Energy and momentum of a photon
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What is a photon?
A photon is a quantum of electromagnetic radiation. It's a discrete packet of energy exhibiting wave-particle duality; it has zero mass and travels at the speed of light in a vacuum.
State the equation that relates the energy of a photon to its frequency.
E = hf, where E is the energy of the photon, h is Planck's constant (6.63 x 10⁻³⁴ Js), and f is the frequency of the electromagnetic radiation.
What is the electronvolt (eV) and how is it related to the joule (J)?
An electronvolt (eV) is the amount of energy gained (or lost) by a single electron when it passes through a potential difference of 1 volt. 1 eV = 1.60 x 10⁻¹⁹ J.
How is the momentum of a photon related to its energy and the speed of light?
The momentum (p) of a photon is given by p = E / c, where E is the energy of the photon and c is the speed of light in a vacuum (approximately 3.0 x 10⁸ m/s).
Explain how electromagnetic radiation exhibits particulate nature.
Electromagnetic radiation, while exhibiting wave-like behavior, also comes in discrete packets of energy (photons). This is demonstrated in phenomena like the photoelectric effect, where electrons are emitted when light of sufficient frequency shines on a metal surface, behaving as discrete particles delivering energy.
A photon has an energy of 4.14 eV. Calculate its momentum.
First convert eV to Joules: 4.14 eV * 1.60 x 10⁻¹⁹ J/eV = 6.624 x 10⁻¹⁹ J. Then, using p = E/c, p = (6.624 x 10⁻¹⁹ J) / (3.0 x 10⁸ m/s) = 2.21 x 10⁻²⁷ kg m/s.
A photon has a wavelength of 500 nm. Calculate its energy in electronvolts.
First, find the frequency using c = fλ: f = c/λ = (3.0 x 10⁸ m/s) / (500 x 10⁻⁹ m) = 6 x 10¹⁴ Hz. Then, E = hf = (6.63 x 10⁻³⁴ Js) * (6 x 10¹⁴ Hz) = 3.978 x 10⁻¹⁹ J. Finally, convert to eV: (3.978 x 10⁻¹⁹ J) / (1.60 x 10⁻¹⁹ J/eV) = 2.49 eV.
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