Why is strontium-90 a particularly dangerous isotope for humans?

Short Answer

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Strontium-90 is particularly dangerous due to its physical and biological properties. As a radioactive isotope, it emits high energy beta particles when it decays. It's chemically similar to calcium, so it can accumulate in the bones when ingested or inhaled, leading to bone disorders, including cancer. Furthermore, it's produced in nuclear reactions and remains in the environment for a long time due to its long half-life, providing persistent potential for exposure.

Step by step solution

01

Understanding Isotopes and Radioactivity

Firstly, we need to understand what an isotope is. An isotope is a variant of a chemical element which differs in neutron number. Strontium-90 is an isotope of Strontium. Strontium-90 is radioactive, which means it has unstable atoms that can release a lot of energy very quickly.
02

Identifying Physical Properties of Strontium-90

Strontium-90 is a soft meta and a pure beta emitter, meaning it gives off beta particles but no alpha particles or gamma rays when it decays. A beta particle is a high-energy, high-speed electron or positron that is emitted by certain types of radioactive nuclei.
03

Biological Implications of Strontium-90

Strontium-90 behaves much like calcium, so if it is ingested or inhaled, it can accumulate in the bones and bone marrow. This can lead to bone disorders, including cancer. Moreover, the beta particles it emits can damage body tissues and DNA.
04

Environmental Presence of Strontium-90

Strontium-90 is produced in nuclear reactions, including those in an atomic bomb explosion or a nuclear reactor meltdown. It can enter the human body through food, water, or air. Strontium-90 has a long half-life (28.8 years), which means it remains in the environment and potential source of exposure for a long time.

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Most popular questions from this chapter

Consider the decay series $$\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}$$ where \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of A, and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

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What are the advantages of a fusion reactor over a fission reactor? What are the practical difficulties in operating a large-scale fusion reactor?

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