Question

The gamma ray emitted when oxygen-19 gives off a beta particle is \(0.197 \mathrm{MeV}\). What is its energy in joules?

Short answer

The energy of the gamma ray in joules is approximately \(3.156 \times 10^{-14} \, \text{J}\).

Step-by-step solution

Understand the energy conversion required

The problem requires converting energy given in mega-electronvolts (MeV) to joules (J). The energy of a gamma ray is provided in MeV, and we need to convert this to J.

Recall conversion factor

1 electronvolt (eV) is equivalent to \[1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\]Since we are dealing with MeV, recall that \[1 \, \text{MeV} = 10^6 \, \text{eV} \]

Convert MeV to eV

Convert the given energy value from MeV to eV by multiplying with \(10^6\):\[0.197 \, \text{MeV} = 0.197 \times 10^6 \, \text{eV}\]

Calculate energy in joules

Now, convert the energy from electronvolts (eV) to joules using the conversion factor:\[0.197 \times 10^6 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 3.15594 \times 10^{-14} \, \text{J}\]

Key concepts

Mega-electronvolts to Joules

Understanding how to convert energy from mega-electronvolts (MeV) to joules (J) is important in many areas of physics, especially nuclear physics. Let's break this down further.Mega-electronvolts, abbreviated as MeV, are a unit of energy used primarily to express particle energies in nuclear physics. One MeV is equal to one million electronvolts (eV), hence the prefix 'mega,' which denotes a factor of one million.Here's how you convert MeV to joules:

• First, remember the conversion factor between eV and J: 1 eV = \(1.602 \times 10^{-19} \) J.
• Since 1 MeV = \(10^6\) eV, multiply the energy in MeV by \(10^6\) to convert it to eV.
• Finally, multiply by the conversion factor \(1.602 \times 10^{-19} \) J/eV to complete the conversion to joules.

For example, to convert 0.197 MeV to J:1. Convert 0.197 MeV to eV: \(0.197 \times 10^6\) eV.2. Then, to find the energy in joules: \(0.197 \times 10^6 \times 1.602 \times 10^{-19}\) J.Now, you understand how to easily convert MeV to J, a useful skill for understanding energy scales in nuclear reactions!

Gamma Ray Energy

Gamma rays are a form of electromagnetic radiation with a very high energy level. They are emitted from the nucleus of an atom, especially during radioactive decay or nuclear reactions. Gamma ray energy is often expressed in terms of electronvolts (eV) or mega-electronvolts (MeV), given the high energy levels involved. For example, in the case of oxygen-19 emitting a beta particle, a gamma ray with energy of 0.197 MeV is released. Why do we use MeV for gamma rays? It's because gamma rays have extremely high energies compared to visible light or X-rays. Gamma rays typically fall in the range from about 100 keV to several MeV. Understanding gamma ray energies helps in various applications:

• Medical applications like cancer radiation therapy, where high-energy gamma rays target cancer cells.
• Understanding cosmic events as gamma rays are also emitted by objects such as neutron stars and supernovae.

Next time you see energy expressed in MeV, you'll know it's measuring a very high-energy phenomenon, such as the powerful gamma rays from nuclear reactions.

Nuclear Physics

Nuclear physics is a fascinating field that studies the components and behavior of atomic nuclei. It covers phenomena such as radioactivity, nuclear decay, and the role of energy in nuclear reactions. At its core, nuclear physics seeks to understand how nuclei are composed, how they interact, and how energy is released in processes like fission and fusion. Key Elements of Nuclear Physics:

• Nuclear Decay: In processes like beta decay, an unstable atom emits particles to reach a more stable configuration. This can result in the emission of gamma rays, carrying energy initially stored in the nucleus.
• Energy Conversion: Nuclear physics frequently deals with converting between energy units, such as from MeV to J, as these are standard in describing many nuclear events.
• Applications: The knowledge garnered from nuclear physics research underpins the development of power generation through nuclear reactors, our understanding of the universe, and technological advancements in various fields including medicine and security.

By studying nuclear physics, we gain insights not only into the smallest nuclear particles but also into the largest structures in the cosmos, bridging the gap between the micro and macro scales of the universe.