
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
Answer
494.1k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 4 Maths: Engaging Questions & Answers for Success

Trending doubts
Give 10 examples of unisexual and bisexual flowers

Draw a labelled sketch of the human eye class 12 physics CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE
