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 and \right. \(\left.1 \mathrm{eV}=1.6022 \times 10^{-19} \mathrm{J} .\right)\) What are the wavelength and frequency of a \(\gamma\) -ray photon with an energy of 1.173 MeV?

Step-by-Step Solution

Answer

The frequency is approximately \(2.834 \times 10^{20}\) Hz and the wavelength is \(1.058 \times 10^{-12}\) m.

Step 1: Convert Energy to Joules

First, convert the energy of the gamma-ray photon from MeV to Joules. Given that \(1 \text{ MeV} = 10^6 \text{ eV}\) and \(1 \text{ eV} = 1.6022 \times 10^{-19} \text{ J}\), the energy in Joules is computed as follows:\[1.173 \text{ MeV} = 1.173 \times 10^6 \times 1.6022 \times 10^{-19} \text{ J} = 1.87838 \times 10^{-13} \text{ J}\]

Step 2: Calculate Frequency

To find the frequency \(f\), use the formula that relates energy (\(E\)), frequency, and Planck's constant \(h\), where \(E = h \cdot f\) and \(h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}\). Solving for \(f\):\[f = \frac{E}{h} = \frac{1.87838 \times 10^{-13}}{6.626 \times 10^{-34}} = 2.834 \times 10^{20} \text{ Hz}\]

Step 3: Calculate Wavelength

Next, calculate the wavelength \(\lambda\) using the speed of light \(c\) and the formula \(c = \lambda \cdot f\). The speed of light \(c\) is approximately \(3 \times 10^8 \text{ m/s}\). Solving for \(\lambda\):\[\lambda = \frac{c}{f} = \frac{3 \times 10^8}{2.834 \times 10^{20}} \approx 1.058 \times 10^{-12} \text{ m}\]

Key Concepts

Cobalt-60

Cobalt-60 is a radioactive isotope widely used in medical applications, particularly in the treatment of cancer. It is a synthetic element, meaning it is not found naturally, but is produced through the irradiation of cobalt-59. Cobalt-60 is useful due to its emission of gamma rays, powerful forms of electromagnetic radiation capable of penetrating tissues. This ability makes Cobalt-60 valuable in radiotherapy, where it helps to target and destroy cancer cells, thereby offering a non-surgical treatment option. Cobalt-60 decays into nickel-60 and emits two potent gamma rays with energies of 1.173 MeV and 1.332 MeV. This transition releases energy as gamma-ray photons, which are then harnessed for both medical treatment and sterilization of medical equipment.

Wavelength Calculation

To determine the wavelength of a gamma-ray photon, understanding the relationship between energy, frequency, and wavelength is crucial. Wavelength (\( \lambda \)) is inversely proportional to frequency (\( f \)), which is expressed mathematically via the speed of light formula: \[\lambda = \frac{c}{f}\]where \( c \) is the speed of light, approximately \( 3 \times 10^8 \) meters per second. After calculating the frequency, the wavelength can be easily found by dividing the speed of light by the frequency.This fundamental concept helps us understand how electromagnetic waves behave, and in this case, allows scientists to accurately quantify the properties of gamma rays emitted by Cobalt-60, enhancing its application in medicine and industry.

Frequency Calculation

To find the frequency of a gamma-ray photon, one must first know its energy. Using Planck's equation, energy (\( E \)) is related to frequency (\( f \)) by:\[E = h \times f\]where \( h \)is Planck's constant (\( 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \)). To solve for frequency:\[f = \frac{E}{h}\]This equation is essential in quantum physics and helps describe the particle-like behavior of electromagnetic waves. In converting the energy of the gamma-ray photon from Cobalt-60 to Joules, the frequency can then be computed, providing a better understanding of the photon's properties and enabling precise adjustments in radiation therapies.

Energy Conversion to Joules

Converting energy from electron-volts (eV) to Joules is crucial for precise scientific calculations. This conversion often applies to photons emitted by radioactive isotopes like Cobalt-60, where energy is initially given in mega-electron volts (MeV). The conversion process involves the following relationships:

• 1 MeV = \(10^6\) eV
• 1 eV = \(1.6022 \times 10^{-19}\) Joules

By multiplying the given energy in MeV by these constants, the energy in Joules is obtained:\[1.173 \text{ MeV} = 1.173 \times 10^6 \times 1.6022 \times 10^{-19} \text{ J}\]Understanding this conversion is essential when performing calculations that integrate with Planck's relation or when determining the properties of electromagnetic radiation.

Planck's Constant

Planck's constant (\( h \)) is a fundamental physical constant that plays a vital role in quantum mechanics, symbolizing the quantization of energy. Its value is \( 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \).Planck's constant helps relate energy to frequency through the equation:\[E = h \times f\]This equation suggests that energy change is proportional to frequency, with \( h \) as the proportionality constant. Understanding Planck's constant is crucial in explaining phenomena at the atomic and subatomic levels, such as photon absorption and emission, and plays a significant role in the study of radioactive isotopes such as Cobalt-60.In the context of gamma rays, Planck's constant is essential to calculating the frequency, thereby revealing more about the photonic characteristics, which help guide their practical applications.