Question

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 \(2.4 \times 10^{-13} \mathrm{~J} / \mathrm{photon} .\)

Short answer

The wavelength of the radiation is 0.828 nm.

Step-by-step solution

Understanding the Relationship Between Energy and Wavelength

The problem requires calculating the wavelength of radiation given the energy of each photon. We know that energy \(E\) and wavelength \(\lambda\) are related by the following equation: \[ E = \frac{hc}{\lambda} \]where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J·s}\), and \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\).

Rearranging the Formula

To find the wavelength \(\lambda\), we rearrange the equation to solve for \(\lambda\):\[ \lambda = \frac{hc}{E} \]

Inserting the Known Values

Insert the values of Planck’s constant \(h\), the speed of light \(c\), and the energy \(E\) into the equation:\[ \lambda = \frac{6.626 \times 10^{-34} \text{ J·s} \times 3.00 \times 10^8 \text{ m/s}}{2.4 \times 10^{-13} \text{ J}} \]

Performing the Calculation

Calculate the wavelength \(\lambda\) using the values inserted:\[ \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{2.4 \times 10^{-13}} \approx 8.28 \times 10^{-13} \text{ meters} \]

Converting Meters to Nanometers

Finally, convert the wavelength from meters to nanometers. Since \(1 \text{ nm} = 10^{-9} \text{ m}\), the conversion is:\[ \lambda = 8.28 \times 10^{-13} \text{ m} \times \frac{1 \text{ nm}}{10^{-9} \text{ m}} = 0.828 \text{ nm} \]

Key concepts

Photon Energy

Photon energy refers to the energy carried by a single photon, the smallest particle of light and electromagnetic radiation. The energy of a photon is directly related to its frequency, and inversely related to its wavelength. This relationship is expressed in the formula:\[ E = \frac{hc}{\lambda} \]Where:- \( E \) is the photon energy.- \( h \) is Planck's constant.- \( c \) is the speed of light.- \( \lambda \) is the wavelength of the photon.Understanding photon energy is crucial when dealing with processes involving radiation, such as medical treatments with radioactive isotopes like cobalt-60. Whenever a gamma ray is emitted, it represents a photon with a specific amount of energy that can influence materials and living tissues. This is why calculating the photon energy can help us predict and study its effects.

Planck's Constant

Planck's constant \( h \) is a fundamental physical constant that plays a central role in quantum mechanics. Its value is approximately \( 6.626 \times 10^{-34} \text{ J·s} \). Planck's constant links the energy of a photon to its frequency and is an essential component of the equation \( E = hf \), where:- \( E \) is the energy of a photon.- \( f \) is the frequency.In many calculations involving electromagnetic waves, especially on the quantum scale, Planck's constant provides the bridge between classical physics and quantum physics. When we use this constant in calculations like those involving gamma-ray wavelengths from cobalt-60 decay, it helps us determine how the energy is transferred at very small scales. Understanding such interactions can contribute significantly to fields like quantum physics, chemistry, and astronomy.

Speed of Light

The speed of light \( c \) is a fundamental constant that represents the maximum speed at which all energy, matter, and information in the universe can travel. Its value is \( 3.00 \times 10^8 \text{ m/s} \).Light speed plays a critical role in many areas of physics and is a key factor in the formula \( \lambda = \frac{hc}{E} \), which allows us to compute the wavelength of photons when their energy is known. Since gamma rays are a form of electromagnetic radiation, they travel at this speed. Knowing this constant helps us understand how radiation such as gamma rays propagates through different media and its implications in experiments and technologies, from radios to medical imaging technologies.

Cobalt-60 Decay

Cobalt-60 decay is a type of radioactive decay where the isotope cobalt-60 emits gamma rays. Gamma rays are very high-frequency electromagnetic waves and are one of the by-products of this decay process.Cobalt-60 is widely used in medical treatments, particularly in cancer therapy, because the gamma rays can effectively target and destroy cancerous cells. Understanding gamma ray emissions from cobalt-60 involves considering the energy of the released photons and their impact, which can be computed using the formula \( \lambda = \frac{hc}{E} \).Calculating the wavelength (or energy) of these emissions assists in devising safer and more effective therapeutic strategies. Recognizing the properties and behavior of gamma rays from cobalt-60 decay also aids in the precise delivery of radiation therapy to maximize its therapeutic benefits while minimizing risks.