Answer
Hint: The relationship between energy and wavelength or frequency can be well understood if you think in terms of a photon. We know that a photon is a discrete unit of quantized energy. Think about where the photon gets its energy from by employing Planck’s quantum theory. Given that photons are emitted in the range of the electromagnetic spectrum, i.e., radiation with increasing frequency and decreasing wavelength or vice versa,
determine the correlation between the two and
energy.
Formula Used:
$c=\nu\lambda$
$E=h\nu$
$E=\dfrac{hc}{\lambda}$
Complete Solution:
Let us begin by looking at the question in the context of photons.
Photons are elementary particles that are essentially quantized electromagnetic radiation. This quantization implies that they hold packets of energy. The energy that they carry depends upon the kind of electromagnetic radiation that they quantized. This quantization of energy is consistent with Planck’s quantum theory which states that
Energy is absorbed or radiated by atoms in discrete packets of energy called quanta, which are now known to be photons.
Each quantum, or photon consists of a specific amount of energy that it can absorb or emit, which is directly proportional to the frequency of radiation.
We know that the electromagnetic spectrum consists of electromagnetic radiation of varying frequencies and wavelengths. Therefore, according to Planck’s quantum
theory we have:
$E \propto \nu \Rightarrow E = h\nu$, where E is the photon energy,
h ($6.626 \times 10^{-34}\;J.s$) is the proportionality (Planck’s) constant and $\nu$ is the frequency of radiation.
Now, all electromagnetic radiation travels in the form of photons at the speed of light $c \approx 3 \times 10^8\;ms^{-1}$.
If $\lambda$ is the wavelength of any radiation, then,
$c=\nu \times \lambda \Rightarrow \nu = \dfrac{c}{\lambda}$
Plugging this back into our energy
expression in place of $\nu$, we get:
$E = h\dfrac{c}{\lambda}$
In essence, energy is directly proportional to the frequency of radiation but is inversely proportional to the wavelength. This means that energy increases with an increase in frequency or decrease in wavelength, and energy decreases with a decrease in frequency or an increase in wavelength.
Note:
Remember that Planck’s quantum theory, contrary to Maxwell's electromagnetic wave theory, suggests that radiant
energy is not absorbed or emitted continuously but discontinuously in the form of small packets of energy called photons. Planck’s quantum theory was able to explain phenomena like the blackbody spectrum and photoelectric effect where Maxwell’s electromagnetic wave theory failed to do so since Maxwell’s theory entailed a continuous energy emission/absorption distribution.
Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy.
Photon energy can be expressed using any unit of energy. Among the units commonly used to denote photon energy are the electronvolt (eV) and the joule (as well as its multiples, such as the microjoule). As one joule equals 6.24 × 1018 eV, the larger units may be more useful in denoting the energy of photons with higher frequency and higher energy, such as gamma rays, as opposed to lower energy photons as in the optical and radio frequency regions of the electromagnetic spectrum.
Formulas[edit]
Physics[edit]
Photon energy is directly proportional to frequency.[1]
whereThis equation is known as the Planck–Einstein relation.
Additionally,
where- E is photon energy
- λ is the photon's wavelength
- c is the speed of light in vacuum
- h is the Planck constant
The photon energy at 1 Hz is equal to 6.62607015 × 10−34 J
That is equal to 4.135667697 × 10−15 eV
Electronvolt[edit]
Energy is often measured in electronvolts.
To find the photon energy in electronvolts using the wavelength in micrometres, the equation is approximately
This equation only holds if the wavelength is measured in micrometers.
The photon energy at 1 μm wavelength, the wavelength of near infrared radiation, is approximately 1.2398 eV.
In chemistry, quantum physics and optical engineering[edit]
See [2]
where
- E is photon energy (joules),
- h is the Planck constant
- The Greek letter ν (nu) is the photon's frequency.
Examples[edit]
An FM radio station transmitting at 100 MHz emits photons with an energy of about 4.1357 × 10−7 eV. This minuscule amount of energy is approximately 8 × 10−13 times the electron's mass (via mass-energy equivalence).
Very-high-energy gamma rays have photon energies of 100 GeV to over 1 PeV (1011 to 1015 electronvolts) or 16 nanojoules to 160 microjoules.[3] This corresponds to frequencies of 2.42 × 1025 to 2.42 × 1029 Hz.
During photosynthesis, specific chlorophyll molecules absorb red-light photons at a wavelength of 700 nm in the photosystem I, corresponding to an energy of each photon of ≈ 2 eV ≈ 3 × 10−19 J ≈ 75 kBT, where kBT denotes the thermal energy. A minimum of 48 photons is needed for the synthesis of a single glucose molecule from CO2 and water (chemical potential difference 5 × 10−18 J) with a maximal energy conversion efficiency of 35%.
See also[edit]
- Photon
- Electromagnetic radiation
- Electromagnetic spectrum
- Planck constant
- Planck–Einstein relation
- Soft photon
References[edit]
- ^ "Energy of Photon". Photovoltaic Education Network, pveducation.org.
- ^ Andrew Liddle (27 April 2015). An Introduction to Modern Cosmology. John Wiley & Sons. p. 16. ISBN 978-1-118-69025-3.
- ^ Sciences, Chinese Academy of. "Observatory discovers a dozen PeVatrons and photons exceeding 1 PeV, launches ultra-high-energy gamma astronomy era". phys.org. Retrieved 2021-11-25.