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A point source of light of power P and wavelength λ is emitting light in all directions. The number of photons presents in a spherical region of radius r to radius r+x with centre at the source is:
(A) \[\dfrac{P\lambda }{4\pi {{r}^{2}}hc}\]
(B) \[\dfrac{P\lambda x}{h{{c}^{2}}}\]
(C) \[\dfrac{P\lambda x}{4\pi {{r}^{2}}hc}\]
(D) none of the above

Last updated date: 25th Jun 2024
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Hint:We are given with a point source that is emitting light in all possible directions. The power of the source is P and the wavelength of the light being emitted is λ. We know point source produces spherical wavefront that goes out.

Complete step by step answer:
We need to find first of all energy density because we have to take into account some specific area. We can write energy in terms of power as \[E=Pt\]. Now the time taken by the light to move a distance from r to (r+x) that is distance x is \[t=\dfrac{x}{c}\], where c is the speed of light. So, E= \[\dfrac{Px}{c}\]
Since light energy is carried by photons and if the energy of one photon is given as \[\dfrac{hc}{\lambda }\]where h is the Planck’s constant and c is the speed of light. So, we can write the energy as \[E=N(\dfrac{hc}{\lambda })=\dfrac{Px}{c}\]
\[N=\dfrac{Px\lambda }{h{{c}^{2}}}\]

So, the correct option is (B).

Note:The point source of light sends away light in all directions and the wavefront is spherical. Also, from Planck’s quantum theory light travels in the form of photons and phonons are the smallest quanta of energy. Photon's rest mass is zero and thus, the photon can be freely created and destroyed. It does not violate the law of conservation of energy.