   Chapter 32, Problem 24PE

Chapter
Section
Textbook Problem

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

To determine

The fraction of the decay energy from 14C in our bodies, which is absorbed by our body.

Explanation

Given info:

The annual radiation dose from C14 in our bodies,

dose=0.01 mSv/y

Average energy of a β particle,

Eβ=0.0750 MeV

Fraction of C14 atoms,

N0=1.3×1012N of normal carbon

Percentage of carbon in a human body,

13%

Average mass of a human body,

M=70.0 kg

Half-life of C14 ,

t1/2=5730 y

Formula used:

The activity of a radioactive sample is given by,

R=0.693N0t1/2

Here, N0 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:

The radiation dose emitted by the human body is 0.01 mSv/y. Express the value in rem.

dose=0.01 mSv/y×0.1 rem1 mSv=103rem/y

dose in rad=dose in remRBE

The RBE for β particle is 1.

Therefore,

dose in rad=dose in remRBE= 10 3rem/y1=103rad/y

Express the dose per year in J/kg.

The average mass of a human being is 70.0 kg. Assuming the radiation emitted is distributed uniformly in the body, the energy emitted in a year is given by,

E=radiation dose×mass of the human=105J/kgy×70.0 kg=7.00×104J/y

Express the energy in MeV/y.

E=7.00×104J/y×1 MeV1.6× 10 13J=4.375×109MeV/y

The energy emitted by a human being in a year is therefore given by 4.375×109MeV/y.

Given that mass of carbon in a human body is 13% of the mass, calculate the mass of carbon in the body.

m=0

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