Nuclear Fuel Cycle

Problem:

Calculate the dose rate at 1 metre from an unshielded Cs-137 source with an activity of 10^12 Bq (1 TBq).

Given data:

• Activity, A = 10^12 Bq

• Distance, r = 1 m

• Specific gamma ray constant for Cs-137: $\Gamma$ = 8.46 x 10^-14 Sv m^2 / (Bq h)

  (Equivalently: $\Gamma$ = 0.0846 uSv m^2 / (MBq h))

Formula:

The dose rate from a point gamma source is given by the inverse square law:

$\dot{D} = \frac{A \times \Gamma}{r^2}$

Where:

• $\dot{D}$ = dose rate (Sv/h)
• A = activity (Bq)
• $\Gamma$ = specific gamma ray constant (Sv m^2 Bq^-1 h^-1)
• r = distance from source (m)

Solution:

Step 1: Write down the known values.

• A = 1 x 10^12 Bq
• $\Gamma$ = 8.46 x 10^-14 Sv m^2 Bq^-1 h^-1
• r = 1 m

Step 2: Substitute into the formula.

$\dot{D} = \frac{1 \times 10^{12} \times 8.46 \times 10^{-14}}{(1)^2}$

Step 3: Evaluate the numerator.

$\dot{D} = \frac{8.46 \times 10^{-2}}{1}$

$\dot{D} = 0.0846 \ \text{Sv/h}$

Step 4: Convert to more practical units.

$\dot{D} = 84.6 \ \text{mSv/h}$

Step 5: Interpret the result.

This is an extremely high dose rate. An unshielded worker at 1 metre from this source would receive the annual occupational dose limit of 20 mSv in approximately 14 minutes. This illustrates why irradiated fuel and reprocessing waste must be handled remotely behind heavy shielding.

Key Point: The inverse square law means that doubling the distance reduces the dose rate by a factor of four. At 2 metres, the dose rate would be 84.6 / 4 = 21.15 mSv/h. Distance is the simplest and most fundamental means of radiation protection.