Physics and Engineering of Radiation Detection

11.7. Radiation Protection 665

Solution:

According to equation 11.7.5, the required thickness is given by

x =

1

μ

ln

(

X 0

X

)

=

1

50

ln

(

500

10

)

=0. 078 cm.

Two terminologies that are commonly used in dosimetry applications. These are
half-value layerorHV Landtenth value layerorTVL.TheHV Lis defined as the
thickness of the shield that reduces the intensity by a factor of two. Similarly aTVL
decreases the thickness by a factor of 10. The values ofHV LandTVLdepend on
the value of the attenuation coefficient. This can be seen by substitutingI=I 0 / 2
(decrease in intensity by a factor of 2) in the above expression of intensity variation
as

I 0

2

= I 0 e−μHV L

⇒HV L =

ln(2)

μ

. (11.7.7)

Similary the expression forTVLcan be obtained by substitutingI = I 0 /10 in
equation 11.7.4, that is

I 0

10

= I 0 e−μT V L

⇒TVL =

ln(10)

μ

. (11.7.8)

The relationship betweenHV LandTVLcanbeobtainedbytakingtheratioof
equations 11.7.7 and 11.7.8, that is

HV L

TVL

=

ln(2)

ln(10)

⇒HV L =0. 3 ×TVL.

Example:

(^137) Csemits 662keV γ-rays. Determine theHV LandTVLfor lead having
attenuation coefficient of 1.15cm−^1.
Solution:
TheHV LandTVLcan be computed from equations 11.7.7 and 11.7.8 as