Question

A photon has momentum of magnitude 8.24 \(\times\) 10\(^{-28}\) kg \(\bullet\) m/s. (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

Short answer

(a) 2.472 × 10⁻¹⁹ J, 1.543 eV. (b) Wavelength: 804 nm, Infrared region.

Step-by-step solution

Recall the Energy-Momentum Relationship

The energy of a photon can be determined using its momentum. We use the equation \( E = pc \), where \( E \) is the energy, \( p \) is the momentum, and \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s).

Calculate the Energy in Joules

Substitute \( p = 8.24 \times 10^{-28} \) kg·m/s and \( c = 3 \times 10^8 \) m/s into the equation:\[E = (8.24 \times 10^{-28}) \times (3 \times 10^8)\]This gives:\[E = 2.472 \times 10^{-19} \text{ J}\]

Convert Energy from Joules to Electron Volts

Use the conversion factor where 1 eV = \( 1.602 \times 10^{-19} \) J. Thus, the energy in eV is:\[E = \frac{2.472 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}}\]This gives:\[E = 1.543 \, eV\]

Use the De Broglie Wavelength Formula

The wavelength \( \lambda \) of a photon is determined by the relationship \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J·s}\).Substitute the values into the equation:\[\lambda = \frac{6.626 \times 10^{-34}}{8.24 \times 10^{-28}}\]This gives:\[\lambda = 8.04 \times 10^{-7} \, \text{m}\]

Determine the Region of the Electromagnetic Spectrum

The calculated wavelength \(8.04 \times 10^{-7} \, \text{m}\) corresponds to \(804 \, \text{nm}\), which falls within the infrared region of the electromagnetic spectrum (700 nm to 1 mm).

Key concepts

Momentum-Energy Relationship

The connection between momentum and energy is pivotal when examining photons. Photons, even though they are massless, possess momentum. This stems from their intrinsic energy as described by the equation \( E = pc \). Here, \( E \) stands for energy, \( p \) is the momentum, and \( c \), the speed of light, is approximately \( 3 \times 10^8 \) meters per second.

• This equation shows that a photon's energy is directly proportional to its momentum.
• Understanding this relationship helps in calculating the energy when the momentum is known and vice versa.

In this exercise, the momentum of the photon was given as \( 8.24 \times 10^{-28} \) kg·m/s. By using the equation \( E = pc \), we can determine the energy of the photon in both joules and electron volts. This transformation helps in understanding the photon's interaction with other particles or fields.
Moreover, this relation underlines a key aspect of quantum mechanics, where particles like photons defy some classical intuitions about mass and momentum because they are energy-packets traveling through space.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, sorted by wavelength or frequency. Radiation ranges from gamma rays with the shortest wavelengths to radio waves, which have the longest.

• Visible light is just one small part of this spectrum, typically from 380 nm to 750 nm.
• Beyond visible light, the spectrum includes ultraviolet, infrared, microwaves, and radio waves.

Both wavelength and frequency are critical for identifying the type of radiation. The energy and properties of radiation depend on its position in the spectrum.
For instance, the calculated wavelength of this photon's electromagnetic radiation was \( 804 \text{ nm} \), placing it in the infrared region. The infrared region typically involves wavelengths from 700 nm to 1 mm and is mostly associated with heat radiation. These infrared waves are longer than visible light, meaning they have lower energy, which is paramount in applications such as thermal imaging and remote controls.

De Broglie Wavelength

The concept of the De Broglie wavelength bridges the gap between classical and quantum physics, proposing that particles exhibit wave-like properties. Louis de Broglie suggested that every particle, including photons, has an associated wavelength \( \lambda \), calculated by \( \lambda = \frac{h}{p} \). Here, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J·s} \) and \( p \) is the momentum.

• For photons, this concept emphasizes their dual nature as both particles and waves.
• Calculating the wavelength provides insight into how these particles interact with matter.

In practical applications, the De Broglie wavelength helps explain patterns in phenomena like diffraction and interference, where even particles traditionally seen as solid show wave-like behavior.
For the given photon momentum of \( 8.24 \times 10^{-28} \) kg·m/s, the De Broglie wavelength calculation resulted in \( 8.04 \times 10^{-7} \text{ m} \). Knowing the wavelength allows us to discern which region of the electromagnetic spectrum the photon belongs to, thus giving insight into its energy and potential uses in technology or natural sciences.