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Chapter 21: Problem 11

Write balanced nuclear equations for the following processes: (a) rubidium-90 undergoes beta decay; (b) selenium-72 undergoes electron capture; (c) krypton-76 undergoes positron emission; (d) radium-226 emits alpha radiation.

Short Answer

Expert verified
a) \(^{90}_{37}\text{Rb} \rightarrow ^{90}_{38}\text{Sr} + ^0_{-1}\beta\) b) \(^{72}_{34}\text{Se} + ^0_{-1}e \rightarrow ^{72}_{33}\text{As}\) c) \(^{76}_{36}\text{Kr} \rightarrow ^{76}_{35}\text{Br} + ^0_{+1}\beta\) d) \(^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^4_2\alpha\)

Step by step solution

01

a) Beta Decay of Rubidium-90

Beta decay occurs when a neutron in a nucleus is transformed into a proton, and an electron (called a beta particle) is emitted. The initial element changes into another element as the atomic number increases by one. In the beta decay of rubidium-90 (\(^{90}_{37}\text{Rb}\)), the resulting element will have an atomic number of 37 + 1 = 38, which is strontium (Sr). The balanced nuclear equation can be written as: \[^{90}_{37}\text{Rb} \rightarrow ^{90}_{38}\text{Sr} + ^0_{-1}\beta\]
02

b) Electron Capture of Selenium-72

Electron capture occurs when a proton in a nucleus captures an inner shell electron, converting into a neutron. This process reduces the atomic number by one, transmuting the initial element into another element. In the electron capture of selenium-72 (\(^{72}_{34}\text{Se}\)), the resulting element will have an atomic number of 34 - 1 = 33, which is arsenic (As). The balanced nuclear equation can be written as: \[^{72}_{34}\text{Se} + ^0_{-1}e \rightarrow ^{72}_{33}\text{As}\]
03

c) Positron Emission of Krypton-76

Positron emission happens when a proton in a nucleus is transformed into a neutron, emitting a positron in the process. The initial element changes into another element with an atomic number reduced by one. In the positron emission of krypton-76 (\(^{76}_{36}\text{Kr}\)), the resulting element will have an atomic number of 36 - 1 = 35, which is bromine (Br). The balanced nuclear equation can be written as: \[^{76}_{36}\text{Kr} \rightarrow ^{76}_{35}\text{Br} + ^0_{+1}\beta\]
04

d) Alpha Radiation Emission of Radium-226

Alpha radiation is the emission of alpha particles, which are essentially helium-4 nuclei containing two protons and two neutrons. As a result, the initial element's atomic number and mass number both decrease: atomic number decreases by two, and the mass number decreases by four. In the alpha radiation emission of radium-226 (\(^{226}_{88}\text{Ra}\)), the resulting element will have an atomic number of 88 - 2 = 86, which is radon (Rn), and a mass number of 226 - 4 = 222. The balanced nuclear equation can be written as: \[^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^4_2\alpha\]

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Most popular questions from this chapter

The half-life of tritium (hydrogen-3) is \(12.3 \mathrm{yr}\). If \(56.2 \mathrm{mg}\) of tritium is released from a nuclear power plant during the course of an accident, what mass of this nuclide will remain after \(12.3\) yr? After 100 yr?

When a positron is annihilated by combination with an electron, two photons of equal energy result. What is the wavelength of these photons? Are they gamma ray photons?

In each of the following pairs, which nuclide would you expect to be the more abundant in nature: (a) \({ }_{48}^{115} \mathrm{Cd}\) or \({ }_{4} 11 \mathrm{Cd}\), (b) \({ }_{13}^{30} \mathrm{Al}\) or \({ }_{13}^{2} \mathrm{Al}\), (c) palladium-106 or palladium113, (d) xenon-128 or cesium-128? Justify your choices.

Potassium-40 decays to argon-40 with a half-life of \(1.27 \times 10^{9} \mathrm{yr}\). What is the age of a rock in which the mass ratio of \({ }^{40} \mathrm{Ar}\) to \({ }^{40} \mathrm{~K}\) is \(4.2 ?\)

Americium-241 is an alpha emitter used in smoke detectors. The alpha radiation ionizes molecules in an air-filled gap between two electrodes in the smoke detector, leading to current. When smoke is present, the ionized molecules bind to smoke particles and the current decreases; when the current is reduced sufficiently, an alarm sounds. (a) Write the nuclear equation corresponding to the alpha decay of americium-241. (b) Why is an alpha emitter a better choice than a gamma emitter for a smoke detector? (c) In a commercial smoke detector, only \(0.2\) micrograms of americium are present. Calculate the energy that is equivalent to the mass loss of this amount of americium due to alpha radiation. The atomic mass of americium- 241 is \(241.056829\) amu. (d) The half-life of americium-241 is 432 years; the half life of americium- 240 is \(2.12\) days. Why is the 241 isotope a better choice for a smoke detector?
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