Question

Give the balanced nuclear equation for each of the following changes: (a) positron emission by carbon-10, (b) alpha emission by curium- 243 , (c) electron capture by vanadium- \(49,\) (d) beta emission by oxygen-20.

Short answer

(a) _{6}^{10}C \rightarrow _{5}^{10}B + _{+1}^{0}e (b) _{96}^{243}Cm \rightarrow _{94}^{239}Pu + _{2}^{4}He (c) _{23}^{49}V + _{-1}^{0}e \rightarrow _{22}^{49}Ti (d) _{8}^{20}O \rightarrow _{9}^{20}F + _{-1}^{0}e

Step-by-step solution

Positron Emission by Carbon-10

Write the nuclear reaction for the positron emission. Identify carbon-10 and represent the positron as 0/+1 e. Since a positron has no mass but a charge of +1, the atomic number will decrease by 1, while the atomic mass will remain the same.

Alpha Emission by Curium-243

Represent alpha particle emission. An alpha particle is represented as 4/2 He, which has a mass number of 4 and an atomic number of 2. When curium-243 emits an alpha particle, its atomic mass will decrease by 4 units, and its atomic number will decrease by 2.

Electron Capture by Vanadium-49

Write the nuclear equation for electron capture by vanadium-49. An electron is captured by the nucleus, so write down the electron as 0/-1 e on the left of the equation. The atomic mass stays the same, but the atomic number decreases by 1.

Beta Emission by Oxygen-20

Write the equation for beta emission. Represent the beta particle as 0/-1 e, which has a mass number of 0 and an atomic number of -1. The atomic mass remains unchanged, but the atomic number increases by 1 since a neutron is converted into a proton and a beta particle.

Key concepts

Positron Emission

Positron emission is a type of radioactive decay where an unstable nucleus releases a positron to become more stable. A positron, often represented as \(0/+1 e\), is a particle with the same mass as an electron but with a positive charge. When a positron is emitted, the atomic number of the element decreases by one, but the atomic mass remains unchanged because the positron itself has negligible mass.

For example, in the case of carbon-10 undergoing positron emission, the nuclear equation is written as follows: \[{}^{10}_{6}\text{C} \rightarrow {}^{10}_{5}\text{B} + {}^{0}_{+1}\text{e}\]. Here, a carbon-10 atom is transformed into a boron-10 atom with the release of a positron.

Alpha Particle Emission

Alpha particle emission is another form of radioactive decay. An alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons, and is denoted by \(\text{He}\) or the symbol \(\alpha\text{.}\) When an atom emits an alpha particle, its atomic mass decreases by four units, and its atomic number decreases by two, as it loses two protons.

In the exercise with curium-243 ({}^{243}_{96}\text{Cm}), the alpha particle emission can be illustrated as: [{}^{243}_{96}\text{Cm} \rightarrow {}^{239}_{94}\text{Pu} + {}^{4}_{2}\text{He}], showing that a plutonium-239 atom results from the decay of curium-243 after the emission of an alpha particle.

Electron Capture

Electron capture is a process where an inner orbital electron is captured by the nucleus of its own atom. This causes a reduction in the atomic number by one because a proton in the nucleus is transformed into a neutron. The captured electron is usually one from the closest orbit to the nucleus, known as the K-shell.

In the context of vanadium-49 undergoing electron capture, we write the nuclear equation as follows: \[{}^{49}_{23}\text{V} + {}^{0}_{-1}\text{e} \rightarrow {}^{49}_{22}\text{Ti}\], indicating that a vanadium atom captures an electron to become a titanium atom. The mass number remains the same, but the atomic number decreases from 23 to 22.

Beta Emission

Beta emission, also referred to as beta decay, is a common type of nuclear reaction where a neutron in an atom's nucleus is transformed into a proton and emits a beta particle, which is a high-speed electron or positron. Specifically, in beta minus decay, an electron denoted by \(0/-1 e\) is emitted, leading to an increase in the atomic number by one while the mass number stays constant.

The beta emission from an oxygen-20 atom is manifested in the nuclear equation as \[{}^{20}_{8}\text{O} \rightarrow {}^{20}_{9}\text{F} + {}^{0}_{-1}\text{e}\]. In this reaction, oxygen-20 transforms into fluorine-20 with the emission of an electron, showcasing how beta emission changes the identity of the original atom.