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

Short Answer

Expert verified
(a) 2.472 x 10^-19 J, 1.543 eV; (b) 804 nm, visible light spectrum.

Step by step solution

01

Energy of the Photon in Joules

The energy of a photon can be calculated using the formula \( E = pc \), where \( p \) is the momentum of the photon and \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \). Given \( p = 8.24 \times 10^{-28} \, \text{kg} \cdot \text{m/s} \), we find:\[ E = (8.24 \times 10^{-28} \text{ kg} \cdot \text{m/s})(3 \times 10^8 \text{ m/s}) = 2.472 \times 10^{-19} \text{ J} \]
02

Convert Energy from Joules to Electronvolts

1 electronvolt (eV) is equal to \( 1.602 \times 10^{-19} \) joules. To convert the energy from joules to electronvolts, use the formula:\[ E_{\text{eV}} = \frac{E_{\text{J}}}{1.602 \times 10^{-19} \text{ J/eV}} \]Substituting the value:\[ E_{\text{eV}} = \frac{2.472 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 1.543 \text{ eV} \]
03

Calculate the Wavelength of the Photon

The wavelength \( \lambda \) of a photon can be found using the formula \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \). Given \( p = 8.24 \times 10^{-28} \, \text{kg} \cdot \text{m/s} \), calculate:\[ \lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{8.24 \times 10^{-28} \text{ kg} \cdot \text{m/s}} = 8.04 \times 10^{-7} \text{ m} \]
04

Determine the Region of the Electromagnetic Spectrum

The calculated wavelength \( 8.04 \times 10^{-7} \text{ m} \) is equivalent to 804 nm, which falls within the visible region of the electromagnetic spectrum, specifically near the infrared boundary.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum and Energy Relationship
The relationship between momentum and energy for a photon is pivotal in understanding how atomic and subatomic processes work. Photons are unique because, unlike particles with mass, they have energy directly related to their momentum. This relationship is governed by the equation \( E = pc \), where \( E \) is the energy, \( p \) is the momentum, and \( c \) is the speed of light. Since a photon is massless, its energy depends entirely on its momentum and the speed of light, rather than mass and velocity as in classical mechanics. This is crucial for calculating certain phenomena in quantum mechanics.

When calculating energy from momentum, we use the constant speed of light (\( 3 \times 10^8 \, \text{m/s} \)), making computations straightforward and reflective of the photon's fundamental properties. This relationship is powerful because it allows the translation of quantum behavior into energetic terms, which are more intuitive to us.
Wavelength Calculation
Determining the wavelength of a photon is an essential part of understanding its properties and behavior. The wavelength \( \lambda \) can be calculated using the formula \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the photon's momentum.

Since photons do not possess a rest mass, their wavelength is a function of Planck’s fundamental constant \( 6.626 \times 10^{-34} \, \text{Js} \). This formula derives from the quantum nature of photons, linking a photon's momentum directly to its corresponding wavelength. Knowing a photon's wavelength can provide insights into its energy type, behavior, and the part of the electromagnetic spectrum it belongs to.

This calculation is significant because it translates a photon’s dynamic state, described by its momentum, into a spatial characteristic, its wavelength. This helps bridge concepts between particle and wave nature, a core aspect of wave-particle duality.
Electromagnetic Spectrum
The electromagnetic spectrum is a comprehensive range of all types of electromagnetic radiation. Each type corresponds to a different wavelength and energy level, placing each type in a unique position on the spectrum.

From gamma rays with the shortest wavelengths to radio waves with the longest, understanding where a photon's wavelength places it provides valuable insights into its properties and potential applications. The visible light spectrum is only a small part of this vast range but is crucial for numerous biological and technological processes.

In our case, a wavelength of 804 nm places the photon in the visible range, near the infrared region. This means the photon is just on the edge of what is perceivable to human vision, bridging visible light and heat radiation. By knowing a photon's place on the spectrum, scientists can predict its interactions with matter and practical uses, such as communications, imaging, and medical therapies.
Planck's Constant
Planck's constant is a fundamental constant of nature that plays a crucial role in quantum mechanics, specifically impacting the quantization of energy. It connects the energy of photons to their frequencies, following the relation \( E = hf \), where \( E \) is energy, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \), and \( f \) is frequency.

In our wavelength calculation context, Planck’s constant helps express the photon's energy in terms of its wavelength instead of frequency. This is vital in many fields such as optics, quantum physics, and even electrical engineering because it allows us to calculate and relate energies in different forms.

Without Planck’s constant, the link between the minute scales of photons and wider observable energy outputs would be much harder to define. This constant is an integral part of understanding modern physics, as it forms the bridge between the particle and wave descriptions of light.

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Most popular questions from this chapter

Removing birthmarks. Pulsed dye lasers emit light of wavelength \(585 \mathrm{nm}\) in \(0.45 \mathrm{~ms}\) pulses to remove skin blemishes such as birthmarks. The beam is usually focused onto a circular spot \(5.0 \mathrm{~mm}\) in diameter. Suppose that the output of one such laser is \(20.0 \mathrm{~W}\). (a) What is the energy of each photon, in eV? (b) How many photons per square millimeter are delivered to the blemish during each pulse?

(a) Calculate the maximum increase in photon wavelength that can occur during Compton scattering. (b) What is the energy (in electronvolts) of the smallest- energy X-ray photon for which Compton scattering could result in doubling the original wavelength?

(a) What is the least amount of energy, in electronvolts, that must be given to a hydrogen atom that is initially in its ground level so that it can emit the \(\mathrm{H}_{\alpha}\) line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the \(n=3\) level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

The predominant wavelength emitted by an ultraviolet lamp is \(248 \mathrm{nm} .\) If the total power emitted at this wavelength is \(12.0 \mathrm{~W}\), how many photons are emitted per second?

A specimen of the microorganism Gastropus hyptopus measures \(0.0020 \mathrm{~cm}\) in length and can swim at a speed of 2.9 times its body length per second. The tiny animal has a mass of roughly \(8.0 \times 10^{-12} \mathrm{~kg} .\) (a) Calculate the de Broglie wavelength of this organism when it is swimming at top speed. (b) Calculate the kinetic energy of the organism (in eV) when it is swimming at top speed.

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