Lesson 5 ◆ 5.8 Dose Calculations from Irradiated Fuel

Problem:

Calculate the thickness of lead shielding needed to reduce the dose rate from the Cs-137 source in Example 1 by a factor of 100 (i.e., from 84.6 mSv/h to 0.846 mSv/h at 1 m).

Given data:

  • Required attenuation factor = 100

  • Linear attenuation coefficient for Cs-137 gamma rays (662 keV) in lead: μ\mu = 1.175 cm^-1

    (Alternatively, the half-value layer (HVL) for 662 keV gammas in lead is approximately 0.59 cm)

Method A: Using the exponential attenuation formula

The transmission of gamma rays through a shielding material is given by:

I=I0×e−μxI = I_0 \times e^{-\mu x}

Where:

  • I = transmitted dose rate
  • I_0 = initial (unshielded) dose rate
  • μ\mu = linear attenuation coefficient (cm^-1)
  • x = thickness of shielding (cm)

We need I/I0=1/100=0.01I / I_0 = 1/100 = 0.01.

Step 1: Set up the equation.

II0=e−μx=0.01\frac{I}{I_0} = e^{-\mu x} = 0.01

Step 2: Take the natural logarithm of both sides.

−μx=ln⁡(0.01)-\mu x = \ln(0.01)

−μx=−4.605-\mu x = -4.605

Step 3: Solve for x.

x=4.605μ=4.6051.175x = \frac{4.605}{\mu} = \frac{4.605}{1.175}

x=3.92 cmx = 3.92 \ \text{cm}

Step 4: State the result.

A lead shield of approximately 3.9 cm (about 39 mm) thickness is required to reduce the dose rate by a factor of 100.

Method B: Using the Half-Value Layer (HVL) approach

The HVL is the thickness of material required to reduce the dose rate by a factor of 2. The number of HVLs, n, required for a given attenuation factor is:

Attenuation factor=2n\text{Attenuation factor} = 2^n

Step 1: Find the number of HVLs required.

100=2n100 = 2^n

n=ln⁡(100)ln⁡(2)=4.6050.693=6.64 HVLsn = \frac{\ln(100)}{\ln(2)} = \frac{4.605}{0.693} = 6.64 \ \text{HVLs}

Step 2: Calculate the total thickness.

x=n×HVL=6.64×0.59 cmx = n \times \text{HVL} = 6.64 \times 0.59 \ \text{cm}

x=3.92 cmx = 3.92 \ \text{cm}

Step 3: State the result.

Again, approximately 3.9 cm of lead is required.

Key Point: In practice, a safety margin is always added to the calculated shielding thickness to account for scatter (build-up factor), non-uniform source geometry, and manufacturing tolerances. The actual shielding specification would also consider the use of concrete (cheaper and structural but much thicker) or composite shields (lead + concrete).