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Chapter 20: Problem 50

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

Short Answer

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

Step by step solution

01

Understand the Notation

In the given notation, \(X(n, \alpha)Y\), \(X\) is the original nucleus that is bombarded by a neutron (\(\mathrm{n}\)), resulting in the emission of an alpha particle (\(\alpha\)) and the formation of a new nucleus \(Y\). Similarly, \(X(p, n)Y\) means \(X\) is bombarded by a proton (\(\mathrm{p}\)), causing emission of a neutron (\(\mathrm{n}\)) and forming nucleus \(Y\).
02

Solve Reaction (a)

For \({ }_{21}^{45} \mathrm{Sc}(\mathrm{n}, \alpha)_{19}^{42} \mathrm{~K}\):Write the equation: \[{}_{21}^{45} \mathrm{Sc} + {}_{0}^{1} \mathrm{n} \rightarrow {}_{2}^{4} \alpha + {}_{19}^{42} \mathrm{~K}\]Ensure total atomic numbers and mass numbers equal on both sides: - Atomic numbers: \(21 = 2 + 19\)- Mass numbers: \(45 + 1 = 4 + 42\)
03

Solve Reaction (b)

For \({ }_{29}^{63} \mathrm{Cu}(\mathrm{p}, \mathrm{n})_{30}^{63} \mathrm{Zn}\):Write the equation: \[{}_{29}^{63} \mathrm{Cu} + {}_{1}^{1} \mathrm{p} \rightarrow {}_{0}^{1} \mathrm{n} + {}_{30}^{63} \mathrm{Zn}\]Ensure total atomic numbers and mass numbers equal on both sides:- Atomic numbers: \(29 + 1 = 0 + 30\)- Mass numbers: \(63 + 1 = 1 + 63\)

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Key Concepts

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

Nuclear Reactions
Nuclear reactions are fundamental processes that involve a change in an atom's nucleus. Unlike chemical reactions, which involve electrons and the outer atomic structure, nuclear reactions can alter the core structure of the atom itself. This happens when particles such as neutrons or protons collide with a nucleus.

There are various types of nuclear reactions, including fusion and fission. In the context of the problem, we're looking at bombardment reactions. Here, a small particle (like a neutron or proton) is introduced to a nucleus, causing changes in the nuclear configuration.
  • Bombardment reactions alter the composition of the nucleus.
  • They can result in the emission of particles, like alpha particles or neutrons.
  • Nuclear equations are used to represent these changes, showing both reactants and products.
Alpha Particle Emission
Alpha particle emission is a common type of radioactive decay in nuclear reactions. An alpha particle consists of two protons and two neutrons, the same as a helium nucleus. When emitted, it reduces the atomic number of the parent nucleus by two and the mass number by four.

For example, in the given reaction a. ( \({ }_{21}^{45} \mathrm{Sc}(\mathrm{n}, \alpha)_{19}^{42} \mathrm{~K}\) ),a neutron bombards scandium, leading to the emission of an alpha particle and the formation of potassium.
  • Alpha particles are represented as i. ( \({}_{2}^{4} \mathrm{He}\) ) or ii. ( \(\alpha\) )li>
  • They reduce both the atomic and mass numbers of the original nucleus.
  • The process can transform one element into another, as seen from scandium to potassium here.
Proton Bombardment
Proton bombardment involves striking a nucleus with protons, which are particles with a positive charge. This technique can induce changes in the nuclear structure, potentially transforming one element into another.

Consider the reaction b. ( \({ }_{29}^{63} \mathrm{Cu}(\mathrm{p}, \mathrm{n})_{30}^{63} \mathrm{Zn}\) ).Here, a copper nucleus is hit by a proton resulting in the expulsion of a neutron and the creation of zinc.
  • Protons are represented as i. ( \({}_1^1 \mathrm{H}\) ) or ii. ( \(\mathrm{p}\) )li>
  • Adding a proton increases the atomic number by one.
  • Simultaneously, losing a neutron maintains the mass number.
Atomic Number Conservation
Atomic number conservation is a principle that must balance in nuclear equations. The atomic number refers to the number of protons in a nucleus, determining an element's identity. During nuclear reactions, the sum of atomic numbers of reactants equals the sum of products.

Take reaction a. ( \({ }_{21}^{45} \mathrm{Sc} + {}_{0}^{1} \mathrm{n} \rightarrow {}_{2}^{4} \alpha + {}_{19}^{42} \mathrm{~K}\) ).Power balance checks:
  • **Left Side**: Scandium has 21 protons plus 0 from the neutron.
  • **Right Side**: Alpha particle has 2 protons, Potassium has 19 protons.
  • Ensuring: 21 equals 19 plus 2 confirms equilibrium.
Mass Number Conservation
Mass number conservation is equally crucial in nuclear reactions, meaning the total number of protons and neutrons (mass number) must remain constant. This ensures mass isn't mysteriously appearing or disappearing in nuclear equations.

Examining reaction b. ( \({ }_{29}^{63} \mathrm{Cu} + {}_{1}^{1} \mathrm{p} \rightarrow {}_{0}^{1} \mathrm{n} + {}_{30}^{63} \mathrm{Zn}\) ).Look at those numbers:
  • **Start**: Copper has mass 63, proton gives it 1 additionally.
  • **Finish**: Neutron's mass is 1, zinc carries mass 63.
  • Confirming: 64 at start equals 64 at end showcases total mass equilibrium.

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Most popular questions from this chapter

Write the nuclear equation for the decay of phosphorus32 to sulfur- 32 by beta emission. A phosphorus- 32 nucleus emits a beta particle and gives a sulfur- 32 nucleus.

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 \(102.1 \mu \mathrm{g}^{87} \mathrm{Rb}\) and \(5.3 \mu \mathrm{g}^{87} \mathrm{Sr}\). What is the age of the rock? The half-life of rubidium- 87 is \(4.8 \times 10^{10} \mathrm{y}\).

Radioisotope thermoelectric generators can be used by satellites to obtain power from radioactive decay of various isotopes, plutonium-238 being the preferred fuel. Plutonium-238 decays via alpha emission and has a half-life of \(87.7\) years. a. Write the nuclear equation for the alpha decay of plutonium-238.

Neutron activation analysis is an analytical technique in which a sample of material is bombarded with neutrons from a fission reactor. When a \(35.0\) -g aluminum can is irradiated, it has an initial activity of \(40.0\) curies (Ci). The safety office won't let you touch anything having an activity in excess of \(0.100 \mathrm{Ci}\). Assuming all the radioactivity is from \({ }^{28} \mathrm{Al}\), which has an halflife of \(2.28\) min, how many minutes do you have to wait after bombardment before you can handle the can safely? a. \(17.0 \mathrm{~min}\) b. \(1.32 \times 10^{3} \mathrm{~min}\) c. \(19.7 \mathrm{~min}\) d. \(478 \mathrm{~min}\) e. \(476 \mathrm{~min}\)

The decay of \(\mathrm{Rb}-87\left(t_{1 / 2}=4.8 \times 10^{10} \mathrm{y}\right)\) to \(\mathrm{Sr}-87\) has been used to determine the age of ancient rocks and minerals. a. Write the balanced nuclear equation for this decay. b. If a sample of rock is found to be \(0.100 \%\) by mass \(\mathrm{Rb}-87\) and \(0.00250 \% \mathrm{Sr}-87\), what is the age of the rock? Assume that there was no Sr-87 present when the rock formed.
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