Calculating Lead Shielding Thickness for Radioactive Sample Transport

In summary, the conversation discusses a problem involving shielding for the transport of radioactive samples. A 100 mCi sample of 212Pb is placed in a lead lined box and the thickness of lead needed to reduce the dose level at the surface to 10mR per day is being calculated. The half-life of 212Pb, gamma disintegration yield, mass energy absorption coefficient for lead, and density are provided for the calculation. The assumption that air does not affect the dose level is made, but it is noted that distance still affects the intensity of radiation according to the inverse square law. A correction is made for the mass energy coefficient of air, which changes the answer to 1.2cm.
  • #1
habman_6
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Hello, i have to do a problem for a written problem set:

1. Shielding. The transport of radioactive samples must follow strict guidelines to ensure that the exposure is very low. A 100 mCi sample of 212Pb is placed at the centre of a cubical lead lined box, each side of which has a length of 30cm. Calculate the thickness of lead required to reduce the dose level at the surface of the box to 10mR per day at most. The half-life of 212Pb is 11 hours, and each gamma disintegration yields 0.24 MeV; the mass energy absorption coefficient for lead at this energy is 0.48 cm2g-1, and its density is 11.4 g.cm-3. (Make the simplest reasonable assumption you can, and provide a full explanation).

Am i right in the "simple assumption" that the air does not affect the dose level, so this way the dose at each point along the surface of the cube would be equal (since distance from the sample would not matter). Then using this surface area i could continue to solve for fluence, etc.
 
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  • #2
habman_6 said:
Am i right in the "simple assumption" that the air does not affect the dose level, so this way the dose at each point along the surface of the cube would be equal (since distance from the sample would not matter). Then using this surface area i could continue to solve for fluence, etc.

You can make the assumtion that the air will not affect the intensity of the radiation. However, this does not mean that the dose at each point on the surface of the cube will be equal. The instenity of radiation varies with distance in accordance to the inverse square law, therefore that distance will still affect the intenisty of the radiation.

-Hoot:smile:
 
  • #3
This is what I did. Can someone confirm so i can settle my feelings? :redface:

Also, I scanned this before i caught a conversion error for the mass energy coefficient of air, it should be 0.0027 instead of 0.027, which changes the answer to 1.2cm.
 

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1. What is nuclear physics shielding?

Nuclear physics shielding is the process of using materials to protect against harmful radiation emitted from nuclear reactions or radioactive materials. This shielding is essential in nuclear power plants, medical facilities, and other environments where radiation is present.

2. How does nuclear physics shielding work?

Nuclear physics shielding works by using materials that can absorb or deflect radiation. These materials, such as lead, concrete, and water, create a barrier between the radiation source and the surrounding environment to prevent harmful exposure.

3. What are the different types of nuclear physics shielding?

The most common types of nuclear physics shielding include passive, active, and biological shielding. Passive shielding uses materials that can block or absorb radiation without any external power source. Active shielding uses an external power source to create a magnetic field that can deflect radiation. Biological shielding uses living organisms, such as plants and bacteria, to absorb radiation.

4. What materials are used for nuclear physics shielding?

Some of the most commonly used materials for nuclear physics shielding include lead, concrete, steel, and water. Lead is often used for its high density and ability to block gamma rays. Concrete and steel are also dense materials that can effectively absorb radiation. Water is used as a shielding material for neutron radiation.

5. Why is nuclear physics shielding important?

Nuclear physics shielding is crucial for protecting people and the environment from the harmful effects of radiation. It is necessary for the safe operation of nuclear power plants, medical procedures involving radiation, and the handling of radioactive materials. Proper shielding can greatly reduce the risk of radiation exposure and ensure the safety of those working with or around nuclear materials.

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