Problem 138 - Chemistry

Question

The radioactive \(\mathrm{Co}-60\) isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequency of an emitted gamma photon having the energy of \(1.29 \times 10^{11} \mathrm{~J} / \mathrm{mol} .\)

Step-by-Step Solution

Answer

The frequency of the emitted gamma photon is approximately \(3.23 \times 10^{20}\) Hz and the wavelength is approximately \(9.28 \times 10^{-14}\) m.

Step 1: Convert given energy from Joules/mol to Joules/photon

Given energy of gamma photon is \(1.29 \times 10^{11}\) J/mol. To convert from J/mol to J/photon, we use the Avogadro's number which is \(6.022 \times 10^{23}\) photons/mol. So, the energy per photon would be \( \frac{1.29 \times 10^{11}}{6.022 \times 10^{23}} \) J/photon, which is approximately \(2.14 \times 10^{-13}\) J/photon.

Step 2: Calculate frequency using Planck-Einstein relation

Using Planck's constant \(h = 6.626 \times 10^{-34}\) Js, we can find the frequency using Planck-Einstein relation \(E = hv\). Therefore, frequency \(v = \frac{E}{h} = \frac{2.14 \times 10^{-13}}{6.626 \times 10^{-34}}\) s^-1, which calculates to approximately \(3.23 \times 10^{20}\) Hz.

Step 3: Calculate wavelength using wave equation

Finally, using speed of light \(c = 299792458\) m/s, we can find the wavelength using the wave equation \(v = c/λ\). Therefore, wavelength \(λ = \frac{c}{v} = \frac{299792458}{3.23 \times 10^{20}}\) m , which calculates to approximately \(9.28 \times 10^{-14}\) m.

Key Concepts

Frequency Calculation Using Planck's Constant

To find the frequency of an emitted gamma photon, we can utilize Planck's constant in a formula known as the Planck-Einstein relation: \( E = h u \). In this equation, \( E \) is the energy of a single photon, \( h \) is Planck's constant, and \( u \) represents the frequency of the photon. To determine the frequency, we rearrange the equation to solve for \( u \): \( u = \frac{E}{h} \).

- **Given Energy:** Start with the energy per photon, \(2.14 \times 10^{-13}\) J/photon (calculated from the given problem's energy of \(1.29 \times 10^{11}\) J/mol by using Avogadro's number).
- **Planck's Constant:** This value is \(6.626 \times 10^{-34}\) Js.
- **Calculating Frequency:** Plugging these numbers into the formula gives us \( u = \frac{2.14 \times 10^{-13}}{6.626 \times 10^{-34}} \) Hz, resulting in approximately \(3.23 \times 10^{20}\) Hz.

This frequency indicates how many wavelengths fit into one second, which is important in understanding the energetic impact of gamma photons.

Wavelength Calculation Using Wave Equation

Gamma photons travel at the speed of light, and their wavelength can be calculated using the wave equation. The wave equation is expressed as \( u = \frac{c}{\lambda} \), where \( u \) is the frequency we calculated earlier, \( c \) is the speed of light, and \( \lambda \) is the wavelength.

- **Frequency of Photon:** We calculated this as \(3.23 \times 10^{20}\) Hz in the previous step.
- **Speed of Light:** The constant \(c\) is approximately \(299,792,458\) m/s.
- **Calculating Wavelength:** Rearrange the equation to \( \lambda = \frac{c}{u} \) and substitute the known values: \( \lambda = \frac{299,792,458}{3.23 \times 10^{20}} \) m, which results in a wavelength of approximately \(9.28 \times 10^{-14}\) m.

Gamma rays have very short wavelengths, emphasizing their high energy levels, able to penetrate deeply and interact with matter, including biological tissues.

Radioactive Isotopes in Nuclear Medicine

Radioactive isotopes play an essential role in nuclear medicine, primarily for diagnostic and therapeutic purposes. Specific isotopes such as Cobalt-60, the one discussed in this exercise, are used in treating cancer by emitting gamma photons.

- **Cobalt-60 Use:** In medical treatments, Cobalt-60 isotopes release high energy gamma rays. These rays are effective in killing or damaging cancerous cells while minimizing exposure to healthy tissues.
- **Role of Gamma Photons:** Gamma photons used in these treatments are of specific frequencies and wavelengths, allowing targeted therapy for tumors.
- **Benefits:** Nuclear medicine enables precise treatment, less invasiveness compared to surgical options, and often leads to better outcomes with fewer side effects.

Thus, understanding the energy, frequency, and wavelength of gamma photons is crucial in optimizing the medical uses of isotopes like Cobalt-60. These treatments harness the penetrating power of gamma rays to ensure effective cancer therapy.