Question

How do you calculate the energy, in eV, of a photon of light of wavelength \[490nm\]?

Answer 1

Hint:The photon energy is the energy possessed by a photon in an electromagnetic field. Photons are elementary particles which are considered as massless and are moving with a speed of light in vacuum.

Complete step by step answer:The photon energy is defined as the energy carried by the photon which is moving constantly in an electromagnetic field. The energy of a photon is directly proportional to the electromagnetic frequency of the photon.
In this regard, Max Planck, a German physicist, gave an expression relating the energy and the frequency of a photon. The equation is
$E = h\nu $, where \[E\] is the energy of a photon of light, \[h\] is the Planck’s constant and \[\nu \] is the frequency of the photon.
The value of the Planck’s constant is\[6.63 \times {10^{ - 34}}Js\] .
The frequency of a photon is related to the wavelength of photon by the relation
$\nu = \dfrac{c}{\lambda }$, where \[c\]is the velocity of light which is \[3 \times {10^8}m/s\] , and \[\lambda \] is the wavelength of the photon.
Inserting the value of frequency the equation of energy becomes
$E = h\nu = \dfrac{{hc}}{\lambda }$.
Given that the wavelength of the photon of light is \[490nm\].
$\lambda = 490nm = 490 \times {10^{ - 9}}m$.
 Thus the energy of photon is
\[E = \dfrac{{6.63 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{{490 \times {{10}^{ - 9}}m}}\]
$E = 4.05 \times {10^{ - 9}}J$
The unit joule is the SI unit of work. It is equal to the work done to move one coulomb of charge through a potential difference of 1 Volt. The electron volt is related to the joule of work by the relation
\[1eV = 1.60 \times {10^{ - 9}}J\]
Thus the energy in eV is = \[\dfrac{{4.00 \times {{10}^{ - 19}}}}{{1.60 \times {{10}^{ - 19}}}} = 2.5eV\]
Hence the energy, in eV, of a photon of light of wavelength \[490nm\] is \[2.5eV\].

Note:
The photon of light exhibits several important effects like photo electric effect and Compton Effect where the photon energy is related to the electric field and the magnetic field of an emitted electron.