What is the effective mass of a photon having wavelength λ?
Answer
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Hint: Photon is a component of the electromagnetic wave and possesses all its properties and behaves like it. Using them we can relate it to the physical quantities and find the value of its effective mass.
Complete step-by-step answer:
A photon’s energy spectrum is line and discrete and it is the component of gamma rays.
A Photon has zero mass but since it can do work it is done in terms of kinetic energy. As it possesses kinetic energy, using Planck’s constant, speed of light and wavelength we can obtain a relation for calculating the effective mass.
The energy is in terms of kinetic energy for a photon which is given by,
$K = \dfrac{1}{2}m{v^2}$
Here, m is the mass and v is the speed of the particle,
This energy is also
$\dfrac{{ch}}{\lambda }$
Where, h is the Planck’s constant
Since c is considered as speed of light, on substitution, we get
$m = \dfrac{{2h}}{{c\lambda }}$
Hence, this is the effective mass of a photon of wavelength λ.
Note: For the wavelength of 1nm, the effective mass using the de-broglie formula
$
\lambda = \dfrac{h}{{mv}} \\
m = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{{{10}^{ - 9}} \times 3 \times {{10}^8}}} = 2.2 \times {10^{ - 33}}kg \\
$
The photons are used in the study of quantum mechanics and used in optical imaging and communication in quantum cryptography. It has applications in photochemistry, microscopy and also to measure molecular distances.
Complete step-by-step answer:
A photon’s energy spectrum is line and discrete and it is the component of gamma rays.
A Photon has zero mass but since it can do work it is done in terms of kinetic energy. As it possesses kinetic energy, using Planck’s constant, speed of light and wavelength we can obtain a relation for calculating the effective mass.
The energy is in terms of kinetic energy for a photon which is given by,
$K = \dfrac{1}{2}m{v^2}$
Here, m is the mass and v is the speed of the particle,
This energy is also
$\dfrac{{ch}}{\lambda }$
Where, h is the Planck’s constant
Since c is considered as speed of light, on substitution, we get
$m = \dfrac{{2h}}{{c\lambda }}$
Hence, this is the effective mass of a photon of wavelength λ.
Note: For the wavelength of 1nm, the effective mass using the de-broglie formula
$
\lambda = \dfrac{h}{{mv}} \\
m = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{{{10}^{ - 9}} \times 3 \times {{10}^8}}} = 2.2 \times {10^{ - 33}}kg \\
$
The photons are used in the study of quantum mechanics and used in optical imaging and communication in quantum cryptography. It has applications in photochemistry, microscopy and also to measure molecular distances.
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