Chapter 19: Problem 86
Cobalt- 60 is an isotope used in diagnostic medicine and cancer treatment. It decays with \(\gamma\) ray emission. Calculate the wavelength of the radiation in nanometers if the energy of the \(\gamma\) ray is \(2.4 \times 10^{-13} \mathrm{~J} /\) photon.
Short Answer
Expert verified
The wavelength of the radiation is approximately 0.83 nanometers.
Step by step solution
01
Identify Given Variables
The problem provides us with the energy of the gamma ray which is \(2.4 \times 10^{-13}\) Joules per photon.
02
Formula Recall
Use the relation between the energy of a photon and its wavelength from the Photoelectric Effect. According to Planck's quantum theory, the energy (E) of a photon is given by the equation: \(E = hf\), where h is Planck's constant (\(6.63 \times 10^{-34} Js\)) and f is the frequency of the light. We can relate the frequency to the wavelength (\(\lambda\)) through the speed of light (c): \(c = f\lambda\). Hence the equation can be rewritten as: \(E = \frac{hc}{\lambda}\). Solving for \(\lambda\) we get: \(\lambda = \frac{hc}{E}\)
03
Substituting Values
Substitute the given energy into the equation, with \(h = 6.63 \times 10^{-34} Js\), \(c = 3.00 \times 10^{8} ms^{-1}\), and \(E = 2.4 \times 10^{-13} Joules\).
04
Calculating for \(\lambda\)
Perform the calculation for \(\lambda\). This will give the wavelength in meters.
05
Unit Conversion
Convert meters to nanometers by multiplying the result in meters by \(1 \times 10^{9}\) (since 1 meter is equal to \(1 \times 10^{9}\) nanometers).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cobalt-60 in Medicine
Cobalt-60, an isotope of the element cobalt, plays a critical role in the medical field, especially when it comes to cancer treatment and diagnostics. It is a radioactive substance that emits gamma rays during its decay process. These gamma rays have the ability to penetrate deeply into the body, making them effective in targeting and destroying cancer cells in radiotherapy. In addition, Cobalt-60 is utilized in medical imaging to help diagnose health conditions, as the gamma rays it emits provide detailed images of the internal body structure.
Furthermore, because of its predictable decay rate and potent radiation, Cobalt-60 is valuable in sterilizing medical equipment. By exposing instruments to the gamma rays, any existing bacteria or viruses are eradicated, ensuring that medical tools are safe for use in surgical procedures or other medical applications.
Furthermore, because of its predictable decay rate and potent radiation, Cobalt-60 is valuable in sterilizing medical equipment. By exposing instruments to the gamma rays, any existing bacteria or viruses are eradicated, ensuring that medical tools are safe for use in surgical procedures or other medical applications.
Planck's Constant
Planck's constant is a fundamental constant in the field of physics, identified with the symbol 'h'. It is central to quantum theory and was first discovered by Max Planck, a pioneering scientist of quantum mechanics. Planck's constant has the value of approximately \(6.63 \times 10^{-34} \) Joule-seconds (Js). It serves as a bridge between the macroscopic and quantum worlds, linking energy and frequency of electromagnetic waves.
The importance of Planck's constant lies in its role in determining the energy of a photon, which is the quantum of electromagnetic radiation. This constant allows us to quantify the discrete nature of energy at the quantum scale, a discovery that contributed to significant shifts in our understanding of physics and eventually led to the development of several modern technologies, such as lasers and semiconductors.
The importance of Planck's constant lies in its role in determining the energy of a photon, which is the quantum of electromagnetic radiation. This constant allows us to quantify the discrete nature of energy at the quantum scale, a discovery that contributed to significant shifts in our understanding of physics and eventually led to the development of several modern technologies, such as lasers and semiconductors.
Photoelectric Effect
The photoelectric effect is a phenomenon where electrons are ejected from the surface of a material when light is shone upon it. It was Albert Einstein who explained this effect by proposing that light, or electromagnetic radiation, can be thought of as being made up of particles called photons. Each photon carries a quantum of energy that is proportional to its frequency, a relationship described by Planck's equation \(E = hf\), where 'E' is the energy of the photon, 'h' is Planck's constant, and 'f' is the frequency.
This effect provided one of the first pieces of evidence for quantum theory, showing that energy is transferred in discrete amounts, contradicting the classical wave theory of light which assumed energy transfer could be continuous. The understanding of the photoelectric effect is not only vital for fundamental physics but also has practical implications, such as in the development of photovoltaic cells which convert light into electricity.
This effect provided one of the first pieces of evidence for quantum theory, showing that energy is transferred in discrete amounts, contradicting the classical wave theory of light which assumed energy transfer could be continuous. The understanding of the photoelectric effect is not only vital for fundamental physics but also has practical implications, such as in the development of photovoltaic cells which convert light into electricity.
Energy of a Photon
The energy of a photon is a concept at the heart of quantum mechanics, often symbolizing the marriage between energy and electromagnetism in quantum terms. According to Planck's equation, the energy of a single photon is directly proportional to its frequency and is given by the formula \(E = hf\), with 'E' representing the energy, 'h' being Planck's constant, and 'f' the frequency of the corresponding electromagnetic wave.
In practical applications like medical imaging or radiation therapy using Cobalt-60, understanding the energy of photons allows us to calculate the wavelength of gamma rays emitted, which in turn helps in tailoring treatments for patients or producing clear diagnostic images. Since the energy of gamma rays is significantly higher than that of visible light, the resulting wavelengths are notably shorter, positioning gamma rays at the extreme end of the electromagnetic spectrum with powerful implications for medical technology.
In practical applications like medical imaging or radiation therapy using Cobalt-60, understanding the energy of photons allows us to calculate the wavelength of gamma rays emitted, which in turn helps in tailoring treatments for patients or producing clear diagnostic images. Since the energy of gamma rays is significantly higher than that of visible light, the resulting wavelengths are notably shorter, positioning gamma rays at the extreme end of the electromagnetic spectrum with powerful implications for medical technology.