Problem 63 - Chemistry and Chemical Reactivity

Question

Cobalt-60 is a radioactive isotope used in medicine for the treatment of certain cancers. It produces \(\beta\) particles and \(\gamma\) rays, the latter having energies of 1.173 and \(1.332 \mathrm{MeV} .\left(1 \mathrm{MeV}=10^{6}\) electron-volts \right. and \(\left.1 \mathrm{eV}=1.6022 \times 10^{-19} \mathrm{J} .\right)\) What are the wavelength and frequency of a \(\gamma\) -ray photon with an energy of \(1.173 \mathrm{MeV} ?\)

Step-by-Step Solution

Answer

The wavelength is \(1.0584 \times 10^{-12}\) meters and the frequency is \(2.835 \times 10^{20}\) Hz.

Step 1: Convert Energy to Joules

First, convert the energy from MeV to Joules. Given that 1 MeV equals \(10^{6}\) electron-volts (eV) and 1 eV is \(1.6022 \times 10^{-19} \text{ J}\), we can calculate the energy in Joules: \[1.173 \text{ MeV} = 1.173 \times 10^{6} \text{ eV = 1.173} \times 10^{6} \times 1.6022 \times 10^{-19} \text{ J} = 1.879 \times 10^{-13} \text{ J}\].

Step 2: Calculate Wavelength using Energy

Use the energy-wavelength relationship for photons: \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \) is Planck's constant and \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light. Rearrange to find the wavelength: \[ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1.879 \times 10^{-13}} = 1.0584 \times 10^{-12} \text{ meters}.\]

Step 3: Calculate Frequency using Energy

Use the energy-frequency relationship: \( E = h u \), where \( u \) is the frequency. Rearrange to solve for \( u \): \[ u = \frac{E}{h} = \frac{1.879 \times 10^{-13}}{6.626 \times 10^{-34}} = 2.835 \times 10^{20} \text{ Hz}.\]

Key Concepts

Radioactive Isotopes in Medicine

Radioactive isotopes, like Cobalt-60, play a crucial role in medicine. Cobalt-60 is especially known for its application in radiotherapy to treat certain types of cancer. This isotope is a source of gamma rays and beta particles.
The gamma rays from Cobalt-60 have high energies, which makes them effective in targeting and destroying cancer cells.
Cobalt-60's ability to emit these high-energy photons helps medical professionals focus radiation doses on tumor sites while minimizing damage to surrounding tissues.
These qualities make radioactive isotopes essential in modern medical treatments, helping to save lives and improve the quality of healthcare.

Energy-Wavelength Relationship

The energy-wavelength relationship in photons is one of the fundamental principles in physics. It relates the energy of a photon to its wavelength through the equation: \[ E = \frac{hc}{\lambda} \]
Here, \(E\) represents the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. This equation tells us that higher energy photons have shorter wavelengths.

Understanding this relationship is crucial when working with gamma rays, as it allows us to calculate one property if the others are known.

• Gamma rays, with high energy, have correspondingly short wavelengths.
• Adjusting the wavelength can change how the electromagnetic waves interact with matter, making this equation vital for applications in medicine and science.

Planck's Constant

Planck's constant is a fundamental constant in physics, denoted by \(h\), and has the approximate value of \(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\).
It plays a significant role in quantum mechanics, serving as the scale at which energy quanta are quantified.
In the context of photons, Planck's constant helps define the energy and frequency of electromagnetic radiation. For example, in the equation \(E = h u\), it shows how energy is proportional to frequency and how this proportionality is governed by Planck's constant.

This constant is key to understanding phenomena at the quantum level, shaping our understanding of light and matter interactions.

Speed of Light

The speed of light, denoted by \(c\), is a core concept in physics and is approximately \(3.00 \times 10^8 \text{ meters per second}\).
It's not just a basic characteristic of light but a universal constant that holds a unique place in our understanding of the universe.

In calculations involving photons, the speed of light appears in the energy-wavelength and energy-frequency relationships.

• In the equation \(E = \frac{hc}{\lambda}\), \(c\) is used to relate energy and wavelength.
• Through these relationships, \(c\) helps describe how electromagnetic waves propagate.

The constancy of the speed of light is pivotal in the theories of relativity and underpins much of modern physics.

MeV to Joules Conversion

MeV, or Mega-electronvolts, is a unit of energy often used in particle physics. To work with this energy unit in practical calculations, we need to convert it to Joules, the SI unit of energy.
The conversion factor here is: \[ 1 \text{ MeV} = 1 \times 10^6 \text{ eV} \] \[ 1 \text{ eV} = 1.6022 \times 10^{-19} \text{ Joules} \]
So, to convert from MeV to Joules, multiply the energy in MeV by \(1.6022 \times 10^{-13}\).

This conversion is crucial when calculating real-world applications of particle energy, such as determining the photon energy in treatments involving gamma rays or in high-energy physics experiments. Converting effectively facilitates the integration of different measurement systems, improving our understanding and application of particle energy.