Positron Emission: Videos & Practice Problems

Positron emission is a fascinating nuclear process where an unstable nucleus releases a positron, which is the antiparticle of the electron. While an electron carries a negative charge, a positron has a positive charge, making it essentially a "positive electron." This concept may seem unusual, but it is a key aspect of nuclear reactions.

In the context of positron emission, the term "emission" refers to the decay of the nucleus, indicating that the positron is a product of this decay process. To illustrate this, consider the element Einsteinium, represented by the symbol Es on the periodic table, with an atomic number of 99. When we examine the isotope ^{253}_{99}Es, the emission of a positron occurs. Since the atomic mass of the positron is 0, the mass number remains 253. However, the atomic number decreases by 1, as the positron emission transforms the original element into a new one. Thus, we need to find a number that, when added to 1, equals 99. This leads us to 98, resulting in the formation of ^{253}_{98}Cf, or Californium.

This example of positron emission highlights the transformation of elements during nuclear decay, showcasing the unique behavior of subatomic particles in nuclear chemistry.

In nuclear chemistry, positron emission is a type of radioactive decay where a positron (β⁺) is released from a nucleus. This process results in the transformation of a proton into a neutron, effectively decreasing the atomic number of the element by one while keeping the mass number unchanged.

For the first example, we consider uranium-235 (U-235). Uranium has an atomic number of 92. When it undergoes positron emission, the balanced nuclear equation can be represented as:

U-235:
^{235}_{92}U \rightarrow ^{235}_{91}Pa + \beta^+

Here, the uranium-235 emits a positron and transforms into protactinium (Pa), which has an atomic number of 91.

Next, we analyze radon-222 (Rn-222). Radon has an atomic number of 86. Upon emitting a positron, the balanced nuclear equation is:

Rn-222:
^{222}_{86}Rn \rightarrow ^{222}_{85}Ac + \beta^+

In this case, radon-222 emits a positron and changes into actinium (Ac), which has an atomic number of 85. These equations illustrate the process of positron emission and the resulting transformation of elements.

In this scenario, we analyze the decay process of Thorium-225, which undergoes a series of transformations through alpha and beta decays, as well as a gamma emission. Thorium has an atomic number of 90, and we will track the changes in both atomic mass and atomic number as it decays.

First, we consider the effects of the three alpha decays. Each alpha decay emits an alpha particle, which consists of 2 protons and 2 neutrons, effectively reducing the atomic mass by 4 and the atomic number by 2. Therefore, after three alpha decays, the changes can be calculated as follows:

• Mass change: \(225 - 3 \times 4 = 213\)
• Atomic number change: \(90 - 3 \times 2 = 84\)

Next, we account for the four beta decays. Each beta decay involves the emission of a beta particle (an electron), which increases the atomic number by 1 while leaving the mass unchanged. Thus, the changes from the beta decays are:

• Mass remains: \(213\)
• Atomic number change: \(84 + 4 = 88\)

Finally, the gamma emission does not affect the atomic mass or atomic number, but it indicates that the resulting nucleus is in an excited state. Therefore, after all decays, we find that the final product is Radium-213, represented as \(^{213}_{88}\text{Ra}^*\), where the asterisk denotes the excited state.

In summary, the decay process of Thorium-225 through three alpha decays and four beta decays results in Radium-213 in an excited state. Understanding the nature of these particles—alpha particles, beta particles, and gamma emissions—is crucial for grasping the underlying principles of nuclear decay and transformation.