/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 Write nuclear equations to repre... [FREE SOLUTION] | 91Ó°ÊÓ

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Write nuclear equations to represent (a) the decay of \(^{214} \mathrm{Ra}\) by \(\alpha\) -particle emission (b) the decay of \(^{205}\) At by positron emission (c) the decay of \(^{212} \mathrm{Fr}\) by electron capture (d) the reaction of two deuterium nuclei (deuterons) to produce a nucleus of \(\frac{3}{2} \mathrm{He}\). (e) the production of \({243}_{97} \mathrm{Bk}\) get by the \(\alpha\) -particle bombardment of\({241}_{95} \mathrm{Am}\) (f) a nuclear reaction in which thorium-232 is bombarded with \(\alpha\) particles, producing a new nuclide and four neutrons.

Short Answer

Expert verified
The nuclear equations corresponding to the tasks are respectively: \n(a) \(^{214}_{88}\mathrm{Ra} \rightarrow ^{210}_{86}\mathrm{Rn} + ^{4}_{2}\mathrm{He}\), \n(b) \(^{205}_{85}\mathrm{At} \rightarrow ^{205}_{84}\mathrm{Po} + ^{0}_{1}\mathrm{e^{+}}\), \n(c) \(^{212}_{87}\mathrm{Fr} + ^{0}_{-1}\mathrm{e^-} \rightarrow^{212}_{86}\mathrm{Rn}\), \n(d) \(^{2}_{1}\mathrm{H} + ^{2}_{1}\mathrm{H} \rightarrow ^{3}_{2}\mathrm{He} + ^{1}_{0}\mathrm{n}\), \n(e) \(^{241}_{95}\mathrm{Am} + ^{4}_{2}\mathrm{He} \rightarrow^{245}_{97}\mathrm{Bk}\) and \n(f) \(^{232}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He} \rightarrow^{232}_{92}\mathrm{U} + 4^{1}_{0}\mathrm{n}\).

Step by step solution

01

Nuclear equation for alpha decay of Radon-214

In alpha decay, an alpha particle, which is a helium nucleus (\(^{4}_2\mathrm{He}\)), is emitted. Therefore, the original nucleus loses 2 protons and 2 neutrons. The equation for this reaction is \(^{214}_{88}\mathrm{Ra} \rightarrow ^{210}_{86}\mathrm{Rn} + ^{4}_{2}\mathrm{He}\).
02

Nuclear equation for positron emission of Astatine-205

In positron emission, a particle with the same mass as an electron but with a positive charge is released. This corresponds to one proton changing into a neutron. The equation for this reaction is \(^{205}_{85}\mathrm{At} \rightarrow ^{205}_{84}\mathrm{Po} + ^{0}_{1}\mathrm{e^{+}}\).
03

Nuclear equation for electron capture by Francium-212

During electron capture, an electron from the electron shell of the atom is captured by the nucleus, converting a proton into a neutron. The equation for this reaction is \(^{212}_{87}\mathrm{Fr} + ^{0}_{-1}\mathrm{e^-} \rightarrow^{212}_{86}\mathrm{Rn}\).
04

Nuclear equation for Deuteron-Deuteron Fusion

In a fusion reaction, two or more smaller nuclei combine to form a larger nucleus. The equation for this reaction is \(^{2}_{1}\mathrm{H} + ^{2}_{1}\mathrm{H} \rightarrow ^{3}_{2}\mathrm{He} + ^{1}_{0}\mathrm{n}\).
05

Nuclear equation for Alpha-particle bombardment of Americium-241

In a reaction with alpha particle bombardment, the target nucleus absorbs the alpha particle. The equation for this reaction is \(^{241}_{95}\mathrm{Am} + ^{4}_{2}\mathrm{He} \rightarrow^{245}_{97}\mathrm{Bk}\).
06

Nuclear equation for a nuclear reaction with Thorium-232

The equation for this reaction (where Thorium-232 is bombarded with an alpha particle, producing a new nuclide and four neutrons) is \(^{232}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He} \rightarrow^{232}_{92}\mathrm{U} + 4^{1}_{0}\mathrm{n}\).

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Key Concepts

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

Alpha Decay
Alpha decay is a type of radioactive decay where an atomic nucleus ejects an alpha particle. An alpha particle is essentially a helium nucleus, containing 2 protons and 2 neutrons. This decay process results in the transformation of the original atom into a new atom that is lighter by 4 atomic mass units and has a reduction of 2 in its atomic number.

For example, when Radium-214 undergoes alpha decay, it emits an alpha particle and is transformed into Radon-210. The nuclear equation for this process is:
\[ ^{214}_{88}\mathrm{Ra} \rightarrow ^{210}_{86}\mathrm{Rn} + ^{4}_{2}\mathrm{He} \]
This equation shows that the new element Radon has 2 fewer protons than Radium and is correspondingly lighter.
Positron Emission
Positron emission is a decay process where a positron is emitted from an unstable nucleus. A positron has the same mass as an electron but carries a positive charge. This decay occurs when there are too many protons in a nucleus, making it unstable.
To simplify, when a proton transforms into a neutron, a positron is emitted to conserve charge.

The nuclear equation for positron emission in Astatine-205 can be expressed as:
\[ ^{205}_{85}\mathrm{At} \rightarrow ^{205}_{84}\mathrm{Po} + ^{0}_{1}\mathrm{e^{+}} \]
Here, a proton in Astatine changes to a neutron, and the emission of a positron results, creating Polonium-205.
Electron Capture
Electron capture is a process where an inner atomic electron is captured by the nucleus, leading to the conversion of a proton into a neutron. This generally happens when an atom is proton-rich and needs to increase its neutron-proton ratio.
During this capture, the atomic number decreases by one, as a proton is transformed, but the mass number remains unchanged.

An example of electron capture is seen in Francium-212, which captures an electron and converts into Radon-212. The equation that describes this is:
\[ ^{212}_{87}\mathrm{Fr} + ^{0}_{-1}\mathrm{e^-} \rightarrow ^{212}_{86}\mathrm{Rn} \]
This illustrates a proton changing into a neutron with the help of the captured electron.
Nuclear Reactions
Nuclear reactions involve the alteration of an atomic nucleus, resulting in new elements or isotopes. Unlike chemical reactions, which involve electrons, nuclear reactions alter the number of protons or neutrons in a nucleus, leading to more significant energy changes.
One kind of nuclear reaction is the bombardment of nucleons, such as in the reaction of Americium-241 with an alpha particle to create Berkelium-245.

The nuclear equation for this process is:
\[ ^{241}_{95}\mathrm{Am} + ^{4}_{2}\mathrm{He} \rightarrow^{245}_{97}\mathrm{Bk} \]
This shows the alpha particle combining with Americium, resulting in a new isotope of Berkelium.
Deuterium Fusion
Deuterium fusion is a form of nuclear fusion involving deuterium, a heavier isotope of hydrogen with one proton and one neutron. During fusion, two deuterium nuclei combine to form a larger nucleus, such as helium-3, with the simultaneous release of energy.

This type of fusion is represented by the following equation:
\[ ^{2}_{1}\mathrm{H} + ^{2}_{1}\mathrm{H} \rightarrow ^{3}_{2}\mathrm{He} + ^{1}_{0}\mathrm{n} \]
Fusion processes are the power source for stars, including our sun, where such reactions release vast amounts of energy. This energy production occurs as light nuclei merge, creating heavier nuclei and releasing binding energy.

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Most popular questions from this chapter

Of the following nuclides, the highest nuclear binding energy per nucleon is found in (a) \(_{1}^{3} \mathrm{H} ;\) (b) \(_{8}^{16} \mathrm{O} ;\) (c) \(_{26}^{56} \mathrm{Fe}\); (d) \(_{92}^{235} \mathrm{U}\).

Iodine-129 is a product of nuclear fission, whether from an atomic bomb or a nuclear power plant. It is a \(\beta^{-}\) emitter with a half-life of \(1.7 \times 10^{7}\) years. How many disintegrations per second would occur in a sample containing \(1.00 \mathrm{mg}^{129} \mathrm{I} ?\)

In some cases, the most abundant isotope of an element can be established by rounding off the atomic mass to the nearest whole number, as in \(^{39} \mathrm{K},^{85} \mathrm{Rb}\), and \(^{88} \mathrm{Sr}\). But in other cases, the isotope corresponding to the rounded-off atomic mass does not even occur naturally, as in \(^{64} \mathrm{Cu}\). Explain the basis of this observation.

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?

A lunar rock was analyzed for argon by mass spectrometry and for potassium by atomic absorption. The results of these analyses showed that the sample contained \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon and \(0.083 \%\) of potassium. The half-life of potassium- 40 is \(1.248 \times\) \(10^{9} \mathrm{y} \cdot\) Calculate the age of the lunar rock.

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