How to Find Energy Released in Alpha Decay: A Comprehensive Guide

How to Find Energy Released in Alpha Decay

Alpha decay is a fascinating phenomenon that occurs in certain radioactive isotopes. In this blog post, we will delve into the concept of alpha decay and explore how to calculate the energy released during this process. So, let’s get started!

Understanding the Concept of Alpha Decay

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle. An alpha particle consists of two protons and two neutrons, making it identical to a helium-4 nucleus. This emission of an alpha particle from the nucleus leads to the formation of a new nucleus, known as the daughter nucleus.

The primary reason behind alpha decay is the desire of the parent nucleus to achieve better stability. By emitting an alpha particle, the parent nucleus reduces its size and becomes more stable. This reduction in size is accompanied by the release of energy in the form of kinetic energy of the alpha particle and the recoil energy of the daughter nucleus.

The Role of Energy in Alpha Decay

Energy plays a crucial role in the process of alpha decay. It is the energy released during the decay that allows us to understand and study this phenomenon. By determining the energy released in alpha decay, we can gain insights into the stability of nuclei, the behavior of radioactive isotopes, and various nuclear reactions.

To find the energy released in alpha decay, we rely on a formula specifically designed for this purpose. This formula takes into account the masses of the parent and daughter nuclei and the mass of the alpha particle.

What is the Alpha Decay Formula

The alpha decay formula provides a direct way to calculate the energy released during alpha decay. It is expressed as:

$Q = (m_{\text{parent}} - m_{\text{daughter}} - m_{\alpha}) \cdot c^2$

Where:
– $Q$ represents the energy released in alpha decay.
– $m_{\text{parent}}$ is the mass of the parent nucleus.
– $m_{\text{daughter}}$ is the mass of the daughter nucleus.
– $m_{\alpha}$ is the mass of the alpha particle.
– $c$ is the speed of light.

By plugging in the respective masses into this formula, we can calculate the energy released during alpha decay.

Energy Released in Decay Formula

The energy released in decay formula allows us to determine the energy released in alpha decay accurately. It takes into consideration the atomic masses and the speed of light, which plays a significant role in converting mass into energy.

The formula shows that the energy released is directly related to the difference in masses between the parent and daughter nuclei, as well as the mass of the alpha particle. The greater the mass difference, the more energy is released.

How to Calculate Energy Released in Alpha Decay

Calculating the energy released in alpha decay involves a step-by-step process. Let’s walk through the process together:

Determine the masses of the parent and daughter nuclei. These values can be found in the periodic table or other reliable sources.
Find the mass of the alpha particle. Since an alpha particle consists of two protons and two neutrons, its mass is approximately equal to four atomic mass units (u).
Plug these values into the alpha decay formula $Q = (m_{\text{parent}} - m_{\text{daughter}} - m_{\alpha}) \cdot c^2$ .
Calculate the difference in masses $(m_{\text{parent}} - m_{\text{daughter}} - m_{\alpha})$ .
Multiply the difference in masses by the speed of light squared $(c^2)$ to obtain the energy released $Q$ .

By following these steps and using the alpha decay formula, you can accurately calculate the energy released in alpha decay.

Worked Out Examples on Calculating Energy Released in Alpha Decay

Let’s now put our knowledge into practice with a couple of worked-out examples:

Example 1:
Suppose we have a parent nucleus with a mass of 220 atomic mass units (u) and a daughter nucleus with a mass of 216 atomic mass units (u). The mass of the alpha particle is approximately 4 atomic mass units (u). Using the alpha decay formula, let’s calculate the energy released in this decay.

$Q = (220 - 216 - 4) \cdot c^2$

The value of $c$ is approximately $3 \times 10^8 \, \text{m/s}$ .
Plugging in the values, we have:

$Q = 0 \cdot (3 \times 10^8)^2$

Simplifying, we find that the energy released in this alpha decay is 0 Joules.

Example 2:
Consider a different scenario, where the parent nucleus has a mass of 238 atomic mass units (u), the daughter nucleus has a mass of 234 atomic mass units (u), and the alpha particle has a mass of 4 atomic mass units (u). Let’s determine the energy released in this alpha decay.

$Q = (238 - 234 - 4) \cdot c^2$

Plugging in the values, we get:

$Q = 0 \cdot (3 \times 10^8)^2$

Again, the energy released in this alpha decay is 0 Joules.

These worked-out examples demonstrate that the energy released in alpha decay can vary depending on the specific isotopes involved. Some alpha decays release significant amounts of energy, while others release none at all.

Factors Influencing the Amount of Energy Released in Alpha Decay

Several factors influence the amount of energy released in alpha decay. These factors include the difference in masses between the parent and daughter nuclei, the specific isotopes involved, and the stability of the resulting daughter nucleus.

In general, heavier parent nuclei tend to release more energy during alpha decay. This is because the larger mass difference leads to a greater energy release. Additionally, highly unstable isotopes are more likely to undergo alpha decay and release more energy in the process.

Practical Examples Demonstrating the Amount of Energy Released in Alpha Decay

Let’s consider a couple of practical examples to illustrate the amount of energy released in alpha decay:

Uranium-238 is a radioactive isotope that undergoes alpha decay to form Thorium-234. This decay process releases approximately 4.27 million electron volts (MeV) of energy.
Polonium-210, another radioactive isotope, decays through alpha decay to form Lead-206. This decay process releases approximately 5.41 MeV of energy.

These examples highlight that alpha decay can result in the release of significant amounts of energy, which is crucial to understand for various applications in nuclear physics and energy production.

Numerical Problems on how to find energy released in alpha decay

Problem 1:

A radioactive element undergoes alpha decay, resulting in the emission of an alpha particle. If the mass of the radioactive element before decay is 220 atomic mass units (amu) and the mass of the resulting nucleus after decay is 216 amu, calculate the energy released in the process.

Solution:

Given:
Mass of radioactive element before decay = 220 amu
Mass of resulting nucleus after decay = 216 amu

The energy released in alpha decay can be calculated using the equation:

$E = \Delta m \cdot c^2$

Where:
$\Delta m$ is the change in mass, and
$c$ is the speed of light $\(3 \times 10^8 \, \text{m/s}$ )

The change in mass $\(\Delta m$ ) is given by:

$\Delta m = \text{Mass of radioactive element before decay} - \text{Mass of resulting nucleus after decay}$

Substituting the given values:

$\Delta m = 220 \, \text{amu} - 216 \, \text{amu} = 4 \, \text{amu}$

Now, substituting the values of $\Delta m$ and $c$ into the equation for energy:

$E = 4 \, \text{amu} \times (3 \times 10^8 \, \text{m/s})^2$

Simplifying:

$E = 4 \, \text{amu} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2$

Therefore, the energy released in the alpha decay process is $3.6 \times 10^{17} \, \text{J}$ .

Problem 2:

A certain radioactive material undergoes alpha decay and emits an alpha particle. If the change in mass during the decay process is 0.004 kg, calculate the energy released in the process.

Solution:

Given:
Change in mass $\(\Delta m$ ) during decay = 0.004 kg

Using the equation for energy released in alpha decay:

$E = \Delta m \cdot c^2$

Where:
$\Delta m$ is the change in mass, and
$c$ is the speed of light $\(3 \times 10^8 \, \text{m/s}$ )

Substituting the given values:

$E = 0.004 \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2$

Simplifying:

$E = 0.004 \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2$

Therefore, the energy released in the alpha decay process is $3.6 \times 10^{14} \, \text{J}$ .

Problem 3:

An alpha particle is emitted during the alpha decay of a radioactive material. If the mass of the alpha particle is 4 atomic mass units (amu) and the energy released during the decay is 3.2 MeV (million electron volts), calculate the change in mass during the decay process.

Solution:

Given:
Mass of the alpha particle = 4 amu
Energy released during decay = 3.2 MeV

The energy released in alpha decay can be calculated using the equation:

$E = \Delta m \cdot c^2$

Where:
$\Delta m$ is the change in mass, and
$c$ is the speed of light $\(3 \times 10^8 \, \text{m/s}$ )

We need to rearrange the equation to solve for $\Delta m$ :

$\Delta m = \frac{E}{c^2}$

Substituting the given values:

$\Delta m = \frac{3.2 \times 10^6 \, \text{eV}}{(3 \times 10^8 \, \text{m/s})^2}$

Converting electron volts (eV) to joules (J):

$\Delta m = \frac{3.2 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J}}{(3 \times 10^8 \, \text{m/s})^2}$

Simplifying:

$\Delta m = \frac{3.2 \times 1.6 \times 10^{-13} \, \text{J}}{9 \times 10^{16} \, \text{m}^2/\text{s}^2}$

Therefore, the change in mass during the alpha decay process is $3.57 \times 10^{-30} \, \text{kg}$ .

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