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College Physics

1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000

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BuyFindarrow_forward

College Physics

1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000
Textbook Problem

(a) A cancer patient is exposed to γ rays from a 5000−Ci 60Co 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, it the average γ energy per decay is 1.25 MeV? None of the β s from the decay reach the patient. (b) Is the dose consistent with stated therapeutic doses?

(a)

To determine

The dose absorbed in the tumor when a patient is exposed to ? rays from a 5000 Ci 60Co source.

Explanation

Given:

Activity of the C60o source   R=5000 Ci

Time of exposure  t=32.0 s

Energy per decay  E0=1.25 MeV

Mass of the tumor  m=1.50 kg

Amount of gamma rays striking the patient = 1% of the total γrays

Amount of gamma rays absorbed by the tumor = 20% of the rays striking the person

Calculation:

Express the activity of the C60o source in Bq.

  R=5000 Ci×1 Bq2.7× 10 11Ci=1.85185×1014Bq=1.85185×1014 diintegrations/s

Since 1 Bq is defined as 1 disintegration per second, the number of disintegrations in a time t is given by,

  n=Rt=(1.85185× 10 14Bq)(32.0 s)=5.92593×1015

Each disintegration releases 1 gamma ray, which has an energy of 1.25 MeV. Therefore, n disintegrations releases an amount of energy E which is given by,

  E=nE0=(5.92593× 10 15)(1.25 MeV)=7.407×1015MeV

Express the energy in joules

(b)

To determine

If the radiation dose of 158 rem calculated in (a) consistent with the stated therapeutic doses.

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