Question

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

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

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: \[^{90}_{37}\text{Rb} \rightarrow ^{90}_{38}\text{Sr} + ^0_{-1}\beta\]

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: \[^{72}_{34}\text{Se} + ^0_{-1}e \rightarrow ^{72}_{33}\text{As}\]

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: \[^{76}_{36}\text{Kr} \rightarrow ^{76}_{35}\text{Br} + ^0_{+1}\beta\]

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: \[^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^4_2\alpha\]