Question

Consider a radio wave with a wavelength of \(10^{7} \mu \mathrm{m}\) and a \(\gamma\)-ray with a wavelength of \(10^{-7} \mu \mathrm{m}\). Determine the photon energies of the radio wave and the \(\gamma\)-ray, and the photon energy ratio of the \(\gamma\)-ray to the radio wave.

Short answer

Answer: The energy ratio of gamma-ray to radio wave is 10^13.

Step-by-step solution

Identify given values and constants

First, let's list the given values and the constants that we need in this problem: 1. Wavelength of radio wave: \(\lambda_{1} = 10^{7} \mu m\) 2. Wavelength of gamma-ray: \(\lambda_{2} = 10^{-7} \mu m\) 3. Planck's constant: \(h = 6.63 \times 10^{-34} Js\) 4. Speed of light: \(c = 3.0 \times 10^8 m/s\) Next, we will convert the given wavelengths in micrometers to meters.

Convert wavelengths to meters

1 micrometer (μm) is equal to \(10^{-6}\) meters. Therefore, we can convert the given wavelengths to meters as follows: 1. \(\lambda_{1} = 10^{7} \mu m \times 10^{-6} m/\mu m = 10 m\) 2. \(\lambda_{2} = 10^{-7} \mu m \times 10^{-6} m/\mu m = 10^{-13} m\) Now, we have the wavelengths in meters.

Calculate the photon energies

We will use Planck's equation, \(E = \frac{hc}{\lambda}\), to find the photon energies: 1. Photon energy of radio wave: \(E_{1} = \frac{h \times c}{\lambda_{1}} = \frac{6.63 \times 10^{-34} Js \times 3.0 \times 10^{8} m/s}{10 m} = 1.989 \times 10^{-24} J\) 2. Photon energy of gamma-ray: \(E_{2} = \frac{h \times c}{\lambda_{2}} = \frac{6.63 \times 10^{-34} Js \times 3.0 \times 10^{8} m/s}{10^{-13} m} = 1.989 \times 10^{-11} J\) Now, we have the photon energies of both the radio wave and gamma-ray.

Calculate the energy ratio

Finally, we will calculate the energy ratio of gamma-ray to radio wave as follows: Energy ratio = \(\frac{E_{2}}{E_{1}} = \frac{1.989 \times 10^{-11} J}{1.989 \times 10^{-24} J} = 10^{13}\) The energy ratio of gamma-ray to radio wave is \(10^{13}\).

Key concepts

Planck's Equation

Planck's equation is fundamental in the field of quantum mechanics and is given by the formula \(E = \frac{hc}{\lambda}\). In this equation, \(E\) represents the energy of a single quantum of electromagnetic radiation, or a photon; \(h\) is Planck’s constant (\(6.63 \times 10^{-34} Js\)); \(c\) denotes the speed of light in a vacuum (\(3.0 \times 10^8 m/s\)); and \(\lambda\) is the wavelength of the electromagnetic wave. This relationship tells us that the energy of a photon is inversely proportional to its wavelength: shorter wavelengths correspond to higher energy photons. It’s helpful to remember that Planck’s equation ties together the wave-like and particle-like properties of light, illustrating the dual nature of light as both a wave and a stream of particles.

Understanding Planck's equation allows us to calculate the energy of electromagnetic waves of any wavelength, from long-radio waves to short-gamma rays. The exercise we are looking into has us calculating the energy of both a radio wave and a gamma-ray. The exercise demonstrates the significant difference in energy between photons of different wavelengths.

Electromagnetic Wave Wavelength

Electromagnetic wave wavelength, denoted as \(\lambda\), is a measure of the distance between two consecutive peaks (or troughs) of a wave. Electromagnetic waves are a form of energy that travels through space at the speed of light. The wavelength determines the type of electromagnetic wave, ranging from radio waves with very long wavelengths, through visible light, to gamma rays with extremely short wavelengths. The wavelength is typically measured in meters (m), but can also be represented in other units like micrometers (\(\mu m\)), which are commonly used for smaller wavelengths.

To visualize this, imagine waves on the surface of water where each wave crest is a peak. The distance from one crest to the next is analogous to wavelength in light. In our exercise, we see a vast difference in wavelength—and thus in energy—between radio waves (\(10^{7} \mu m\)) and gamma rays (\(10^{-7} \mu m\)), reflecting the broad range of the electromagnetic spectrum.

Speed of Light

The speed of light, symbolized as \(c\), is a constant value of approximately \(3.0 \times 10^8 m/s\) in a vacuum. It's one of the most fundamental constants in physics and is pivotal in many areas, including relativity and electromagnetism. The speed of light connects space and time, and its constancy underpins Einstein’s theory of relativity. In the context of photon energy calculation, the speed of light is essential for determining how fast energy travels as electromagnetic waves through space.

In our application of Planck’s equation, the speed of light is used along with Planck’s constant to calculate the energy of a photon given a specific wavelength. The invariance of light's speed across all observers, regardless of the light source's motion, was a groundbreaking discovery that led to new insights about the nature of spacetime.

Energy of Gamma Rays

Gamma rays possess the shortest wavelengths and thus the highest energy within the electromagnetic spectrum. They are produced by the most energetic events in the universe, such as nuclear explosions, supernovae, and processes involving radioactive decay. On Earth, they are used in medical treatments for their ability to kill cancer cells and in industrial radiography to inspect heavy machinery and welds.

The energy of gamma rays can be calculated using Planck's equation, just as with any other type of electromagnetic wave. In our example, the energy of the gamma-ray was found to be \(1.989 \times 10^{-11} J\), which is many orders of magnitude greater than that of the radio wave, resulting in an enormous energy ratio of \(10^{13}\). This high energy is why gamma rays are incredibly penetrating and can be dangerous due to their potential to cause damage to living cells.