transillumination unit containing 5.00 kCi of 60 Co ?

4.43 g

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Large amounts of 65 Zn are produced in copper exposed to accelerator beams. While machining contaminated copper, a physicist ingests 50.0 μCi of 65 Zn . Each 65 Zn decay emits an average γ -ray energy of 0.550 MeV, 40.0% of which is absorbed in the scientist’s 75.0-kg body. What dose in mSv is caused by this in one day?

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Naturally occurring 40 K is listed as responsible for 16 mrem/y of background radiation. Calculate the mass of 40 K that must be inside the 55-kg body of a woman to produce this dose. Each 40 K decay emits a 1.32-MeV β , and 50% of the energy is absorbed inside the body.

0.010 g

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(a) Background radiation due to 226 Ra averages only 0.01 mSv/y, but it can range upward depending on where a person lives. Find the mass of 226 Ra in the 80.0-kg body of a man who receives a dose of 2.50-mSv/y from it, noting that each 226 Ra decay emits a 4.80-MeV α particle. You may neglect dose due to daughters and assume a constant amount, evenly distributed due to balanced ingestion and bodily elimination. (b) Is it surprising that such a small mass could cause a measurable radiation dose? Explain.

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The annual radiation dose from 14 C in our bodies is 0.01 mSv/y. Each 14 C decay emits a β averaging 0.0750 MeV. Taking the fraction of 14 C to be 1.3 × 10 –12 N of normal 12 C , and assuming the body is 13% carbon, estimate the fraction of the decay energy absorbed. (The rest escapes, exposing those close to you.)

95%

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If everyone in Australia received an extra 0.05 mSv per year of radiation, what would be the increase in the number of cancer deaths per year? (Assume that time had elapsed for the effects to become apparent.) Assume that there are 200 × 10 4 size 12{"200" times "10" rSup { size 8{ - 4} } } {} deaths per Sv of radiation per year. What percent of the actual number of cancer deaths recorded is this?

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Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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