Question

Define the intensity of radiation on the basis of the photon picture of light. Write its S.I unit.

Answer 1

Hint:The intensity, in general, refers to the measurement of energy flux over an area. Hence, it measures the amount of energy travelling through a certain region during a certain period of time. Hence, the S.I unit will be \[W{{m}^{-2}}\].


Complete step by step solution:
The intensity usually is a measure of the energy flux flowing or passing through an area over a period of time. In the photon picture, a light wave consists of multiple photons. Each of these photons is considered to be energy packets carrying a certain amount of energy.

The energy of a photon is given by: \[E=hv\], where (E) is the energy of the photon, (h) is the Planck’s constant, and ($v$) is the frequency of the photon.
The Standard unit of energy is Joule.

A light wave contains multiple energy packets of photons. Hence the energy of a light wave is given by: \[E=nhv\], where (n) is any natural number.

Further, as per the definition of Intensity, the amount of energy flux passing over an area over a period of time is the Intensity. Hence, let’s consider a surface area of $1{{m}^{2}}$and the amount of flux passing through it is; \[E=nhv\]. Let the amount of time in which the light wave hits the surface area be 1 second.

Hence, the Intensity becomes: $I=\dfrac{E\ J}{(Area\ {{m}^{2}})(Time\ s)}=\dfrac{nhv\ J}{(1\ {{m}^{2}})(1\ s)}$. The Intensity’s unit hence becomes:$I=\dfrac{J}{s}\dfrac{1}{{{m}^{2}}}$and we know that$\dfrac{J}{s}=W$, where W is the unit Wattage signifying energy per unit time.

Therefore, the S.I unit of Intensity becomes: $I=\dfrac{W}{{{m}^{2}}}$, since all the S.I units have been used for the relations.


Note:
The energy of a single photon given by: \[E=hv\], is the smallest amount of energy possible also known as a quanta.
A monochromatic light wave consists of photons, all of which are of the same frequency.