FE Physics
Formulas & Explanation

The half value thickness d1/2  is the thickness of the material at which the intensity of the λ-radiation entering it is reduced by one –half.

The half value thickness( also half value layer) depends on the energy of the γ-photons and the material

The intensity of  λ-radiation after passing a material with thickness x:

1.                 I(x) = I(0)  (1/2)x/d1/2

I(0)  incident intensity in J.s-1 m-2

I(x)  intensity  after passing the material in J.s-1 m-2

Remark :  x en d1/2  in cm

2.                      I(x) = I(0)  e- μ x

μ = ln2/d1/2

μ        absorption coefficient  in cm-1

Example

A gamma ray falls perpendicular on a wall of concrete.  Thickness of the wall is 30 cm

The number of photons that is falling on the wall amounts 2.0 x104 .

The energy of the photons is 2.0 MeV

The half value thickness of concrete is 6.6 cm for photons of 2.0 MeV

i    How many photons  per second are transmitted by the wall ?

ii   At what thickness of the wall is the number of the transmitted photons per second 2.0 103

Resolution I

i    I(x) = 2.0 x 104 (1/2) 30/6.6 = 8.6 x 102  photons/s

ii    2.0 x 103= 2.0 x 104 (1/2) x/d1/2

0.1= (1/2) x/d1/2

log 0.1 = x/d1/2 log (1/2)

-1 =  x/d1/2 (- 0.301)

3.32 = x/d1/2

x = 3.32 d1/2 = (3.3) (6,6 )= 21.9 cm

Resolution II

i     μ = ln2/d1/2

μ = 0.693/6.6 = 0.105 cm -1

I(x) = I(0) e- μ x

I(x) = 2.0 x 104 e- 0.105 x 30 =( 2.0 x 104)( 0.043) = 8.6 x 102 photons/s

ii     2.0 x 103 = 2.0 x 104 e- μ x

0.1   = e- μ x

ln 0.1 = - μ x

-2.302 = - 0.105 x

x = -2.302/-0.105 = 21.9 cm

Remark : Intensity is proportional to the number of photons per second.