Question

We can say that the energy of a photon of frequency $ f $ is given by $ E = hf $ , where $ h $ is Planck's constant. The momentum of a photon is $ P = \dfrac{h}{\lambda } $ where $ \lambda $ is the wavelength of the photon. Then we may conclude that velocity of light is equal to:
(A) $ \sqrt {\left( {\dfrac{E}{P}} \right)} $
(B) $ \dfrac{E}{P} $
(C) $ EP $
(D) $ {\left( {\dfrac{E}{P}} \right)^2} $

Answer 1

Hint: Speed of any wave is $ v = \lambda f $ . Where, $ \lambda $ is the wavelength and $ f $ is the frequency of the wave.
Find the ratio of the given energy and momentum formula. From the ratio speed of light can be obtained.

Complete Step By Step Answer:
It is given that the energy of a photon is $ E = hf $ .
The momentum of the photon $ P = \dfrac{h}{\lambda } $ .
Where, $ h $ is the Planck’s constant
 $ f $ is the frequency of the photon
 $ \lambda $ is the wavelength of the photon
It is required to find the speed of light in terms of energy $ E $ and momentum $ P $ .
We know that speed of light $ c = \lambda f $
From the energy equation, we have $ h = \dfrac{E}{f} $ .
Substitute the above obtained value of $ h $ in the momentum equation.
 $ P = \dfrac{E}{{\lambda f}} $
Substitute \[\lambda f = c\]
 $ \Rightarrow P = \dfrac{E}{c} $
Or $ c = \dfrac{E}{P} $
Hence, the correct option is (B) $ \dfrac{E}{P} $ .

Additional Information:
A photon is an elementary particle of light defined as an energy packet of light. Some properties of photons are discussed below;
-It behaves likes a particle as well as a wave simultaneously
-The rest mass is zero
-Travels with a constant speed of light
-The energy depends on its frequency. The higher the frequency, the more energy it has.

Note:
Light is an electromagnetic wave. The speed of light is constant in free space and is equal to $ c = 3 \times {10^8}m{s^{ - 1}} $ . All the electromagnetic waves travel at a speed equal to the speed of light. The speed of light in a medium of refractive index $ \mu $ is $ v = \dfrac{c}{\mu } $ .