Question

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

Short answer

The balanced nuclear equations for the given processes are: a) \(^{90}Rb \rightarrow ^{90}Sr + ^0_-1\beta\) b) \(^{72}Se + ^0_-1e \rightarrow ^{72}As\) c) \(^{76}Kr \rightarrow ^{76}Br + ^0_+1e\) d) \(^{226}Ra \rightarrow ^{222}Rn + ^4_2\alpha\)

Step-by-step solution

Identify the initial nucleus

Before the beta emission, we have a rubidium-90 nucleus, which can be represented as \(^{90}Rb\). The atomic number of rubidium (Rb) is 37.

Write the beta particle

During beta emission, a nucleus emits a beta particle, represented as \(^0_-1\beta\).

Determine the final nucleus

In beta emission, a neutron is converted into a proton. The mass number remains the same (90), but the atomic number increases by 1. The final nucleus will therefore have a mass number of 90 and an atomic number of 38, indicating a strontium isotope: \(^{90}Sr\).

Write the balanced nuclear equation

We can now write the balanced nuclear equation for the beta emission of rubidium-90: \(^{90}Rb \rightarrow ^{90}Sr + ^0_-1\beta\). #b) Electron Capture of Selenium-72#

Identify the initial nucleus

Before the electron capture, we have a selenium-72 nucleus, which can be represented as \(^{72}Se\). The atomic number of selenium (Se) is 34.

Write the captured electron

During electron capture, a nucleus captures an electron, represented as \(^0_-1e\).

Determine the final nucleus

In electron capture, a proton is converted into a neutron. The mass number remains the same (72), but the atomic number decreases by 1. The final nucleus will therefore have a mass number of 72 and an atomic number of 33, indicating an arsenic isotope: \(^{72}As\).

Write the balanced nuclear equation

We can now write the balanced nuclear equation for the electron capture of selenium-72: \(^{72}Se + ^0_-1e \rightarrow ^{72}As\). #c) Positron Emission of Krypton-76#

Identify the initial nucleus

Before the positron emission, we have a krypton-76 nucleus, which can be represented as \(^{76}Kr\). The atomic number of krypton (Kr) is 36.

Write the positron

During positron emission, a nucleus emits a positron, represented as \(^0_+1e\).

Determine the final nucleus

In positron emission, a proton is converted into a neutron. The mass number remains the same (76), but the atomic number decreases by 1. The final nucleus will therefore have a mass number of 76 and an atomic number of 35, indicating a bromine isotope: \(^{76}Br\).

Write the balanced nuclear equation

We can now write the balanced nuclear equation for the positron emission of krypton-76: \(^{76}Kr \rightarrow ^{76}Br + ^0_+1e\). #d) Alpha Emission of Radium-226#

Identify the initial nucleus

Before the alpha emission, we have a radium-226 nucleus, which can be represented as \(^{226}Ra\). The atomic number of radium (Ra) is 88.

Write the alpha particle

During alpha emission, a nucleus emits an alpha particle, represented as \(^4_2\alpha\).

Determine the final nucleus

In alpha emission, both the mass number and the atomic number decrease. The mass number decreases by 4, and the atomic number decreases by 2. The final nucleus will therefore have a mass number of 222 and an atomic number of 86, indicating a radon isotope: \(^{222}Rn\).

Write the balanced nuclear equation

We can now write the balanced nuclear equation for the alpha emission of radium-226: \(^{226}Ra \rightarrow ^{222}Rn + ^4_2\alpha\).

Key concepts

Understanding Beta Emission in Nuclear Reactions

Beta emission is a type of radioactive decay in which an unstable nucleus transforms to a more stable state. During this process, a neutron in the nucleus is converted into a proton and a beta particle—an electron—is released. For example, rubidium-90, with an atomic number of 37, emits a beta particle, resulting in a new atom, strontium-90, with an atomic number of 38. This is because the loss of a beta particle effectively increases the proton count by one. The balanced equation for this process is:
\( ^{90}_{37}Rb \rightarrow ^{90}_{38}Sr + ^0_{-1}\beta \).
Understanding this process is crucial for grasping various nuclear transformations and their implications in both natural and engineered systems.

Electron Capture: A Unique Nuclear Phenomenon

Electron capture is a fascinating process where an atom's nucleus captures one of its own electrons and combines it with a proton, forming a neutron and a neutrino. This process decreases the atomic number by one but leaves the mass number unchanged. Taking selenium-72 as an example, it undergoes electron capture, turning into arsenic-72. The balanced nuclear equation reflects this subtle internal exchange:
\( ^{72}_{34}Se + ^0_{-1}e \rightarrow ^{72}_{33}As \).
This transformation is significant in many fields, including astrophysics and medical diagnostics, where electron capture is used in the imaging technique known as positron emission tomography (PET).

Positron Emission: Matter's Mirror Image

Positron emission is the mirror opposite of beta decay. Instead of an electron, a positron—which is an anti-electron—is emitted when a proton in the nucleus transforms into a neutron. The atom's atomic number goes down by one, while the mass number remains stable. In our textbook example, krypton-76 releases a positron and becomes bromine-76. The reaction is detailed as follows:
\( ^{76}_{36}Kr \rightarrow ^{76}_{35}Br + ^0_{+1}e \).
This concept is also integral to PET scans, where the emitted positrons encounter electrons, leading to annihilation and the release of gamma rays—key to imaging internal bodily functions.

Alpha Radiation: A Heavyweight in Nuclear Decay

Alpha radiation involves the ejection of an alpha particle, which is essentially a helium nucleus composed of two protons and two neutrons, from an unstable atom. Consequently, this type of emission reduces both the mass and atomic numbers — by four and two, respectively. Radium-226, for instance, undergoes alpha decay to become radon-222. The balanced equation for alpha decay of radium-226 is:
\( ^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^4_2\alpha \).
Alpha particles, due to their relatively large mass, have a short range and can be stopped by a sheet of paper. However, if ingested or inhaled, they pose significant health risks due to their high ionization power.