Buildup and Shielding

(1)

To discuss shielding for photon beams and the increase in photon transmission through a shield resulting from buildup

Buildup and Shielding

(2)

HVL and TVL

•  The amount of shielding required to reduce the incident radiation levels by ½ is called the “half- value layer” or HVL

•  The HVL is dependent on the energy of the photon and the type of material.

•  Similarly, the amount of shielding required to

reduce the incident radiation levels by 1/10 is

called the “tenth-value layer” or TVL.

(3)

HVL and TVL

HVL (cm) TVL (cm)

Isotope Photon E (MeV)

Concrete Steel Lead Concrete Steel Lead

137

Cs 0.66 4.8 1.6 0.65 15.7 5.3 2.1

60

Co 1.17, 1.33

6.2 2.1 1.2 20.6 6.9 4

198

Au 0.41 4.1 0.33 13.5 1.1

192

Ir 0.13 to 1.06

4.3 1.3 0.6 14.7 4.3 2

226

Ra 0.047 to 2.4

6.9 2.2 1.66 23.4 7.4 5.5

(4)

HVL and TVL

The half value layer (HVL) and tenth value layer (TVL) are mathematically related as follows:

HVL = ln(2)

µ TVL = ln(10)

µ

TVL = HVL

ln(10) µ ln(2)

µ

= ln(10)

ln(2) = 2.303

0.693 = 3.323 TVL = 3.323 x HVL

and

(5)

Shielding

Ø  Shielding is intended to reduce the radiation level at a specific location Ø  The amount of shielding (thickness)

depends on:

Ø  the energy of the radiation Ø  the shielding material

Ø  the distance from the source

(6)

Inverse Square Law

Ø  If the radiation emanates from a point source, the radiation follows what is commonly known as the

“Inverse Square Law” or ISL.

Ø  Most real sources which are considered to be

“point” sources are actually not “point” sources.

Most sources such as a

⁶⁰

Co teletherapy source

have finite dimensions (a few cm in each direction).

These sources appear to behave like point sources at some distance away but as one gets closer to the source, the physical dimensions of the source

result in a breakdown of the ISL.

(7)

Inverse Square Law

Ø  If the source of the radiation is not a point but is a line, a flat surface (plane) or a finite volume, the ISL does not apply.

Ø  However, if one gets far enough away from a finite line, plane or volume, they appear to be a point and the ISL applies with some acceptable error.

Ø  For the remaining discussion let’s assume we have a point source. The ISL predicts that the intensity of the radiation will decrease as distance from the

source increases even without any shielding.

(8)

These photons should not strike the

individual. But due to scatter, they do, so the calculated value is too low. It needs to be

increased by the buildup factor.

Scatter

(9)

Photon Attenuation and Absorption

•  Absorption refers to the total number of photons absorbed by the material (dark blue arrows)

Ø  Attenuation refers to total number of

photons removed from incident beam

(absorbed + scattered) (dark blue and

light blue arrows)

(10)

Photon Attenuation

I _x = I _o e ^-µx = I _o e

where:

I

_x

= photon intensity after traversing x cm of some material

I

_o

= initial or incident photon intensity x = thickness of material (cm)

µ = linear attenuation coefficient (cm

^-1

) ρ = density (g/cm

³

)

µ/ρ = mass attenuation coefficient (cm

²

/g)

µ

ρ (ρx) -

(11)

Attenuation and Buildup

I

_o

IxB

I

_o

I

where B ≥1

(12)

I = I _o B e ^(-µx)

primary photons + scattered photons primary photons

B =

If there are no scattered photons, then B = 1 If there are scattered photons, then B > 1

Buildup

(13)

Build Up function

•  Build up functions in water and concrete

(14)

What amount of lead shielding is needed to reduce the dose rate beyond the shield from 1 mSv/hr to 0.02 mSv/hr for a 1 MeV photon beam?

The equation for x with the B function

cannot be solved analytically: numerical/

iterative approach

Sample Buildup

(15)

Sample Buildup

Photon

Energy Material

(16)

Sample Buildup

The mass attenuation coefficient (µ/ρ) for a 1 MeV photon incident on a lead shield is:

(µ/ρ) = 0.0708 cm

²

/g

The density of lead = 11.35 g/cm

³

The linear attenuation coefficient is:

(µ/ρ) x ρ = 0.0708 cm

²

/g x 11.35 g/cm

³

= 0.804 cm

^-1

(17)

I = I

_o

B e

^(-µx)

I = 1 mSv/hr

I

_o

= 0.02 mSv/hr

B = 1 (assumed) -> first iteration µ = 0.804 cm

^-1

Solve for “µx”

Sample Buildup: iterative method

(18)

Sample Buildup

ln(0.02) = ln[e

^(-µx)

] -3.91 = -µx

Although it is not required:

x = -3.91/-0.804 = 4.86 cm

This would be the calculated value of the lead shield if scatter was not considered.

0.02 mSv/hr

1 mSv/hr = e

^(-µx)

(19)

Sample Buildup

µx = 3.91

(20)

Sample Buildup

µx = 3.91 Let’s interpolate:

when µx = 4, B = 2.26 when µx = 2, B = 1.69

1.27 = 0.09/(2.26 - x)

(2.26 - x) = 0.09/1.27 = 0.071 2.26 – 0.071 = x = 2.19

(4-2)

(2.26-1.69)

(4-3.91)

(2.26 - x)

=

(21)

Buildup and Shielding

To discuss shielding for photon beams and the increase in photon transmission through a shield resulting from buildup

Buildup and Shielding

HVL and TVL

• The amount of shielding required to reduce the incident radiation levels by ½ is called the “half- value layer” or HVL

• The HVL is dependent on the energy of the photon and the type of material.

• Similarly, the amount of shielding required to

reduce the incident radiation levels by 1/10 is

called the “tenth-value layer” or TVL.

HVL and TVL

HVL (cm) TVL (cm)

Cs 0.66 4.8 1.6 0.65 15.7 5.3 2.1

Co 1.17, 1.33

6.2 2.1 1.2 20.6 6.9 4

Au 0.41 4.1 0.33 13.5 1.1

Ir 0.13 to 1.06

4.3 1.3 0.6 14.7 4.3 2

Ra 0.047 to 2.4

6.9 2.2 1.66 23.4 7.4 5.5

HVL and TVL

The half value layer (HVL) and tenth value layer (TVL) are mathematically related as follows:

HVL = ln(2)

µ TVL = ln(10)

µ

TVL = HVL

ln(10) µ ln(2)

µ

= ln(10)

ln(2) = 2.303

0.693 = 3.323 TVL = 3.323 x HVL

and

Shielding

Ø Shielding is intended to reduce the radiation level at a specific location Ø The amount of shielding (thickness)

depends on:

Ø the energy of the radiation Ø the shielding material

Ø the distance from the source

Inverse Square Law

Ø If the radiation emanates from a point source, the radiation follows what is commonly known as the

“Inverse Square Law” or ISL.

Ø Most real sources which are considered to be

“point” sources are actually not “point” sources.

Most sources such as a

Co teletherapy source

have finite dimensions (a few cm in each direction).

These sources appear to behave like point sources at some distance away but as one gets closer to the source, the physical dimensions of the source

result in a breakdown of the ISL.

Inverse Square Law

Ø If the source of the radiation is not a point but is a line, a flat surface (plane) or a finite volume, the ISL does not apply.

Ø However, if one gets far enough away from a finite line, plane or volume, they appear to be a point and the ISL applies with some acceptable error.

Ø For the remaining discussion let’s assume we have a point source. The ISL predicts that the intensity of the radiation will decrease as distance from the

source increases even without any shielding.

These photons should not strike the

individual. But due to scatter, they do, so the calculated value is too low. It needs to be

increased by the buildup factor.

Scatter

Photon Attenuation and Absorption

• Absorption refers to the total number of photons absorbed by the material (dark blue arrows)

Ø Attenuation refers to total number of

photons removed from incident beam

(absorbed + scattered) (dark blue and

light blue arrows)

Photon Attenuation

I x = I o e -µx = I o e

where:

I

= photon intensity after traversing x cm of some material

I

= initial or incident photon intensity x = thickness of material (cm)

µ = linear attenuation coefficient (cm

) ρ = density (g/cm

)

µ/ρ = mass attenuation coefficient (cm

/g)

Attenuation and Buildup

I

IxB

I

I

where B ≥1

I = I o B e (-µx)

•  The amount of shielding required to reduce the incident radiation levels by ½ is called the “half- value layer” or HVL

•  The HVL is dependent on the energy of the photon and the type of material.

•  Similarly, the amount of shielding required to

Ø  Shielding is intended to reduce the radiation level at a specific location Ø  The amount of shielding (thickness)

Ø  the energy of the radiation Ø  the shielding material

Ø  the distance from the source

Ø  If the radiation emanates from a point source, the radiation follows what is commonly known as the

Ø  Most real sources which are considered to be

Ø  If the source of the radiation is not a point but is a line, a flat surface (plane) or a finite volume, the ISL does not apply.

Ø  However, if one gets far enough away from a finite line, plane or volume, they appear to be a point and the ISL applies with some acceptable error.

Ø  For the remaining discussion let’s assume we have a point source. The ISL predicts that the intensity of the radiation will decrease as distance from the

•  Absorption refers to the total number of photons absorbed by the material (dark blue arrows)

Ø  Attenuation refers to total number of

I _x = I _o e ^-µx = I _o e

I = I _o B e ^(-µx)

•  Build up functions in water and concrete

Solve for µx using B = 2.2 I = I ₀ B e ^(-µx)

0.02 mSv/hr = (1 mSv/hr) (2.2) e ^(-µx) ln(0.02/2.2) = ln[e ^(-µx) ]