Question

The energy of photon of wavelength $\lambda$ is
[$h$=Planck's constant, c = speed of light]
A. $hc\lambda $
B. $\dfrac{{h\lambda }}{c}$
C. $\dfrac{\lambda }{{hc}}$
D. $\dfrac{{hc}}{\lambda }$

Answer 1

Hint: Each quantum of light has energy $hv$, where $h$ is Planck's constant and $v$ is the frequency of the light. Additionally, the physical quantities wavelength and frequency are inversely proportional to each other.

Complete step by step answer:
On observing the phenomena of photoelectric effect, it is found that when light interacts with matter, it behaves as if made of discrete packets of energy. These packets of energy are called quanta. Each quantum of light has energy $hv$, where $h$ is Planck's constant and $v$ is the frequency of the light. Additionally, it is associated with momentum also of value $\dfrac{{hv}}{c}$. The fact that a light quantum has definite energy as well as momentum allows us to associate it with a particle. This particle is called a photon. The rest mass of a photon is zero.
Now, Frequency and wavelength are both physical quantities associated with waves. These two are inversely proportional to each other.

Mathematically:
$\lambda \propto \dfrac{1}{v}$
But speed is defined as the ratio of distance travelled over the time taken to travel the same distance. In case of a wave, the speed is velocity of light denoted by c, distance travelled is wavelength and time taken is the Time period, T of wave, so we have:
$Speed,c = \dfrac{{{\text{wavelength, }}\lambda }}{{{\text{Time Period, T}}}}$
Since frequency is the reciprocal of time period only, i.e, $v = \dfrac{1}{T}$
$\eqalign{
  & \therefore c = \lambda v \cr
  & \Rightarrow v = \dfrac{c}{\lambda } \cr} $
Substituting the value of $v$, in equation of energy, E= $hv$, we get:
$\eqalign{
  & E = hv \cr
  & \Rightarrow E = \dfrac{{hc}}{\lambda } \cr} $
Therefore, the correct option is D. i.e., $\dfrac{{hc}}{\lambda }$

Note: Photons are electrically neutral particles and are hence not deflected by electric and magnetic fields. Additionally, all photons of a particular frequency and wavelength will have the same energy and momentum independent of the intensity of radiation.