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Nuclear reactors are usually surrounded by lead and concrete. Which of the following is this safety precaution particularly for? (a) neutron (b) \(\alpha\) particle (c) \(\beta\) particle (d) \(\gamma\) rays

Short Answer

Expert verified
(d) \(\gamma\) rays

Step by step solution

01

Identify the Purpose of Shielding

The primary function of surrounding nuclear reactors with lead and concrete is to shield against harmful radiation.
02

Consider the Properties of Radiation Types

Identify each type of radiation: - **Neutron:** Neutral radiation, can penetrate deeply, needs materials like water or polyethylene coupled with boron or lead for effective shielding.- **Alpha Particle (\(\alpha\)):** Heavy, positively charged, can be stopped by a sheet of paper or skin.- **Beta Particle (\(\beta\)):** Light, negatively charged, requires materials like plastic or glass for shielding.- **Gamma Rays (\(\gamma\)):** High-energy electromagnetic radiation, very penetrative, requires dense materials like lead or thick concrete for effective attenuation.
03

Analyze the Shielding Materials

Lead is known for its high density, making it excellent at attenuating high-energy rays such as gamma rays. Concrete adds additional mass and is effective against a broad range of radiation types, but is most commonly used in combination with lead to effectively absorb or slow down gamma rays and neutrons.
04

Select the Type of Radiation

Given the use of lead and concrete, materials commonly used to protect against very penetrative radiation, the primary purpose is to shield against \(\gamma\) rays, which require dense materials to reduce their intensity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radiation Shielding
Radiation shielding refers to the practice of using materials to reduce exposure to harmful radiation. This is vital in nuclear reactors to ensure the safety of both workers and the environment. Larger structures such as nuclear reactors are equipped with layers of shielding to prevent the escape of radiation. They are commonly surrounded by materials like lead and concrete, which serve as barriers to various types of radiation.
Lead is particularly effective due to its high density. This allows it to effectively absorb high-energy particles. Concrete, while less dense than lead, offers extensive bulk and additional protective qualities. Combining these materials provides a robust defense against radiation, particularly gamma rays, which can penetrate most materials. Shielding is essential in minimizing radiation exposure for both humans and other living organisms, protecting them from the harmful effects of ionizing radiation. Understanding the properties of various materials helps engineers build effective barriers.
Gamma Rays
Gamma rays are a type of electromagnetic radiation known for their high energy and deep penetration capabilities. Unlike neutrons, alpha particles, and beta particles, gamma rays do not carry a charge and travel at the speed of light. This makes them particularly dangerous as they can penetrate most materials, including human tissue, potentially causing damage to cells.
Gamma rays often require dense materials for effective shielding. Lead is one of the best options due to its high density, which helps in reducing the intensity of these rays. This is why it is commonly used around nuclear reactors and in medical facilities where radiation exposure might occur.
The use of lead in conjunction with other materials like concrete enhances the shielding effect, effectively reducing the potential hazards posed by gamma rays. By understanding the nature of gamma rays, safety protocols can be developed to mitigate exposure.
Properties of Radiation Types
Radiation comes in several forms, each possessing unique characteristics and hazards. Understanding these properties can help in crafting appropriate safety measures for various environments.
- **Neutrons:** These are neutral particles and can penetrate materials deeply. Water, polyethylene, combined with boron or lead, are often used for their effective shielding capacities.
- **Alpha particles ():** These are heavy and positively charged, but can be easily stopped by something as simple as paper or a layer of skin. Despite their low penetration power, they can be harmful if ingested or inhaled.
- **Beta particles ():** Lighter than alpha particles, these negatively charged particles require denser materials such as plastic or glass for effective shielding.
- **Gamma Rays ():** As high-energy electromagnetic waves, gamma rays can travel long distances and penetrate materials better than other types of radiation. Dense materials such as lead are essential for blocking these rays.
Different radiation types demand specific materials for effective shielding. By analyzing these properties, better protective measures can be implemented, ensuring public safety and minimizing health risks.

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Most popular questions from this chapter

The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduce to (a) \(1 / 2 \mathrm{~g}\) (b) \(1 / 4 \mathrm{~g}\) (c) \(1 / 8 \mathrm{~g}\) (d) \(1 / 16 \mathrm{~g}\)

Consider the following nuclear reactions \({ }_{92} \mathrm{M}^{238} \longrightarrow \mathrm{y}^{\mathrm{x}}+2{ }_{2} \mathrm{He}^{4}\) \({ }_{\mathrm{y}} \mathrm{N}^{\mathrm{x}} \longrightarrow{ }_{\mathrm{B}} \mathrm{L}^{\mathrm{A}}+2 \beta^{+}\) The number of neutrons in element \(L\) is (a) 146 (b) 144 (c) 142 (d) 140

A radioactive form of phosphorus undergoes Y-decay. What will the radioactivity level (R) versus time graph for the decay process appear?

Which of the following notations shows the product incorrectly? (a) \({ }_{5} \mathrm{~B}^{10}(\alpha, \mathrm{n}){ }_{7} \mathrm{~N}^{13}\) (b) \({ }_{96} \mathrm{Cm}^{242}(\alpha, 2 \mathrm{n}){ }_{97} \mathrm{BK}^{243}\) (c) \({ }_{7} \mathrm{~N}^{14}(\mathrm{n}, \mathrm{p}){ }_{6} \mathrm{C}^{14}\) (d) none of these

Which of the following is a fusion reaction? (a) \({ }_{98} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{56} \mathrm{Ba}^{141}+{ }_{36} \mathrm{Kr}^{92}+3_{0} \mathrm{n}^{1}\) (b) \({ }_{92} \mathrm{Fe}^{239}+{ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{92} \mathrm{U}^{239}+\mathrm{Y}\) (c) \({ }_{26} \mathrm{Fe}^{55}+{ }_{-1} \mathrm{e}^{0} \longrightarrow{ }_{25} \mathrm{Mn}^{55}\) (d) \({ }_{1} \mathrm{H}^{1}+{ }_{1} \mathrm{H}^{1}+2{ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{2} \mathrm{He}^{4}+\) energy

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