Question

Cobalt-60 is an artificial radioisotope that is produced in a nuclear reactor and is used as a gamma-ray source in the treatment of certain types of cancer. If the wavelength of the gamma radiation from a cobalt- 60 source is \(1.00 \times 10^{-3} {nm},\) calculate the energy of a photon of this radiation.

Short answer

The energy of a photon of the gamma radiation from a cobalt-60 source is \( 1.988 \times 10^{-13} \, \text{J} \).

Step-by-step solution

- Convert Wavelength to Meters

The given wavelength is \( \lambda = 1.00 \times 10^{-3} \, \text{nm} \). Convert this wavelength to meters (m) by using the conversion factor \( 1 \, \text{nm} = 1 \times 10^{-9} \, \text{m} \). \[ \lambda = 1.00 \times 10^{-3} \times 10^{-9} = 1.00 \times 10^{-12} \, \text{m} \]

- Use Planck's Equation

Planck's equation relates the energy of a photon \( E \) to its wavelength \( \lambda \). The equation is \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \) and \( c \) is the speed of light \( (3.00 \times 10^8 \, \text{m/s}) \).

- Substitute the Values

Substitute the values for \( h \), \( c \), and \( \lambda \) into the equation: \[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.00 \times 10^8 \, \text{m/s})}{1.00 \times 10^{-12} \, \text{m}} \]

- Calculate the Energy

Perform the calculation: \[ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{1.00 \times 10^{-12}} = 1.988 \times 10^{-13} \, \text{J} \]

Key concepts

radioisotopes

Radioisotopes are atoms that have an unstable nucleus and emit radiation to attain a more stable form. They are widely used in various fields like medicine, industry, and research. Cobalt-60, a common radioisotope, is artificially produced in nuclear reactors. It’s particularly notable for its use in cancer treatments due to its emission of gamma radiation, which can effectively target and destroy cancer cells.
Understanding radioisotopes helps in fields like nuclear medicine and radiotherapy, ensuring precise and safe handling of these powerful tools.

gamma radiation

Gamma radiation is a highly penetrating type of electromagnetic radiation. It consists of high-energy photons and is produced by the radioactive decay of atomic nuclei. Unlike alpha and beta radiation, gamma rays can penetrate deeply into materials, making them useful for medical treatments like radiotherapy, as utilized with Cobalt-60 sources.
The ability to penetrate and damage cells makes gamma rays effective for targeting cancer cells, allowing for their destruction without invasive surgery. However, this same property requires careful control and shielding to protect healthy tissues and ensure safety.

Planck's equation

Planck's equation is fundamental in quantum mechanics, connecting the energy of a photon to its frequency or wavelength. The equation is:
\[ E = \frac{hc}{\lambda} \]
where:

• \( E \) is the energy of the photon
• \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) Js)
• \( c \) is the speed of light (\( 3.00 \times 10^8 \) m/s)
• \( \lambda \) is the wavelength of the radiation

By using Planck's equation, we can calculate the energy of the gamma photon emitted by Cobalt-60 when we know its wavelength, providing crucial data for applications in both medical and physical sciences.

wavelength conversion

Wavelength conversion is the process of changing the units of a wavelength to match those required for further calculations. In this exercise, we converted the wavelength of gamma radiation from nanometers to meters to use in Planck's equation.
To convert nanometers (nm) to meters (m), remember that:
\[ 1 \, \text{nm} = 1 \times 10^{-9} \, \text{m} \]
For example, converting a wavelength of \( 1.00 \times 10^{-3} \) nm to meters involves:
\[ 1.00 \times 10^{-3} \, \text{nm} \times 10^{-9} = 1.00 \times 10^{-12} \, \text{m} \]
This conversion is vital for correct calculations in many scientific contexts where standard SI units are required for consistency and accuracy.