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A long-cherished dream of alchemists was to produce gold from cheaper and more abundant elements. This dream was finally realized when \({ }_{80}^{198} \mathrm{Hg}\) was converted into gold by neutron bombardment. Write a balanced equation for this reaction.

Short Answer

Expert verified
The balanced nuclear equation describing the conversion of mercury into gold by neutron bombardment is ${ }_{80}^{198}\mathrm{Hg}$ + ${ }_{0}^{1}n$ -> ${ }_{81}^{199}\mathrm{Au}$ + ${ }_{-1}^{0}β$.

Step by step solution

01

Identify the Target and Product Nuclei

The target nucleus given is ${ }_{80}^{198}\mathrm{Hg}$ (mercury) and it's being converted into gold. However, the exact isotope of gold is not provided, so it's not immediately clear what changes are occurring in the atomic and mass numbers.
02

Understand the Process of Neutron Bombardment

Neutron bombardment refers to the process of firing neutrons at a target nucleus. The neutron is represented as ${ }_{0}^{1}n$. When a neutron is absorbed by a nucleus, it increases only the mass number of the target nucleus (since neutron carries no charge and does not change the atomic number).
03

Determine the Product Nucleus

Once the mercury nucleus absorbs a neutron, it becomes an isotope of mercury with mass number 199 (${ }_{80}^{199}\mathrm{Hg}$). This isotope is unstable and undergoes beta decay, where a neutron in the nucleus is converted into a proton, increasing the atomic number by one and producing an electron (or beta particle, ${ }_{-1}^{0}β$) in the process. This results in the formation of the nucleus with atomic number 81 that is an isotope of gold ${ }_{81}^{199}\mathrm{Au}$.
04

Write Initial and Final Reactions

The initial reaction is ${ }_{80}^{198}\mathrm{Hg}$ + ${ }_{0}^{1}n$ -> ${ }_{80}^{199}\mathrm{Hg}$. Then the decay process can be written as ${ }_{80}^{199}\mathrm{Hg}$ -> ${ }_{81}^{199}\mathrm{Au}$ + ${ }_{-1}^{0}β$. These two steps can be combined to write the overall balanced nuclear equation.
05

Write the Overall Balanced Nuclear Reaction

The final balanced equation that describes the entire process is ${ }_{80}^{198}\mathrm{Hg}$ + ${ }_{0}^{1}n$ -> ${ }_{81}^{199}\mathrm{Au}$ + ${ }_{-1}^{0}β$.

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Most popular questions from this chapter

Complete these nuclear equations and identify \(\mathrm{X}\) in each case: (a) \({ }_{12}^{26} \mathrm{Mg}+{ }_{1}^{1} \mathrm{p} \longrightarrow{ }_{2}^{4} \alpha+\mathrm{X}\) (b) \({ }_{27}^{59} \mathrm{Co}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{27}^{60} \mathrm{Co}+\mathrm{X}\) (c) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{36}^{94} \mathrm{Kr}+{ }_{56}^{139} \mathrm{Ba}+3 \mathrm{X}\) (d) \({ }_{24}^{53} \mathrm{Cr}+{ }_{2}^{4} \alpha \longrightarrow{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (e) \({ }_{8}^{20} \mathrm{O} \longrightarrow{ }_{9}^{20} \mathrm{~F}+\mathrm{X}\)

Bismuth-214 is an \(\alpha\) -emitter with a half-life of 19.7 min. A 5.26 -mg sample of the isotope is placed in a sealed, evacuated flask of volume \(20.0 \mathrm{~mL}\) at \(40^{\circ} \mathrm{C}\). Assuming that all the \(\alpha\) particles generated are converted to helium gas and that the other decay product is nonradioactive, calculate the pressure (in \(\mathrm{mmHg}\) ) inside the flask after 78.8 min. Use 214 amu for the atomic mass of bismuth.

(a) Calculate the energy released when a \({ }^{238} \mathrm{U}\) isotope decays to \({ }^{234} \mathrm{Th} .\) The atomic masses are given by: \(^{238} \mathrm{U}: 238.0508 \mathrm{amu} ;{ }^{234} \mathrm{Th}: 234.0436 \mathrm{amu} ;{ }^{4} \mathrm{He}\) 4.0026 amu. (b) The energy released in (a) is transformed into the kinetic energy of the recoiling \({ }^{234} \mathrm{Th}\) nucleus and the \(\alpha\) particle. Which of the two will move away faster? Explain.

Consider the decay series \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}\) where \(A, B,\) and \(C\) 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.

How do nuclear reactions differ from ordinary chemical reactions?

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