Question

If we express the energy of a photon in $KeV$ and the wavelength in angstroms, then energy of a photon can be calculated from the relation.
(A) $E = 12.4\,hv$
(B) $E = 1.2\,h/\lambda$
(C) $E = 12.4/\lambda$
(D) $E = hv$

Answer 1

Hint we find the energy of a photon, first we multiply Planck's constant by the speed of light, then divide by the photon's wavelength. Here we know the value of the energy photon wavelength then we calculate the relation of the photon based on the unit.

Useful formula
Energy of photon,
$E = \dfrac{{hc}}{\lambda }$ $ev$
Where,
$\lambda$ is wavelength of radiation,
$h$ is planck constant,
$c$ is the speed of light,

Complete step by step procedure
Given by,
express the energy of a photon in $KeV$
we can find the energy of photon,
If the photon's frequency f is understood, then we use the formula $E{\text{ }} = {\text{ }}hf.$ This equation was first proposed by Max Planck and is therefore referred to as the equation of Planck.
Similarly, if the photon's wavelength is known, the photon 's energy can be determined using the $E{\text{ }} = {\text{ }}hc{\text{ }}/{\text{ }}\lambda$ formula.
According to the energy photon,
$E = \dfrac{{hc}}{\lambda }(joules)$ is equal to $E = \dfrac{{hc}}{{e\lambda }}(eV)$
Here,
We know the value,
$hc = 12400\,eV$
So, we can be written as,
$E = \dfrac{{12400eV}}{\lambda }$
Therefore,
$E = \dfrac{{12.4}}{\lambda }KeV$
therefore
We get,
$E = 12.4/\lambda$
Hence,
The energy of a photon can be calculated from the relation is $12.4/\lambda$

Thus, option C is the correct answer.

Note Here we calculate the wavelength of photon relation. When a photon is a light particle and also is an electromagnetic radiation packet. The photon's energy depends on how easily the electric field and magnetic field wiggle. The frequency of the photon is high.