   Chapter 32, Problem 23PE

Chapter
Section
Textbook Problem

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

To determine

(a)

The mass of R226a that must be inside 80.0 kg body of a man to produce a dose of 2.50 mSv/y.

Explanation

Given info:

dose=2.50 mSv/y

Mass of the man's body,

m=80.0 kg

Energy of the emitted a particle,

Eα=4.80 MeV

Half-life of R226a ,

t1/2=1600y

RBE for a particle,

RBE=20

Formula used:

The activity of a radioactive sample is given by,

R=0.693Nt1/2

Here, N is the number of atoms of the radioactive substance present.

The dose in rem is related to the dose in rad as follows:

dose in rem= dose in rad×RBE

Here RBE is the relative biological effectiveness.

Calculation:

Express the radiation dose in rem.

dose=2.50 mSv/y× 10 3Sv1 mSv ×100 rem1 Sv=0.250 rem/y

dose in rad=dose in remRBE

The RBE for a particle is 20.

Therefore,

Express the dose per year in J/kg.

The energy deposited in the man for the radiation of the energy for an year as calculated above is given by,

E=radiation dose×mass of the man

Therefore,

E=radiation dose×mass of the man=1.25× 10 4J/kgy80.0 kg=0.01 J/y

Express the energy in MeV/y

To determine

(b)

How such a small mass of R226a ( 1.128×108g ) calculated in (a) could cause a measurable amount of radiation:

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