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Write the nuclear equations for the following bombardment reactions. a \({ }_{21}^{45} \mathrm{Sc}(\mathrm{n}, \alpha)_{19}^{42} \mathrm{~K}\) b) \({ }_{29}^{63} \mathrm{Cu}(\mathrm{p}, \mathrm{n})_{30}^{6} \mathrm{Zn}\)

Short Answer

Expert verified
For (a), \(_{21}^{45} \mathrm{Sc} + _{0}^{1} \mathrm{n} \rightarrow _{19}^{42} \mathrm{K} + _{2}^{4} \alpha\); For (b), \(_{29}^{63} \mathrm{Cu} + _{1}^{1} \mathrm{p} \rightarrow _{30}^{64} \mathrm{Zn} + _{0}^{1} \mathrm{n}\).

Step by step solution

01

Identify Initial and Final Nuclei

For reaction (a), the initial nucleus is scandium-45 \(_{21}^{45}\mathrm{Sc}\), and for reaction (b), the initial nucleus is copper-63 \(_{29}^{63}\mathrm{Cu}\). For reaction (a), the final nucleus after the reaction is potassium-42 \(_{19}^{42}\mathrm{K}\), and in reaction (b), the final nucleus is zinc-64 \(_{30}^{64}\mathrm{Zn}\).
02

Determine the Incoming Particle

In reaction (a), the incoming particle is a neutron \((\mathrm{n})\), and for reaction (b), the incoming particle is a proton \((\mathrm{p})\).
03

Determine the Emitted Particle

In reaction (a), an alpha particle \((\alpha)\) is emitted, which is a helium nucleus \(_{2}^{4}\mathrm{He}\). In reaction (b), a neutron \((\mathrm{n})\) is emitted.
04

Balance Reaction Equation

For each reaction, ensure that the sum of atomic (proton) numbers and mass numbers are the same on both sides of the equation. - Reaction (a):\[{ }_{21}^{45} \mathrm{Sc} + { }_{0}^{1} \mathrm{n} \rightarrow { }_{19}^{42} \mathrm{K} + { }_{2}^{4} \mathrm{He}\]- Reaction (b):\[{ }_{29}^{63} \mathrm{Cu} + { }_{1}^{1} \mathrm{p} \rightarrow { }_{30}^{64} \mathrm{Zn} + { }_{0}^{1} \mathrm{n}\]
05

Verify Conservation Laws

Verify that the sum of atomic numbers (protons) and mass numbers (nucleons) before and after the reaction are equal for both reactions. - Reaction (a): Atomic numbers: 21 + 0 = 19 + 2, Mass numbers: 45 + 1 = 42 + 4. - Reaction (b): Atomic numbers: 29 + 1 = 30 + 0, Mass numbers: 63 + 1 = 64 + 0.

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Key Concepts

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

Bombardment Reactions
In the intriguing world of nuclear chemistry, bombardment reactions are key. These reactions involve the collision of a target nucleus with a smaller particle, like a proton or neutron. This collision often results in a nuclear reaction that transforms the original nuclei into different elements or isotopes.

A great example of this is when a neutron is "shot" at scandium-45, resulting in the formation of potassium-42 and an alpha particle in reaction (a). Similarly, firing a proton at copper-63 transforms it into zinc-64 while releasing a neutron, as seen in reaction (b).

Bombardment reactions are essential for discovering new isotopes and elements. They also play a vital role in applications like producing medical isotopes for imaging and treatment. Simplifying these complex transformations helps us understand how elements interact and change at the atomic level.
Nuclear Chemistry
Nuclear chemistry is like the cornerstone of all things atomic. It explores how particles within an atom interact and transform, primarily focusing on the nucleus where protons and neutrons reside. This field helps us understand natural and artificial processes involving atomic nuclei.

These reactions affect our daily lives in ways we might not realize, from energy production in nuclear reactors to the functioning of smoke detectors using radioisotopes. By delving deep into nuclear chemistry, we unveil the mysteries of matter, offering insights into phenomena such as radioactive decay, fission, and fusion.
  • Radioactive Decay: The process where unstable nuclei release energy to become stable.
  • Nuclear Fission: Splitting a heavy nucleus into lighter ones, releasing energy.
  • Nuclear Fusion: Combining light nuclei to form a heavier nucleus, as seen in the sun.

By deciphering the interactions and transformations of nuclei, we harness nuclear chemistry's power for innovative technologies and better comprehension of our universe.
Conservation of Mass and Atomic Numbers
One hallmark of nuclear reactions, including bombardment reactions, is the conservation of mass numbers and atomic numbers. This conservation law dictates that just as mass and charge are preserved in chemical reactions, nuclear reactions must equally maintain these numbers.

In our exercise examples, we see this conservation in action. For reaction (a), with scandium transforming into potassium and emitting an alpha particle, the balance is demonstrated through equivalence of atomic and mass numbers. Atomic numbers: 21 (Sc) + 0 (n) = 19 (K) + 2 (He), while mass numbers: 45 + 1 = 42 + 4.

Similarly, in reaction (b), copper becomes zinc, emitting a neutron, with matching atomic and mass numbers ensuring the equation is balanced: Atomic: 29 (Cu) + 1 (p) = 30 (Zn) + 0 (n); Mass: 63 + 1 = 64 + 0.
  • Conservation of Atomic Numbers: Ensures the same total protons pre- and post-reaction.
  • Conservation of Mass Numbers: Total nucleons (protons and neutrons) remain constant throughout.

Understanding and applying these principles ensures accurate representation of nuclear transformations, vital for accuracy in scientific endeavors.

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Most popular questions from this chapter

Radon-222 gas can be found seeping from granite that contains uranium-238. Radon-222 is a nuclide in the radioactive decay series of uranium-238. Radon is an element with a half-life of 3.82 days. a Predict the most likely particle emitted when radon222 undergoes nuclear decay and write the nuclear equation for the decay. What is the decay constant \((k)\) for radon- 222 . c) The U.S. Environmental Protection Agency recommends that the concentration of radon gas in dwellings not exceed 4 piCi per liter of air. What mass of \(\mathrm{Rn}-222\) would be in a l-L sample of air that had a decay rate of 4 piCi?

Briefly describe neutron activation analysis.

When considering the lifetime of a radioactive species, a general rule of thumb is that after 10 half-lives have passed, the amount of radioactive material left in the sample is negligible. The disposal of some radioactive materials is based on this rule. What percentage of the original material is left after 10 half-lives? b) When would it be a bad idea to apply this rule?

The naturally occurring isotope rubidium-87 decays by beta emission to strontium- 87 . This decay is the basis of a method for determining the ages of rocks. A sample of rock contains \(102.1 \mu \mathrm{g}{ }^{87} \mathrm{R} \mathrm{b}\) and \(5.0 \mu \mathrm{g}{ }^{87} \mathrm{Sr}\). What is the age of the rock? The half-life of rubidium- 87 is \(4.8 \times 10^{10} \mathrm{y}\)

The first isotope of plutonium discovered was plutonium- \(238 .\) It is used to power batteries for heart pacemakers. A sample of plutonium- 238 weighing \(2.8 \times 10^{-6} \mathrm{~g}\) decays at the rate of \(1.8 \times 10^{6}\) disintegrations per second. What is the decay constant of plutonium- 238 in reciprocal seconds \((/ \mathrm{s}) ?\)

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