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Write nuclear equations to represent the formation of an isotope of element 111 with a mass number of 272 by the bombardment of bismuth-209 by nickel-64 nuclei, followed by a succession of five \(\alpha\) -particle emissions.

Short Answer

Expert verified
First, the nuclear equation representing the bombardment of Bismuth-209 by Nickel-64 is \(^{209}_{83}Bi + ^{64}_{28}Ni ⟶ ^{273}_{111}X\). Second, the nuclear equation representing the subsequent five α-particle emissions is \(^{273}_{111}X ⟶ 5(^{4}_{2}He) + ^{253}_{103}X\).

Step by step solution

01

Form the equation for Bismuth bombardment by Nickel

During the bombardment process, the Bismuth atom \(^{209}_{83}Bi\) absorbs the Nickel nucleus \(^{64}_{28}Ni\), resulting in the formation of a heavier isotope. According to conservation of charge and mass, the atomic number and mass number of the resulting isotope should respectively be the sum of the atomic numbers and mass numbers of Bismuth and Nickel. Add the atomic and mass numbers together to get the equation:\(^{209}_{83}Bi + ^{64}_{28}Ni ⟶ ^{273}_{111}X\) Here, \(X\) represents the new isotope formed.
02

Form the equation for the Alpha particle emissions

In an α-particle emission, the nucleus loses 2 protons and 2 neutrons, represented by \(^{4}_{2}He\). To reduce the atomic and mass numbers of the heavy isotope formed in the bombardment process. As the problem states that there are five α-particle emissions, carry out the process five times for the new isotope:\(^{273}_{111}X ⟶ 5(^{4}_{2}He) + ^{253}_{103}X\)Here, \(X\) represents the final isotope formed after the α-particle emissions.

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Key Concepts

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

Isotope Formation
Isotope formation is a key concept in nuclear chemistry. It involves the creation of different forms of the same element, with the same number of protons but differing numbers of neutrons. This results in isotopes having the same atomic number but different mass numbers.
In the context of the given exercise, isotope formation occurs through the bombardment of the bismuth nucleus by a nickel nucleus. When Bismuth, which is represented as \( ^{209}_{83}Bi \), is bombarded by Nickel \( ^{64}_{28}Ni \), the two nuclei combine. This interaction forms a new element with mass and atomic numbers that are a sum of the two participating atomic nuclei:
  • Mass number is the total of protons and neutrons, \( 209 + 64 = 273 \).
  • Atomic number, illustrating the amount of protons, is summed, \( 83 + 28 = 111 \).
This combined element, represented as \( ^{273}_{111}X \), is an isotope of element 111. This demonstrates how isotopes can be artificially made by nuclear bombardment methods.
Alpha Particle Emission
Alpha particle emission is a type of radioactive decay where a nucleus emits an alpha particle. Alpha particles are comprised of 2 protons and 2 neutrons, making it similar to the helium nucleus \( ^{4}_{2}He \). This process decreases both the mass number and the atomic number of the nucleus.When a nucleus like our newly formed isotope \( ^{273}_{111}X \) undergoes alpha particle emission, it loses part of its mass and its proton count. This explains the decrease in the isotope's atomic number and mass number:
  • Each emission reduces the mass number by 4 (due to 2 protons and 2 neutrons).
  • The atomic number is lowered by 2 (due to the loss of 2 protons).
As mentioned in the exercise, the isotope undergoes five alpha particle emissions. This means its final mass number is \( 273 - 5 \times 4 = 253 \), and final atomic number is \( 111 - 5 \times 2 = 101 \). The resulting atom, represented as \( ^{253}_{101}X \), is a significantly different isotope than the one initially formed.
Nuclear Equations
Nuclear equations are like mathematical equations that describe the changes occurring during nuclear reactions. They help keep track of atomic and mass numbers before and after reactions, ensuring the law of conservation is followed.
In nuclear equations:
  • Reactants are shown on the left, products on the right.
  • Mass and atomic numbers must be balanced on both sides of the equation.
In the exercise, the bombardment of Bismuth by Nickel is represented by the equation:\[ ^{209}_{83}Bi + ^{64}_{28}Ni ⟶ ^{273}_{111}X \]This expresses the formation of the new isotope. Following this, the alpha emissions' equation:\[ ^{273}_{111}X ⟶ 5(^{4}_{2}He) + ^{253}_{101}X \]demonstrates how the new isotope undergoes changes after emitting five alpha particles, reflecting the initial and final states of the reaction. Understanding these equations is crucial for visualizing and calculating the changes in nuclear processes.

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Most popular questions from this chapter

The overall change in the radioactive decay of \({238}_{92} \mathrm{U}\)to 206 \(\mathrm{Pb}\) is the emission of eight \(\alpha\) particles. Show that if \(_{82}^{206} \mathrm{Pb}\)this loss of eight \(\alpha\) particles were not also accompanied by six \(\beta^{-}\) emissions, the product nucleus would still be radioactive.

For medical uses, radon-222 formed in the radioactive decay of radium-226 is allowed to collect over the radium metal. Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn. The procedure can be continued indefinitely. The process is somewhat complicated by the fact that radon-222 itself undergoes radioactive decay to polonium- 218 , and so on. The half-lives of radium-226 and radon-222 are \(1.60 \times 10^{3}\) years and 3.82 days, respectively.(a) Beginning with pure radium- \(226,\) the number of radon-222 atoms present starts at zero, increases for a time, and then falls off again. Explain this behavior. That is, because the half-life of radon-222 is so much shorter than that of radium- \(226,\) why doesn't the radon-222 simply decay as fast as it is produced, without ever building up to a maximum concentration?(b) Write an expression for the rate of change \((d \mathrm{D} / d t)\) in the number of atoms (D) of the radon- 222 daughter in terms of the number of radium- 226 atoms present initially ( \(\mathrm{P}_{0}\) ) and the decay constants of the parent \(\left(\lambda_{\mathrm{p}}\right)\) and daughter \(\left(\lambda_{\mathrm{d}}\right)\) (c) Integration of the expression obtained in part (b) yields the following expression for the number of atoms of the radon-222 daughter (D) present at a time \(t\).$$\mathrm{D}=\frac{\mathrm{P}_{0} \lambda_{\mathrm{p}}\left(\mathrm{e}^{-\lambda_{\mathrm{p}} \times t}-\mathrm{e}^{-\lambda_{\mathrm{d}} \times t}\right)}{\lambda_{\mathrm{d}}-\lambda_{\mathrm{p}}}$$,Starting with \(1.00 \mathrm{g}\) of pure radium- \(226,\) approximately how long will it take for the amount of radon222 to reach its maximum value: one day, one week, one year, one century, or one millennium?

Explain why the rem is more satisfactory than the rad as a unit for measuring radiation dosage.

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The carbon-14 dating method is based on the assumption that the rate of production of \(^{14} \mathrm{C}\) by cosmic ray bombardment has remained constant for thousands of years and that the ratio of \(^{14} \mathrm{C}\) to \(^{12} \mathrm{C}\) has also remained constant. Can you think of any effects of human activities that could invalidate this assumption in the future?

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