EBK INTRODUCTION TO HEALTH PHYSICS, FIF
EBK INTRODUCTION TO HEALTH PHYSICS, FIF
5th Edition
ISBN: 9780071835268
Author: Johnson
Publisher: VST
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Chapter 5, Problem 5.14P

(a)

To determine

To plot:

The data given in the table and suggest the type of radiation based on curve.

(a)

Expert Solution
Check Mark

Answer to Problem 5.14P

Beta and gamma radiation

Explanation of Solution

Calculation:

EBK INTRODUCTION TO HEALTH PHYSICS, FIF, Chapter 5, Problem 5.14P

The graph is plotted for counting per min to the thickness of the absorber. The count for beta particle decreases rapidly and then after certain range it decreases slowly. The counts for beta particle approach zero. But, in the above diagram the count is not approaching zero because of gamma radiations.

Thus, the types of radiation according to the above curve is Beta and gamma radiation.

Conclusion:

The type of radiation suggested by the curve is Beta radiation.

(b)

To determine

The energy of the beta radiations.

(b)

Expert Solution
Check Mark

Answer to Problem 5.14P

The energy of the beta radiations is, E=0.985MeV

Explanation of Solution

Given:

Range of beta particle, R=0.4g/cm2

Formula used:

The Energy of beta particle is,

  E=0.85E+0.245

Where,

  E =Energy of beta particle

Calculation:

From the graph plotted between the thickness and the count per min curve. The range can be estimated as,

  R=0.4g/cm2

The Energy of beta particle is,

  E=0.85E+0.245E=0.985MeV

Conclusion:

From the above graph the energy of beta radiation is E=0.985MeV

(c)

To determine

Whether gamma rays are present, the energy of gamma rays if present.

(c)

Expert Solution
Check Mark

Answer to Problem 5.14P

Yes, the energy is 1.5 MeV.

Explanation of Solution

The attenuation of gamma ray can be described as,

  I=I0eμxμ=1xln( I 0 I)

As intensity observed initially is 1000. So,

  I0=1000

For x=0.2mm= 2×104m

  μ=10.02ln( 1000 576)μ=27.58cm1

For x=0.4mm= 4×102cm

  μ=14× 10 2ln( 1000 348)μ=26.38cm1

For x=0.6mm= 6×102cm

  μ=16× 10 2ln( 1000 230)μ=24.49cm1

For x=0.8mm= 8×102cm

  μ=18× 10 2ln( 1000 168)μ=22.29cm1

For x=1mm= 0.1cm

  μ=10.1ln( 1000 134)μ=20.09cm1

For x=1.2mm=0.12m

  μ=10.12ln( 1000 120)μ=17.66cm1

For x=1.4mm= 0.14cm

  μ=10.14ln( 1000 107)μ=15.96cm1

For x=1.6mm= 0.16m

  μ=10.16ln( 1000 96)μ=14.64m1

For x=2mm= 0.2m

  μ=10.2ln( 1000 95)μ=11.76cm1

For x=4mm= 0.4m

  μ=10.4ln( 1000 90)μ=6.01m1

For x=8mm= 0.8cm

  μ=10.8ln( 1000 82)μ=3.12cm1

For x=15mm= 1.5cm

  μ=11.5ln( 1000 68)μ=1.79cm1

For x=20mm= 2cm

  μ=12ln( 1000 60)μ=1.406cm1

For x=280mm= 2.8cm

  μ=12.8ln( 1000 50)μ=1.06cm1

So, mean is,

  μ=13.78cm1

Thus, the energy corresponding to it is 1.5 MeV.

(d)

To determine

The isotope compatible with absorption data.

(d)

Expert Solution
Check Mark

Answer to Problem 5.14P

The isotope compatible with above absorption data is 198Au.

Explanation of Solution

The isotope compatible with above absorption data is 198Au.

(e)

To determine

The equation fits the absorption data.

(e)

Expert Solution
Check Mark

Answer to Problem 5.14P

  I=900e24t+100e0.25t

Explanation of Solution

The equation fits the absorption data is,

  I=I0eσxI=900e24t+100e0.25t

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