If a maintenance task is to be undertaken close to a source

If a maintenance task is to be undertaken close to a source of ionizing radiation, and the health physicist directs you to provide shielding that will reduce the dose by a factor of 10, what thickness of lead shield will be required? A patient is to be injected with 0.15 Curies of 1-131, which emits a 0.35 MeV gamma ray. What dose rate is the nurse, who is administering the injection, receiving if the hypodermic needle that she is holding is about 1.5 feet from her body?

Solution

Actul energy is not given. The shielding thickness depends on the energy of the gama photons or the nature of the particles.

For safe side we consider C0-50 gama rays of 1.12 Mev which is significantly high

Half value thicknes of lead for Co-60 = 12.5mm

I = I₀ exp(-0.693*x/12.5), x is the thickness of lead required in mm

we want the dose to be reduced by a factor of 10

I₀/I = 10 = exp(0.693*x/12.5)

x = 41.53 mm require to reduce the dose by a factor of 10

I-131 activity = 0.15 Cu

photon energy E = 0.35 Mev

Total energy emitted from the source = 0.35Mev*0.15*3.7e+10*1.6e-13 J/s

                                                            = 0.31 mJ/s

This is the total energy emitted from the source in all directions

at 1.5 feet approx. 50 cm

energy per unit area = 0.31e-3/4*3.14*(0.5)² = 0.97 e-3

we assume approx. 0.1 sq.m of the body is exposed

energy exposed = 0.97e-3*0.1

                           = 0.97e-4

we assume the weight of the person aprrox. 50 Kg

Dose rate exposed =  0.97e-4/50

                            = 1.94 e-6 Gy/s